Introduction to Differential Equations

2. Time-Independent Schrödinger Equation

Background A course covering modern physics A course in intermediate mechanics Harris, Non-Classical Physics Paul A. Tipler, Ralph A. Llewellyn, Modern Physics Hermite polynomial and function handout Definite integrals handout Fourier transform section of the integral transform handout Concepts of primary interest: Schrödinger equation Separation Ladder operations – raising and lowering Expectation values Phase and group velocities Motivation of the momentum operator Sample calculations: SC1 - Griffiths Example 2.2; eigenstate expansion [SE(ti).7] SC2: Square well mixed state probability density [SE(ti).8]

SC3: Raising o to find 1 [SE(ti).22] SC4: Expectation values using ladder operators [SE(ti).25]

SC5: Bound state for a delta potential: - (x – xo) [SE(ti).26] SC6.) Free particle scattering by a delta potential [SE(ti).30] SC7.) The step potential [SE(ti).33] SC8.) The high step potential [SE(ti).36] SC9.) The finite square well [SE(ti).37] SC10.) Barrier penetration [SE(ti).41] Helpful handouts: Hermite Polynomials Tools of the trade:

This handout is keyed to Griffiths Introduction to Quantum Mechanics, 2nd Ed. It is not designed to be used independently.

The wave function: A quantum system is described by a wave function which contains all the information that can be known about the system. It is a function of the generalized coordinates for the system and time. In 1D, the function might have the form (x,t). It is a complex-valued function.

for use with Griffiths QM Contact: [email protected] Schrödinger proposed the equation that governs the time-development of the wave function.1

22 ˆ  i (,) xt H (,) xt2 V () x  (,) xt [SE(ti).1] tmx2 22  The operator 2 Vx() is an example hamiltonian for a one dimensional 2mx system. At least in simple cases, the hamiltonian represents the energy of the system. Expressing [SE(ti).1] as an eigenvalue equation yields: 22 2 Vx() nnnn (,) xt E (,) xt i (,) xt [SE(ti).2] 2mx t

To invoking the separation hypothesis, assume: n(x,t) = n(x) Tn(t). 1122 2 Vx() nn () x  E i Tt()  n ()xmx 2  Ttt() 22 Time independent SE: 2 Vx()nnn (,) xt  E () x 2mx

En  it   Time dependence: E Tt() i Tt () Tt() e nn t n n

En it   Energy eigenfunctions: n(x, t) = n(x) e 

Each energy eigenvalue solution has a stationary probability distribution.

In summary, the hamiltonian is the time development operator, and it is the energy operator for essentially for all systems that we shall study. After separation, the focus of each new problem exploration is to find the physically acceptable solutions of the time-independent Schrödinger equation.

1 See the article by Felix Bloch in the December 1976 issue of Physics Today.

9/17/2010 SP425 Notes –Time Independent Schrödinger Equation Ch2-2 22 2 Vx()nn (,) xt  En () x [SE(ti).3] 2mx

The set of values of the constant En for which the equation has an acceptable solution are the energy eigenvalues for the problem, and the corresponding states described by the n(x) are the eigenstates. When combined with the time dependences, the full

En i  t wavefunctions n(x, t) = n(x) e are a complete (basis) set for the expansion of any permissible solution to the problem. That is: for a general (x, t),

 (x ,0)axkk ( ,0)  akk (x ) kk11

 itk (,)xt akk  (,) xt  akk () xe [SE(ti).4] kk11 The solutions corresponding to distinct eigenvalues are necessarily orthogonal, and those corresponding to degenerate eigenvalues can be chosen to be orthogonal (or made orthogonal using a method like Gram-Schmidt).

  *()xxdx () [SE(ti).5] km k m km

Note that [SE(ti).5] also states that (x) is normalized. Beware: we can only normalize bound-state wavefunctions. A plane wave like A eikx does not represent a bound state, and it cannot be normalized(using our bound-state convention). You will generalize the definition of normalized when you reach graduate school. The problem disappears.  Given an orthogonal basis, the expansion coefficients can be found using projection.

To find ap, take the inner product of the corresponding behavior p and [SE(ti).4] using the orthogonality [SE(ti).5].

  p ()xx (,0) pk()x  (,0)xax p ()k ()x  ap  [SE(ti).6] k1 pp()x ()x

9/17/2010 SP425 Notes –Time Independent Schrödinger Equation Ch2-3 In the case of quantum mechanics, the abstract inner product represents

 a concrete integration. ()x  (,0)xx [ ()](,0)* xdx. pp

upper BRA KET [] BRA* KET dx lower

Erwin Rudolf Josef Alexander Schrödinger (August 12, 1887 – January 4, 1961) was an Austrian - Irish physicist who achieved fame for his contributions to quantum mechanics, especially the Schrödinger equation, for which he received the Nobel Prize in 1933. In 1935, after extensive correspondence with personal friend Albert Einstein, he proposed the Schrödinger's cat thought experiment. The huge Schrödinger crater on the far side of the Moon was posthumously named after him by the IAU. The Erwin Schrödinger International Institute for Mathematical Physics was established in Vienna in 1993. http://en.wikipedia.org/wiki/Schroedinger

The infinite square well: A few observations about the character of the solution of the infinite square well that are common to quantum problems are to be noted as well as one our two that are peculiar to problems like the infinite square well.

forbidden 0inside Vx()   outside forbidden

The energy must be greater than zero (or, in general, E > Vmin) in order that an acceptable (normalizable) solution exist.

9/17/2010 SP425 Notes –Time Independent Schrödinger Equation Ch2-4 The spatial part of an energy eigenvalue solution can be expressed as a real-valued function of an oscillatory form in the classically allowed regions. (Oscillatory form means that the function is concave toward zero.)

Interlaced Zeros Property: In one dimension, if a reference spatial eigenfunction has n internal* zero crossings, the state with the next higher eigen-energy has n + 1 zero crossings with one crossing located between each adjacent pair of zero crossings of the state with n crossing.

* Internal: those not at an endpoint. Here, crossings for xmin < x < xmax. Endpoint zeros may be common to all allowed states for a problem.

The infinite square well potential is symmetric about the center of the well. The eigenstates are alternately even and odd about the center of the well.

The eigenstates are normalizable, and they are non-zero only inside the well. As they are continuous functions, the eigenstates have a zero at each edge of the well. The eigenstates form a complete, orthogonal basis for the expansion of normalizable spatial functions in the region inside the well (in the region in which they are non-zero).

The expansion coefficients can be found using projection. For a general (x,0), you

take the inner product with k| to find ck.

(,0)x cxmm ()

 k()xx (,0) cmkm () x () xwherex km() () x km

*  ()xx (,0)  k ()xxdx (,0) cxk inside  *() (,0)xdx k * inside k km()xx () ()xxdx () inside kk

We assume that the functions k(x) are normalized.

9/17/2010 SP425 Notes –Time Independent Schrödinger Equation Ch2-5 If, at time t = 0, the wave function of the system is (x, 0), then

Em (,)x tcxe ()it() where c* () xx  (,0)dx  mm kinside k

Exercise: Consider each fact presented bout the infinite symmetric square well. Discuss the validity of each for a finite symmetric square well problem.

Properties of acceptable wave functions: The presence of the second derivative with respect to position in the energy operator  (the hamiltonian) requires that (x) is continuous, and it normally requires that x is continuous. The wavefunctions for the infinite well problem are continuous as is  always required. A discontinuous first derivative x leads to a singular (ill-defined, infinite) second derivative. The infinite well hamiltonian has a singular potential, and its wavefunction can have a discontinuous first derivative at points for which the potential is singular. Not that the potential has an infinite discontinuities at the points  for which x is discontinuous.

 n 2m Vx() E () x xx  2 nn For a particular infinite depth square well with the zero-potential region running from 0 to a, the eigenfunctions are:

En 22 (xt , )2 sin[nx ]ei(  )t where E n2   nnaa  2ma2  The particles are confined in a region of width a, and the spacing between the energy -2 levels is proportional to a . A narrower confinement region leads to greater spacing between allowed energies. In the limit of large a (no confinement), there would be a continuum of allowed energies rather than a spectrum of discrete values. The n2

9/17/2010 SP425 Notes –Time Independent Schrödinger Equation Ch2-6 dependence of the energy on the quantum number is characteristic of a particle confined in a region of fixed width.

Sample Calculation SC1 - Griffiths Example 2.2: Infinite Square Well Problem Consider a square well with the zero-potential region running from 0 to a. At time 0, the spatial wave function is: (x, t) = A x (a – x). [SE(ti).7] Note that this wave function respects the conditions required for an admissible wavefunction. It meets the boundary conditions as it vanishes at 0 and at a. It is normalizable. (Evil things can happen when the wavefunction has no respect!) You should always verify that a wavefunction to be studied is normalizable and that it meets the boundary conditions for the problem. If it does not, you may proceed, but you should be prepared for nonsense to follow. a.) Normalize the function. b.) Expand the function in terms of the eigenfunctions.

Em (,)x tcxe ()it() where c* () xx  (,0)dx  mm kinside k 2 2 c.) Comment on the value of  |ck| . d.) Compute  Ek |ck| . e.) Compute H. (x, 0) = A x (a – x) for 0 xa. Normalize:

aaa *2222(,0)(,0)x x dx  1 A22 xaxdxA (  )  ( ax  2 axxdx34 ) 000

225 121 a5 30 1 Aa34 5 A30  Aa5

Em Find (x, t): (,)x tcxe ()it() where c* () xx  (,0)dx  mm kinside k

2 kx  30 2 cak aasin[ ]5 (xx )dx inside a

 815 for k odd kx   33k Use the change of variable u  a and integrate by parts. ck    0 for k even

9/17/2010 SP425 Notes –Time Independent Schrödinger Equation Ch2-7 (2mt 1)22   i 3 2 (,)xt 30 21sin(2mx 1) e 2ma aa  (2m 1)3  m0 2 Probabilities: The value |ck| is the probability initial energy measurements on an ensemble of states prepared using the recipe (x,t) is energy of the eigenstate k. The 2 sum of the probabilities for all possible results ck| should be 1. This conjecture can

 be verified for this problem as 1   6 , a fact established in a Fourier series  (2m 1)6 960 m0

 problem using Parseval’s identity as was 1   4 .  (2m 1)4 96 m0

Expectation value of the energy: 2 Using our new results, ck| Ek.

 22 2  64 (15)   (2m 1)   6(2m 1)6  2ma2  m0 22 10   1.013E 1.013 E  2 2ma2 1 GS The energy is just a little greater than the  ground state energy E1. A quick comparison plot shows that (x, 0) is very similar to the ground state wavefunction 1(x). We should expect expectation values to be similar for states that are very similar. A more direct calculation yields: 

aa ˆˆ* 2 2 2 H HxHxd  (,0)(,0) xAxaxxa  ()()()  2 xd x  00 2m x 

a 30 222330 11 10  22 55( ) (ax  x )  2 dx  (2 ) a  2 2 1.013EGS () aa22mm0 23   2ma 

 2 Exercise: Show the steps necessary to evaluate: E   cEkk for the case above.  k1

9/17/2010 SP425 Notes –Time Independent Schrödinger Equation Ch2-8  0 forx  0 2  Exercise: Show that  (,0)x  Aa ( x ) 2 a ( x  a ) for  x a . x2    0 for xa

Repeat the calculation of H. Sometimes you get lucky. You should always check!

SC2 – Square Well Mixed State: A probability density that is not static:

iE(/)1  t The wavefunction for an energy eigenstate has the form 11(,)xt () xe . The single frequency imaginary exponential time dependence is associated with a time- independent probability density.

* * iE(/)11 t iE (/) t 2 Pxt(,)11 (,) xt  (,) xt  1 () xe 1 () xe |  1 ()| x For this reason, the eigenstates of energy as also called stationary states. The single frequency character leads to their designation as pure states. A state that contains contributions for multiple eigenstates with distinct energies is a mixed state. For a first example, a mixture of two states is to be discussed.

iE(/)12 t iE( /) t mix (,)xt a121 (,) xt  b (,) xt  a () xe b 2() xe [SE(ti).8]

* Pxt(,) mix (,) xt mix (,) xt 22 22* it21 * it21 axbxabxxebaxxe121221() () () ()  () ()

The wavefunctions 1 and 2 have been chosen to be real, and the difference frequency

1 21  [E2E 1]. The result can be written as:

2222 Pxt(,) a1 () x b 2 () x 1 () x 2 () x C cos( 21 t ) D sin( 21 t ) Mixed states have probability densities that are the sums of stationary pieces and pieces that oscillate at the difference frequencies.

Pure states are stationary; mixed states are where something is happening. Mixed states have time dependent probability densities. However, the probability that an energy

9/17/2010 SP425 Notes –Time Independent Schrödinger Equation Ch2-9 measurement return a particular energy eigenvalue is not time dependent. The behavior of mixed states alone does not describe transitions from level to level.

Exercise: Find the values of C and D in the expression above. (a and b are complex)

Consider the infinite well mixed state (x, t) = 1 exexit1 (,0)i (,0) , 2  12 Probability plots for the 1, 2 mixed state as a function of relative phase :

      = 0  = .25  = 0.5   = .75   = 1 

See the box state movies at: http://usna.edu/Users/physics/tank/TankMovies/

1 Note that  = ( -  )t = (4  -  )t = 3  t for (x, 0) =  (,0)xx (,0). 2 1 1 1 1 2  12

Free particles and Fourier methods: A free particle is one that is not confined or even influenced by a potential. The potential is V(x) = 0. The Schrödinger equation is:

2 2   (,)x tE () x 2m x2 nnn The solutions are: ikx -ikx 22k (x) = A e + B e each with energy eigenvalue: E  2m . There is no confinement so there is no quantum condition to restrict the energies to a discrete spectrum of values. The continuum of values E  0 is allowed. The time dependent solutions have the form:

 kk22 ikx  t ikx t  2m2m k (,)xt Ae  Be [SE(ti).9]

9/17/2010 SP425 Notes –Time Independent Schrödinger Equation Ch2-10 Unfortunately, these wavefunctions fail critical tests. They cannot be normalized, and they do not vanish in the limit that |x|  . A possible remedy would by to use states:1

2  ()x Limit Aeikx Be ikx e x  k  0    We will use the original forms and just remain wary of procedures that might diverge or that might involve integration by parts and surface terms with evaluations at . For now, we will only require normalization for bound state wavefunctions, and we will be on guard whenever we deal with free-particle or other unbound states.

One should note that each term on the right hand side of [SE(ti).9] is an eigenfunction of momentum as well as of energy.

kk22 k2 ikx t ikx t ikx  t ˆ 22mm  2m pkAe i x Ae  Ae

kk22 k2 ikx t ikx t ikx t ˆ 22mm  2m pkBe ix Be Be An eigenfunction of an operator has a precisely defined value of the quantity measured by the operator. For plane waves, the precise value is the momentum (and the kinetic energy). The uncertainty relation (p x½ ) requires that x be infinite if the

momentum is precisely known (p = 0). The functions in [SE(ti).9] have that property as the associated probability density is spread uniformly from - to + so x is infinite.

Comparing with classical wave solutions (x,t) = f(x – vt) + g(x + vt), a speed, the

k p phase velocity, can be identified as /2m = /2m or one half the classical velocity of a

1 Search for the term wavelet for more on this topic. As stated, we will stick with the infinite plane wave functions.

9/17/2010 SP425 Notes –Time Independent Schrödinger Equation Ch2-11 particle. A second speed associated with wave is the group velocity. It is the group velocity that corresponds most closely to the classical velocity of a particle.

“In what limit should quantum mechanics predict or agree with classical behavior?” This limit is called the correspondence limit. Some suggest that it should be in the limit of large quantum numbers. More is required. In order to construct a state that exhibits classical behavior, one needs a superposition of a large number of carefully chosen eigenstates with high quantum number. An energy eigenstate with a high quantum number does not approximate classical behavior. Plot of ||2 for the n = 200 state of a QHO. (Shown for x < 0 only)

The n = 200 probability density is a stationary distribution, and hence nothing like a classical oscillating particle. A careful superposition of a large number of energy eigenstates of even larger eigenvalue number is required to see behavior approaching classical expectations.

9/17/2010 SP425 Notes –Time Independent Schrödinger Equation Ch2-12

Correspondence State Example: Go to http://usna.edu/Users/physics/tank/TankMovies/MovieGuide.htm. Click into the QuantumHO directory for Correspondence 20OscillatorCompFinal.gif. The green curve represents a correspondence state probability density that oscillates at the frequency of the classical oscillator. The animation represents a superposition of about

20 states around quantum number 10.

This section is to be replaced by a treatment that departs from Griffiths conventions slightly.

To gain a better understanding, we must recall that the eikx can be used a basis function for an expansion. It is just the Fourier transform.

1  (,0)x  ()keikx dk Inverse Fourier Transform [SE(ti).10] 2 

1  ()kkx () (,0)edikx x Forward Fourier Transform [SE(ti).11] 2  The time dependence of the wavefunction follows by replacing each plane wave by its time

ikx i[kx -t] E

dependent form (e  e ) where  = /.

2  ikxk t 1  2m  (,)x tk ()edk [SE(ti).12] 2  Compare this result with [SE(ti).4].

Digression: Group velocity and the Fourier transform:

Consider a wave pulse g(x) = f(x) eikox traveling in the +x direction at time t = 0 that is an envelope function f(x) amplitude modulating the plane wave eikox. Given that the

 ikox  Fourier transform of f(x) is f ()k , the Fourier transform of g(x) = f(x) e is f ()kk 0 .

9/17/2010 SP425 Notes –Time Independent Schrödinger Equation Ch2-13 1  f()kfxe ()ikx dx 2 

11 gk() f () xeik00 x eikx dx f () xei() k k x dx f ( k k ) 22  0 The Fourier transform expands f(x) as a sum of pure spatial frequency components

1 eikx  2  After a time t, a component such as the one above will have developed into

ikx[] t 1 e k  2  where k = (k), the value of  as dictated by the Schrödinger equation. ASSUME that 2 the envelope function g(x) varies slowly over a distance o = /ko. The function g(x) can be represented in terms of spatial frequencies near ko. Expand (k) about ko.

dd1 2 2 ()kkkkk00dk ()  2 dk 2 ()  0 ... k0 k0 Next assume that the first two terms are adequate to faithfully represent (k).

d ()kvk0G ( k0 ) where ()k0 0 and vG  dk k0 Recalling the form of the inverse transform,

1  gx() gke ()ikx dk 2  and re-summing the time developed components, we find the shape and position of the wave for time t.

11ikxtvtkk[]  ikxtvtkk[]  gxt(,) gke ()00GGdk fk ( k ) e00dk 22  0

1 ikx t  ik()() k x vt gxt(,) e00 fk ( k ) e0 G dk 2  0

With the change of variable  = k – ko,

9/17/2010 SP425 Notes –Time Independent Schrödinger Equation Ch2-14 1 ikx t  ixvt() ikx t gxt(,) e00 f () eG d f ( x vte ) 00 2  G

gxt(,) eikx 00 t f ( x vt ); v o ; v d Gpk G o dk ko

The result is the time-dependent representative plane wave eikx 00 t modulated by an envelope function with fixed shape and width that translates at speed vG, the group velocity.

d 1.) The pulse envelope translates at the group speed (‘velocity’) vG = dk with its k0 envelope shape undistorted.

2.) The factor eikx 00 t shows that the wave crests inside the envelope propagate at the

 phase velocity vp which is: k . k0

22k k 2 k k p For a free particle, E = 2m so k  2m . The phase velocity is k 22mmor half the classical particle velocity. The probability lump translates at the group velocity dk k p dk mm which agrees with the classical particle velocity. The probability density ||2 is a measurable; the phase of the wavefunction is not.

For an animated example, go to: http://usna.edu/Users/physics/tank/TankMovies/MovieGuide.htm Select: PlaneWaveAndPackets and PacketSpreadCompareNoFrmLbl.gif

As you view the animation, use your finger tip to follow one wave crest. Notice that the wave packet translates faster than does any one of the wave crests.

Conclusion: For a wave packet, the group velocity is analogous to the classical velocity of a particle described by the wave packet.

9/17/2010 SP425 Notes –Time Independent Schrödinger Equation Ch2-15

Some pulses require a broader range of frequencies for their representation. In such

1 d 2 2 cases, the term 2 2 (kk 0 ) must be included, and it leads to distortions of the dk k0 pulse shape. The distortions expected most often are spreading and the degradation of sharp features.

Wave packet example requiring quadratic terms  pulse distortion Initial pulse with sharp features Later time, spread, less sharp

Crude representation, not faithful in detail For cases in which the Taylor’s series expansion of (k) does not converge rapidly, the d pulse shapes will always distort, and the concept of a group velocity /dk is of no d value. (If one finds that /dk > c, the group velocity (first order expansion) approximation is failing rather than Special Relativity.)

Initial pulse pulses with various x’s Pulses after the same elapsed time

9/17/2010 SP425 Notes –Time Independent Schrödinger Equation Ch2-16

A pulse with a small initial spatial spread x has a large p so the pulse spreads very rapidly. A large x permits a small p and slower spreading. End digression.

The (linear) momentum basis set: Consider an infinite plane wave: A ei(kx - t). The wave has the form of a traveling wave propagating in the positive x direction. The wave evolves through a full cycle (2 change of its argument) for each wavelength of travel . We conclude that k = 2. As

h the wavelength is constant, we assign a uniform momentum p = / (= k) to the wave.

We leap to the conclusion that A ei(kx - t) is an eigenfunction of momentum. As we must have a Hermitian operator pˆ for the measurable momentum, we assume that the collection of all the A ei(kx - t) forms a basis set and that A ei(kx - t) has a momentum eigenvalue k.

Normalize A ei(kx - t)

9/17/2010 SP425 Notes –Time Independent Schrödinger Equation Ch2-17 By now, we should know that bound states can be unity normalized and that continuum states like A ei(kx - t) cannot. Let’s just give it a try anyway.

* pp22()p p  A eiixx AedxAeix dxA e i()p pu du    

*  pp 2 A eAedxApiixx 2(   p)1    x (Used u = /.) If we adopt a Dirac delta normalization for continuum basis states and a Kronecker delta normalization for bound basis states,

 kp for bound states kp   (kp ) for contiuum states The Dirac normalized momentum states become:

p  ()xe 1 ixt()  p 2 Any allowed wavefunction can be expended in terms of the basis set.

pp  ()x  ()pedp11ixt()() peixt()dp [SE(ti).13] 22 

 p  **()x  1  (pe ) ixt()  dp [SE(ti).14] 2  Assuming that  is normalized, what else is true?

pp  **()x ()xdx 11()p eix() dp() p eix() dpdx  22  

  pp  *()()p p dpdp1 eixix()() e dx   2 

  **(p )()pdpdp  ( p p )  ()() p pdp   

  1 See a Fourier transform reference for a discussion of the equality: edupip() pu 2(  p).  

9/17/2010 SP425 Notes –Time Independent Schrödinger Equation Ch2-18  ** 1 As ()()x xdx  ()() p  pdp [SE(ti).15] , and * = x(x), the spatial  

probability denstiy, we conclude that *(p)(p) must be p(p), the momentum probability density. It follows that:

 p ppdppppd() *()()p [SE(ti).16] p  The operator for momentum in the momentum (basis) representation is just the momentum p.

Motivate the operator for momentum: Attempt to transform p back to the coordinate representation to see if a coordinate representation for the momentum operator can be found.

 p ppdppppd() *()()p p 

The process is just an inverse Fourier transform. The factors of ½ arise in the change

p of variable required to convert the exponent: i ( /) x  ikx.

pp ()p 11exdxpix()() ;** () exix()()dx [SE(ti).17] 22 

p p pppdppp**()()  () 1 eix() () xdxdp   2  The last factor is to be integrated by parts.

 p ix() ex ()  ix()p  ()x p p * ()p 1  e dxdp  2  x iipp ZERO  Using the standard boundary conditions () = 0, we conclude that the surface term vanishes and that the factors of p cancel.

1 Locally, we refer to relations of this form as Parseval relations.

9/17/2010 SP425 Notes –Time Independent Schrödinger Equation Ch2-19  p p 1 *()p eix() ( i  ) () x dxdp 2   x Next  is replaced using [SE(ti).17]. (Both  and * could have been replaced using [SE(ti).17] at the same time, but then the equation would have swollen to two lines in length before it began to simplify.)

  pp p 11exdxeixdxix() *()ix() (  ) () dp 22    x

  p dx dx* ()( x i  ) () x1 eixx(( )  dp  x 2  

p After choosing u = /, we use the Fourier transform representation of the Dirac delta.

  p dx dx*( x )( i  ) () x ( x x )* ()( x i ) () x dx    xx   

  pxixdx*()xpˆ () xdx  *(())     pˆ i    x x

ˆ   [SE(ti).18] pi x

 Note that the operator – i  x is in the canonical sandwiched position.

Griffiths’ motivation of the momentum operator in chapter one could be considered to be circular. The development above was founded on the de Broglie hypothesis and our expectations for operators representing physical concepts that can be measured.

Alternative approaches for solving the Schrödinger equation: Schrödinger’s equation is a partial differential equation, and it is attacked using standard differential equation techniques in a first course. A second solution method employs an operator algebra, and it can be used effectively. In some cases, the

9/17/2010 SP425 Notes –Time Independent Schrödinger Equation Ch2-20 approach provides insights into the structure and interpretation of quantum mechanics. The first target for this approach is to be the quantum harmonic oscillator.1

k ½ **** Notation alert. The symbol o is to be replaced by c = ( /m) , the classical oscillator frequency. Quantum harmonic oscillator: The standard mass-spring system has a mechanical

2 2 2 energy p / +½ kx2 and the Hamiltonian operator:   ½½kx221 pˆ kx2. 2m 22mmx2

2 p 2 2 2 2 HH(,xp )½½kx ;Hˆ (, x i  )    kx  classical 22mmclassical x x2 One might attempt to simplify the differential equation by factoring the differential k operator into two linear differential operators. For example: c  m

2 2 ?! ˆ 22   mmcc Hmx½ cc2  22x 22mmx  2mxx  cc x It’s a deceptive transformation. It looks good until one recalls that the value of an operator is its action on an arbitrary function. The problem is that the cross-term

mm portion of the expression, ccx  x , is not a null operator. 2222mmccxx Allowing the active components to operate on an arbitrary function returns the negative of that function. The active parts are equivalent to multiplication by -1.    xxx xfx () fx () or  xxx x 1 Continuing on this path, one develops the ladder operation method for the solution of differential equations. An alternative path is to be pursued that begins with the Hamiltonian function studied in classical mechanics and makes the canonical substitutions to form the quantum operator

1 22  . H = 2m pkˆ  ½ x; x  x; p  -i  /x.

1 Search for Ladder Operations in the Hermite Polynomials and Functions handout.

9/17/2010 SP425 Notes –Time Independent Schrödinger Equation Ch2-21 11 H (,xpxc )mxip c  mxip c (classical mech.) 22mmcc

1 22 A detailed calculation shows that this operators action differs from 2m pkˆ  ½ x due to the cross terms.

Factoring out c ensures that the quantity in square brackets is dimensionless.

In order to convert to quantum mechanics the position x is represented by the  coordinate x, and the and the momentum is represented by i x . However, the expression involves a product, and the prescription given in the previous chapter guide was that a product is to be replaced by the symmetric average of all the possible orderings of the operators in the product. That is: substitute ½[AˆBBAˆˆ ˆ ] for the classical product AB.

2 1 H (,x pxc ) ½m c x ip m c x ip m c x ip m c x ip  2mc   The following operators are defined:

11 amxipamxˆˆˆ+-ccand  ipˆ 22mmcc    Substituting for pˆ 

11  amxamxˆˆ+-ccand  22mmcc x   x  The quantum Hamiltonian operator becomes: Hˆ ½½aaˆˆ aa ˆˆ 2 2 m22 x () cc  2m x2 The symmetric averaging of the orderings of operator products appears to have provided us the desired outcome. This point is to be studied more carefully in the formal quantum theory section.

Digression: a study of aaˆˆ-+and .

9/17/2010 SP425 Notes –Time Independent Schrödinger Equation Ch2-22 ˆˆ11  amxipmx+ 22mm cc  x  cc 11 amxipmxˆˆ- cc 22mmccx First, the products that appear in the hamiltonian are investigated.

1 Hˆ i aaˆˆ+- mcc x ipˆ m x ip ˆx, pˆ 22mc    c 

1 Hˆ i aaˆˆ mxipmxipccˆ ˆ x, pˆ  22mc    c  The commutator x, pˆ  is considered by some to be the foundation of quantum mechanics. x, pˆ  = i. Substituting,

11Hˆ aˆˆ+- a mcc x ipˆ m x ip ˆ  2mc    c 2

11Hˆ aaˆˆ mxipmxipccˆ ˆ   2mc    c 2 Comparing, the commutator of our new operators is found.

[aaˆˆ,]1 The clean, simple, dimensionless form of the relation might suggest that it is an expression in the native language of quantum mechanics.

Priority: Study the commutators of all pairs of operators that you encounter. These relations are foundation material of the same import as the divergence and curl of each vector field studied in E&M.

Operators that represent observables must be Hermitian. The new operators are to be tested for that property.

REVIEW - Hermitian property of operators: The Hermitian conjugate of an operator Qˆ is represented by the symbol Qˆ † and it is defined such that:

9/17/2010 SP425 Notes –Time Independent Schrödinger Equation Ch2-23  Qqtqtdˆˆ† (,) (,)  (,) qtQqtd (,) ii i i

Abstract:  QQˆˆ†  for any two functions (qi,t) and (qi,t) that are permissible wavefunctions for the problem.

The minus operator is tested first. (with a multiplicative constant suppressed)

  (qt ,) aˆˆ ( qt ,) d  ( a )† ( qt ,) ( qtd ,)   iiii

  (,)x tmx (  )  (,) xtd  ( mx  )†  (,) xt  (,) xtdx ccxx An integration by parts, moves the derivative to the left with a change of sign. The boundary conditions required of permissible/normalizable wavefunctions ensure that the surface terms vanish.

 (,)x tmx ( ) (,) xtdxmx ( ) (,) xtxtd (,) x ccxx   Everything inside the parentheses is real so: ()()mxcc xx mx *

  (,)(x tmx ) (,) xtdxmx ( ) (,) xtxtd (,)x ccxx After restoring the factor of: 1 , 2mc

 (,)x taˆˆ (,) xtdx a (,) xt (,) xtdx 

† † The conclusion is that aaˆˆ (). It also follows that aaˆˆ (). Neither of the new operators is Hermitian, they are each others Hermitian conjugates.

In chapter 3, we will show that a real scalar multiple of a Hermitian operator is a Hermitian operator

( QQˆˆ). A pure imaginary multiple of a Hermitian operator is anti-Hermitian

9/17/2010 SP425 Notes –Time Independent Schrödinger Equation Ch2-24 ˆˆ 1 1 ( QQ). For amxˆˆ+  c ip and amxˆˆ- c ip, the first 2mc 2mc term in each is Hermitian and the second is anti-Hermitian making them mixed bags.

End digression.

† NOTATION ALERT: Henceforth, aˆ  aˆ and aaˆˆ  ˆˆ11  amxipmx22mm cc  x  cc † 11 amxipmxˆˆcc 22mmccx [aaˆˆ,]† 1

† The operators aˆ  aˆ and aaˆˆ  are called lowering and raising operators or perhaps annihilation and creation operators. We will see that the action of aˆ† on an oscillator eigenstate is to create a state energy raised by c. The action raises the energy or ‘creates’ an additional quantum of excitation.

The commutation relation â↠- â†â = 1 can be used to express the hamiltonian in several forms. ˆ ˆˆ ˆˆ ˆˆ†† ˆˆ ˆˆ †  ˆˆ† Haaaaaaaaaaaa½½cccc   ½½  The action of aˆ† , ‘A-dagger’:

Assume that m(x) is an energy eigenfunction with eigenvalue Em. ˆ  ˆˆ† Hmc()xaaxE½  mmm ()   ()x [SE(ti).19] † Consider the new wavefunction â m(x). (Always remember and never forget that an operator acts on a wavefunction and returns another wavefunction for the same problem.) Use: [aaˆˆ,† ] 1 (Equivalently: â↠= â†â + 1) and that: The â’s are linear operators and hence commute with multiplicative scalar constants.

9/17/2010 SP425 Notes –Time Independent Schrödinger Equation Ch2-25 ˆ ˆˆ††ˆˆ† Consider: H axmc()   aaax½ m()

†† ††† cmcmaaˆˆ½() a ˆ x aaaˆˆˆ ½() a ˆ x

†† † † † cmaaaˆˆˆ a ˆ[, a ˆ a ˆ]½ a ˆ () x

ˆˆ††ˆˆ† ˆ† aacmaa1 ½ () x Emm  cmc () x  E am ()x

Haˆaxˆˆ†††() aaaxEˆˆ ()  ˆ† () x [SE(ti).20] mc½ m m cm 

† The state aˆ  m ()x is an eigenstate of the hamiltonian with eigenvalue Em + c.

† k Compare with [SE(ti).19]. Clearly, this is a ladder operation; the state (â ) m(x) is an eigenstate of the problem with eigenvalue Em + k c.

k An analogous argument shows that the state (â) m(x) is an eigenstate of the problem with eigenvalue Em - k c. Eventually, the laddering down will reach an energy for which E < Vmin in which case no permissible solution can exist. The down ladder must terminate at o(x), the lowest rung that still has positive energy (Require E > Vmin to have a normalizable state). The state o(x) must be annihilated by another lowering. (again, a multiplicative constant is suppressed.)

 âo(x) = 0   mxipccˆ 00()xx mx x  () 0

mx 2 mx  c 22  ()xxxAeAe c () () 2  ½ x x 000

1 4 mmkc  where:   2

1 mx 2 4  c 1 22 mc 2  2 ½ x After normalizing,  0()xe 1 e.    4

9/17/2010 SP425 Notes –Time Independent Schrödinger Equation Ch2-26 Notation Alert: The value  defined here has dimensions of length inverse, and it agrees with the notation used in the Hermite handout. It is a common, but not universal notation. Do not assume that it matches conventions in your particular text.

The energy of the lowest rung state  is:

 Hdxˆ   [½] aaˆˆ†  dx oo c  o o

  aaˆˆ†  dx [½]dx ½ co o co  o c

Recall that the lowering operator annihilates o; âo = 0. The ground state energy is

½c and the allowed values for all states are (n + ½)c. Comparing this result with

† † the form of the hamiltonian, (â â + ½)c one concludes that â â n = nn. For this reason, â†â is represented as nˆ and is called the number operator. (Is the number operator Hermitian?) 

1 mx 2 4  c 1 22 mc 2  2 ½ x Exercise: Verify that  0()xe  and  o ()x  1 e are    4 normalized properly.

Exercise: Verify that the number operatornˆ (=â†â) is Hermitian. Use the defining property of Hermitian operators.

A direct application of the hamiltonian ½mx 22 2 2 on  (x) shows that it is an c 2m x2 o eigenfunction with eigenvalue ½ c. Stepping up the ladder, one concludes that

† n n(x) = (â ) o(x) is an eigenfunction with energy eigenvalue (n + ½) c.

9/17/2010 SP425 Notes –Time Independent Schrödinger Equation Ch2-27 n ˆˆ†† ˆ ˆˆ †  ccaa½½ a 0 () x  aan() x cn ½n()x In addition to being important as it stands, this result establishes that:

†† aaˆˆnn()x  n ()andxx aa ˆˆ n () ( n1) n ()x. Used: â↠= â↠+ [â, â†] = â↠+ 1 † n One can use these relations to normalize the function (â ) o(x). Assume n(x) is † normalized and that â n(x) = dn+1n+1(x).

 † (1)n  nn()xxdxxaaxdx ()  n ()ˆˆ  n ()    

 †† 2  (1)n  axaxdxdˆˆnnnnn() () 11 () x () xdx   

An operator acting in the ket is equivalent to its Hermitian conjugate acting in the bra.

As the original function n(x) is normalized, the raised function needs its amplitude

½ † ½ divided by a factor (n + 1) to be normalized. (â )n(x) = (n + 1) n+1(x)

After an analogous study of (â)n(x), it follows that:

† axˆˆnnn() n 111 ()and x axn () n () x [SE(ti).21] Hence the normalized wavefunctions are:

††nn []aaˆˆ [] in[½] t  ()xxxtxe (); (,) () c nnnn!!00

SC3 - Sample Calculation of 1(x):

2 1 mx 1 4  c 1 22 mmk 4 mc 2  2 ½ x c   ()xe1 e   [SE(ti).22] 0   4 2     amxipxipxˆˆˆ11 ()1 1 1  2m  c  22    x  c amxipxipxˆˆˆ† 11 ()1 1 1  c 22  2mc  x

† 11 aˆ 111½ 224422x ½ 22x 10()xx () xx ex e 1 22x  

9/17/2010 SP425 Notes –Time Independent Schrödinger Equation Ch2-28 1 1  2 4 ½½22x 2(x ) (x) 2 1()xe2 x e 2    21!1 

This result can be compared with the result in the Hermite handout for the QHO:

2  ()xe Hzn ()  ½(z ) where zx n 2!n n  The factor of  in ||2 is combined with the integration differential dx to yield dimensionless differential dz when the variable is changed from x to z. Hx()  2 (,)xt n e½(x )e inc ( ½)t n 2!n n 

2 3 4 2 H0(z) = 1 H1(z) = 2 z H2(z) = 4 z – 2 H3(z) = 8 z - 12 z H4(z) = 16 z - 48 z + 12

Exercise: Show that the wavefunction 1(x) found just above is normalized.

SHOwavefn[n_,z_]:=HermiteH[n,z] Exp[-(z^2)/2]*(1/Sqrt[2^n n! Sqrt[Pi]]) Plot[{SHOwavefn[0,z], SHOwavefn[1,z],SHOwavefn[2,z]},{z,-4,4} , AspectRatio0.4,PlotStyle  {Thickness[0.004],Thickness[0.004]}]

QHO States: Plot of o(x), 1(x) and 2(x) Computing expectation values using ladder operations:

9/17/2010 SP425 Notes –Time Independent Schrödinger Equation Ch2-29 In preparation, the operators for position and momentum are to be expressed in terms of the ladder operators â and â†.

Ladder operators: amxipxipxˆˆˆ11()1 1 1  2m  c  22    x  c amxipxipxˆˆˆ† 11 ()1 1 1  2m c 22x c  Ladder relations: Dirac notation axˆˆ††() n 1 () x annn 1  1 nn1 axˆˆnn() n1 () x annn 1 Basic operators: xˆˆ  1 1 [aa†  ˆ] [SE(ti).23] 2 pˆˆ i [aa† ˆ] [SE(ti).24]   2

1 4  mmkc    2  mk()()xxdx mk   SC4 - Sample calculation: Ground state expectation values

Coordinate representation: wavefunction: o(x) Abstract Vector Space Form:   1  xx00 0  1 [ aaˆˆ†  ]0 x  ()xx () xdx [SE(ti).25] 2  00 11  x  0[ 11 0] 1 01  0 1 ()x 1 [aaˆˆ† ]  () xdx 22  002

  1  ()[1x 0]dx 2  01

1  1()()  xxdx0 2  01

  † p   ()xpˆ () xdx ppiaa00ˆˆ 0[ ˆ ]0  00 2

9/17/2010 SP425 Notes –Time Independent Schrödinger Equation Ch2-30     ()x i [aaˆˆ† ]  () xdx  ii0[ 11 0] 1 01  0  002 22

 ixx  ()[1 () 0]dx 2  01

 1()()  xxdx0 2  01

 22 22 21 † 2 x  ()xx () xdx xx000 2 [ aaˆˆ  ]0  00 2   †† † † 22 0[aaˆˆ aa ˆˆ aa ˆˆ aa ˆˆ ]0 ()[x 1 aaˆˆ† ]  () xdx 2  002  2 2  0[ 2 2 10 0 0]0   []aaˆˆ†† aa ˆˆ † aa ˆˆ † aa ˆˆ dx 2 2  00 22  202 100   22  22 [2 0 0]dx  22 020

Exercise: Compute the expectation value of p2 in the QHO ground state.

The Dirac delta potential: Consider a potential that is zero almost everywhere and that is strongly attractive in a very small neighborhood of x = 0. This potential can be modeled using the Dirac delta. V(x) = -  (x) [SE(ti).26] The delta function has dimensions of the inverse of its argument (m-1; inverse length) so the constant  has dimensions of (J-m; energy-length). To be bound, a particle must be fenced in by a large forbidden region. As V(x)  0 as |x|  , any bound state must

2 2 have negative energy. Represent that energy as E =  2m  . Substituting in the Schrödinger equation, we find that the allowed form of the wavefunction in the regions with V(x) = 0 is: (x) = AeBex  x . In order to be normalizable, the wavefunction is restricted to the following forms in the specified regions.

9/17/2010 SP425 Notes –Time Independent Schrödinger Equation Ch2-31 Aexx(0 x ) Ae (0 x  )  ()x  xx Be(0) x Ae (0 x ) The wavefunction must be continuous. Continuous at x = 0  A = B. Note that A or B must be non-zero in order that the solution be non-zero somewhere. These conditions require that (x) is an even function.

Clearly, the delta function potential presents a new challenge. Appeal to that which governs the wavefunction, the Schrödinger equation.

2 2 ()x ()xxEx ()  () 2m x2

  22mm ()x ()xx E  () xx  22 The delta function is defined by its action in an integral so integrate it. It dominates the behavior in a small region around each zero of its argument. Integrate from - to + and take the limit that   0. (Note that the partial derivatives could be replaced with total derivatives as (x) is a function of a single variable.)

   dx  22mm ()x () x dxE()x dx xx   x  x 22  

  22mm E 2 22 (0) [2 (0)   (  )] xx  The continuity of (x) has been used, and () 2 represents terms of second and higher order in . In the limit that  vanishes,

   2m  (0) [SE(ti).27]  xx  x 2  

The usual boundary conditions obeyed by the wave function are: i.) The wave function is continuous. ii.) The first derivative of the wavefunction is continuous.

9/17/2010 SP425 Notes –Time Independent Schrödinger Equation Ch2-32 We expect that the wavefunction is continuous in all cases. The Schrödinger equation dictates that the first derivative must be continuous unless the potential is singular (not finite). The second derivative is not defined if the first derivative is not continuous. 2 ()x 2m [Vx () E ] () x x2 2

The second condition is becomes: ii.)  2m  (0) for V(x) = -  (x).  x  2

Exercise: Give another case in which the first derivative of the wavefunction is not continuous. Identify the singular nature of the potential for that case.

SC5: The bound state(s) of an attractive delta potential Find the bound states of the potential V(x) = -  (x). First, as the potential V(x) = -  (x) is symmetric in x, the ground state wavefunction should be an even function of x, and the eigenstates should alternate between being even and odd with each increase in energy eigenvalue. In our case, the required form of the wave function dictates the choice of a symmetric trial wavefunction.

 Ae x (0 x  )

 ()x    x Ae(0 x  )

Plot of (x)

9/17/2010 SP425 Notes –Time Independent Schrödinger Equation Ch2-33 Exercise: In light of the discussion in the previous paragraph, how many bound states are expected for the potential V(x) = -  (x).

The trial wave function meets condition (i); (0+) = (0-). Applying [SE(ti).27],

  22mm A ()AA22  (0) xx 

  Ae x (0 x  ) d  dx   x  Ae(0 x )

  d xx dx  2 A 

d 22mm (0) A  dx  22

(x) Plot of the /x: Note the discontinuity is – 2A

It follows that  = m and E = 2  2m  2. The matching process has yielded a 2 2m 22 single, unique result so the problem has exactly one bound state for any value of  so long as the potential, V(x) = -  (x), is attractive ( > 0).

With  identified, the normalized bound state wave functions is:

 2  m i m 23x  t m  2  (,)xt e e ------end sample calculation Exercise: Attempt to sketch an anti-symmetric normalizable bound state for this   problem. What is the discontinuity in /x at x = 0?

9/17/2010 SP425 Notes –Time Independent Schrödinger Equation Ch2-34 SC6.) Free particle scattering by an attractive1 delta potential V(x) = -  (x) A free particle has an energy greater than V() or positive energy. Represent the

2 2 energy as  2m k and the allowed form of the wavefunction in the free particle (V(x) = 0) regions becomes: AeBeikx  ikx , a superposition of free particle propagation to the right and left. The singular potential at x = 0 requires that solutions form the left and right be matched at that point.  Aeikx Be ikx (0 x  ) ()x [SE(ti).28]    ikx ikx Fe Ge(0 x ) Condition (i) at x = 0 requires that A + B = F + G ([SE(ti).29]). As before, condition (ii) is replaced by [SE(ti).27] because the potential has a delta singularity.

  22mm ikF ikG ikA ikB 22 (0) (AB ) xx  ()A BABF2m ()  G [SE(ti).30] ik 2   Defining   m , the equations become: k 2 A  BFG

A(1 2iBiF ) (1 2 ) G We have two equations and four unknowns so some additional information must be specified if the problem is to be defined. Let’s assume that a particle beam of known

k 2 intensity m A is incident from the left so that the value of A is set. That is, the wave function A eikx is not a normalizable free particle state, but rather represents an infinite region filled with a finite uniform density of particles moving at a constant velocity

k /m to the right. As the particles interact with the potential, there is some probability

1 A repulsive delta potential would be of the form  (x). One can just make  negative in SC6 to treat the repulsive case.

9/17/2010 SP425 Notes –Time Independent Schrödinger Equation Ch2-35 that they will be transmitted past the singularity at x = 0 and some probability that they will be reflected. No particles are being shot in from the right so G = 0. A  BF

A(1 2iBiF ) (1 2 )  The unknown values of B and F can be expressed in terms of the known parameter A. A standard solution technique is addition/subtraction.

B i AF1 A   m [SE(ti).31] 1i 1i k 2 In order to find the probability for reflection and transmission, one should carefully i   *  compute the probability current, Jxt(,)2m  x x  . For this problem, the B 2 F 2 reflection and transmission probabilities are given by R = | /A| and T = | /A| as the probability current is proportional to k |A|2 for a wave of the form A eikx.

22 BF 2 1 RT2222 [SE(ti).32] AA11 

2 2 The results pass the first sanity check, R + T = 1. Recalling that E =  2m k , the final results become: 11 RT222Em1; m 2 1 11 mE 22 EE0  2   For very low positive energies, the reflection probability approaches 1. For very high energies, the transmission probability approaches 1. The high energy result might be expected; the low energy result is not unreasonable.

SC7.) The step potential:

A classic problem is that of a free plane wave incident on a step of height Vo.

9/17/2010 SP425 Notes –Time Independent Schrödinger Equation Ch2-36

2 2 The incident particles are characterized by an energy E =  2m k1 . The corresponding wave function (including right and left traveling particles is:

 (,)xt  AeBeik1 x  ik1 x (region I)

 (,)xt  Feik2 x Ge ik2 x (region II)

V0 where k2 = k1 1  E  . [SE(ti).33]

Condition (i);  cont.: A + B = F + G

d Condition (ii); /dx cont.: k1 A - k1 B = k2 F - k2 G

Consider the case of a beam of particles incident from the left that is partially transmitted and partially reflected at x = 0. For this case, G = 0 (nothing incoming from the right). Condition (i);  cont.: A + B = F

d Condition (ii); /dx cont.: k1 A - k1 B = k2 F 2k k k FAB1 12A [SE(ti).34] kk12  kk 12 In this case the particles have different ‘velocities’ in the two regions so the flux (flow) of particles in each region is related to the probability density times the velocity. Using

9/17/2010 SP425 Notes –Time Independent Schrödinger Equation Ch2-37 i   *  the probability current, Jxt(,)2m  x x  , to find the particle flux ensures

2 ik1x  ||k1 that all the factors are correct. For a wave of the form A e , the current is m A and

2  ||k1 -ik1x it is  m B for a wave of the form B e . The intensity or flux of the transmitted

 ||k2 2 wave is m F . Combining,

2 2 Fk2 4[kk12 k 1 k 2] T2 2; R 2 [SE(ti).35] Ak1 []kk12 [] kk 12

B 2 F 2 Exercise: Verify that T + R = 1 while | /A| + | /A| may not be one.

SC8.) The high step potential:

Consider the case for E < Vo. The region x > 0 is classically forbidden. The parameter

V0 k2 is imaginary and may be represented as i where  = k1  E  1.

2k k i FAB1 1A [SE(ti).36] ki11  ki

Note: These results can be found by replacing k2 by i in the equations [SE(ti).34] derived above. The lesson is that math works. 

i  *  - x Exercise: Find Jxt(,)2m x x  for a wave of the form F e  

Exercise: Show that R = 1 and T = 0 when Vo > E.

ki  ki  i2 The result B 1 A shows that BAAe1 where ki1   ki1  

 V0  arctan arctan E   1 .  k1  

9/17/2010 SP425 Notes –Time Independent Schrödinger Equation Ch2-38

Significance Results: The wave is 100% reflected from a (very long) step up with

V0 Vo > E, and the reflected wave is phase shifted by - 2arctan  E   1 .

 As the reflected wave is of the form e-ikx, the negative phase shift is equivalent to having traveled farther. [Fuzzythink: The wave traveling to the right penetrates a distance into the barrier before it is turned around and sent back.]

 i2 The limiting case, Vo >> E gives  = arctan(huge) or  about 2 , and e  -1.

SC9.) The finite square well: A symmetric finite square well has the potential:  0(xa )  V(x) Vx() V0 (  a x  a ) [SE(ti).37]   0(xa ) Region I Region II Region III x < 0 -a 0 a As the potential is symmetric, the 0 x energy eigenfunctions are alternately even and odd. The ground state is Ckxcos  Ae x  x x -x Ae   x even  A e ; C cos[kx] ; A e . Dkxsin[ ] Ae k  2m ()VE ;    2m E  2 0 2 -VO Once the even or odd symmetry case is chosen and the boundary conditions are met at x = a, the conditions at x = – a are redundant. In order to for a state to be bound, it must

have an energy less than the potential at infinity (E < V; E < 0);.

Matching for symmetric states at a: i.) The wave function is continuous. C cos[ka] = A e-a

9/17/2010 SP425 Notes –Time Independent Schrödinger Equation Ch2-39 ii.) The first derivative of the wavefunction is continuous. -kC sin[ka] = -A e-a

 The continuity requirements are met if x  is continuous at the boundaries which yields a transcendental equation for the eigen-condition. ka tan[ka] = a [SE(ti).38] Review the energy relations; look for the product: ka.

k  2m ()VE ;    2m E  k 22   2mV ;()aVa22 2m  k 2a2 2 0 2 2 0 2 0

Define z = ka and zV2 2m a2 to facilitate analyzing the transcendental equation. o 2 0

z 2 tan[z ] 0 z  1 [SE(ti).39]

Note multiply u values from the plot by . z = u Plot[{Tan[Pi u],UnitStep[4-u] Sqrt[(4 Pi/(Pi u))^2 -1],Sqrt[(6 Pi/(Pi u))^2 -1]},{u,0,6 },PlotRange  {0,12}]

z 2 z Symmetric states: tan(z) and 0 z 1 plotted versus u = /

z The variable u = /was chosen to identify the critical values of z as the half-integer multiples of .

2 2 For the lower curve, u 2 =  V0 = 4; for the upper curve, u 2 =  V0 = 6. o 2ma2 o 2ma2

9/17/2010 SP425 Notes –Time Independent Schrödinger Equation Ch2-40 We see that the well depth Vo set the right endpoint for the curve representing the right

z 2 hand side or 0 z 1 . That curve ‘starts’ at + for z = 0 so the curve is guaranteed to intersect the tan(z) at least once. The number of intersections (even bound states) is

2 1 + Int[  V0 ] where Int[x] is the integer part of x.  2ma2

The kinetic energy of each state is: 22k  22z . The smallest-k state is deepest in the 2m 2ma2 well, and it penetrates the least into the forbidden region. For the ground state,  zGS  2 . For comparison, the ground state energy for an infinite well of width 2a is 22 . The ground state kinetic energy for the finite well is just a little less than 2(2)ma2

22 as the particle can penetrate into the forbidden region just a little making the 2(2)ma2  effective well width just a little larger than 2a and making zGS just a little less than 2 . As the energy of the state approaches zero, the energy deficit in the forbidden region decreases, and the particle can penetrate farther into the forbidden region. Effectively, the wavefunction needs only to turn over and be concave toward zero in order that it can match a near zero energy decaying exponential. Just bending toward zero rather, than reaching zero, means that the wavefunction in the well can be narrower (as measured in wavelengths) by almost a quarter wave at each edge for a weakly bound state. The conclusion is that if the well-depth V  n222  , the well has n + 1 bound 0 2(2)ma2 states. For problems in one dimension, a well of any depth will have at least one bound state. This relation predicts that the attractive delta function potential will have only one bound state (See problem 3).

z 2 Anti-symmetric matching case: cot[z ]  0 z 1

9/17/2010 SP425 Notes –Time Independent Schrödinger Equation Ch2-41 Plot[{Cot[Pi z],-Sqrt[( (Pi/2.55)/(Pi z))2-1],-Sqrt[((Pi)/(Pi z))2-1],-Sqrt[(4 Pi/(Pi z))2- 1]},{z,0,4 },PlotRange  {-8,8}] Note multiply z values from the plot by .

2 z0 Anti-symmetric States cot[z ]  z 1

There are no roots for zo < ½ . Requirement is Vo > E1 for the infinite square well of the same width. The requirement to have at least one anti-symmetric state is that V  22 . Width = 2a. The number of odd states is the integer part of (zo/ + ½). o 2(2)ma2 

A shallow well V  22 has one bound state; its energy is proportional to the 0 2(2)ma2 square of the well depth if the width is held fixed. See problem 32.

 22 Vm02 aV0 2 EV  (shallow-well limit, V  2 ) GS 2 0 2 0 2ma 2ma2 

Wavelengths in the well: A =.95; k = 1;Plot[{UnitStep[2+x] Sin[Pi x] (1 - UnitStep[x - 2]),UnitStep[2+x] Sin[ 0.8Pi x] (1 - UnitStep[x - 2]), (1 - UnitStep[x + 2]) A Exp[k (2 + x)] - UnitStep[x-2] A Exp[- k (-2 + x)]},{x,-3,3}]

9/17/2010 SP425 Notes –Time Independent Schrödinger Equation Ch2-42

Comparison of the fourth infinite square well state* with the fourth finite well state: The finite well state has just turned over at the well edges and can match to a decaying exponential for a very small negative energy. The finite well state has just over 3 (½  in the well as compared to 4 (½  for the infinite well state.* This behavior is for a fourth state with energy that lies near the top of the well. Deep in the well, a finite well state has nearly a full n(½ ) in the well. The yellow-shaded region is classically forbidden.

Comparison of the ground state wavefunction deep in the well with short penetration, the third state with greater penetration and the fifth state with even greater penetration. Note that all of there states are even about the center of the well. What would be true about the second and fourth states?

9/17/2010 SP425 Notes –Time Independent Schrödinger Equation Ch2-43 Exercise: Suppose that the well depth is increased so that the value of the energy deficit in the forbidden region increases by a large factor. What would change about the behavior of the wave function in the forbidden region? How would the number of ‘wavelengths in the well’ change from its current value of just over three? Would the energy of the state relative to the bottom of the well increase or decrease?

Exercise: Repeat the developments above to find the analog of [SE(ti).38] for the odd bound states of the finite well. Continue to find the analog of [SE(ti).39].

Phase method: an alternative development of the eigen-condition ki  Recalling the finite barrier with E < the step height, the result B 1 A shows ki1   ki  1 i2  V0 that BAAe where   arctan arctan E  1 .  k1   ki1  

 x forbidden Ckxcos  Ae forbidden  x Ae   x Dkxsin[ ] Ae

 

Condition for a stationary state: Basically, we have a wave reflecting back and forth between the two walls. Energy eigenstates are stationary. If the state is stationary, the phase change for a complete cycle must be an integer multiple of 2. The infinite

9/17/2010 SP425 Notes –Time Independent Schrödinger Equation Ch2-44 square has a width of d = 2a. Starting at the left edge of the infinite square well, the phase change is kd for the trip across,  for reflection from the infinite barrier, kd for the trip back and  for the left side reflection that gets the wave ready to go across again. Stationary requires:  kd +  + kd +  = m (2) or k = (m – 1) ( /d) n or k = /d for integers n > 0 where d is the well width Note that this phase requirement is identical to the resonance condition that one might apply to a string fixed at the ends or a double closed (or perhaps open) organ pipe. The resonance condition is that after bouncing around for a cycle, the net phase change of the wave is an integer multiple of 2. Exercise: Why is the case n = 0 excluded just above? Would including negative integers add anything?

n Exercise: Confirm that the condition k = /d is equivalent to (d = 2a), the width of the well, being an integer number of half-de Broglie wavelengths.

Exercise: The wavelength of a free particle with kinetic energy equal to the well depth

  h V would be well 2mV . Define the dimensionless parameter  = well/L where L is the full width of the well. Develop expressions for the number of even and odd bound states in terms of .

Exercise: The wavelength of a free particle with kinetic energy equal to the well depth

  h V would be well 2mV . Define the dimensionless parameter  = well/L where L is

9/17/2010 SP425 Notes –Time Independent Schrödinger Equation Ch2-45 the full width of the well. Imagine that V is fixed and L is varied. Show that the uncertainty product in the ground state is a minimum for (not checked!)

V(x)

-a a 0 x E=0 E E Vo E

Vo

K E K E=0 Finite Well Energies Barrier Energies

In the barrier problem, Vo is the height of the step, and E is the kinetic energy of the particle incident on the step. For the well, the kinetic energy of the particle in the well

is Vo + E. Note that E is negative for particles bound in the well.

This same process is applied to the finite well noting that the energy of the particle relative to the bottom of the well is Vo + E and the barrier height is Vo. The phase shift on reflection form the step potential at each edge is 2

VV V arctan00 1 arctan 1 arctan 01 barrier   EK      well   (Vo E)  The resonance condition is that after bouncing around for a cycle, the net phase change of the wave is an integer multiple of 2. travel + reflection + travel + reflection = integer cycles kd +  + kd +  = m (2)

9/17/2010 SP425 Notes –Time Independent Schrödinger Equation Ch2-46 ||E or kd – m = - 2  = - 2 arctan VE   0   Take the tangent of the expression, (tan[ + ½] = - cot[]; …. . )  tan[½kd – m(½ )] = ||E = /  VE0   k

Recalling that the well width is d = 2a,

ka tan[ka – m(½ )] =a [SE(ti).40] Compare this result to [SE(ti).38], the condition for symmetric bound states. What conditions are required for equation [SE(ti).40] to reproduce the eigen-relation for the anti-symmetric bound states in a finite well?

The point: The normal matching process employed to find eigenstates assumes stationary forms when, in fact, the state can only be stationary if the assumed state has a correct energy eigenvalue. If the energy is not correct, there is a contradiction leading to non-physical behavior. The wave function matches to growing exponentials in the forbidden regions and all the probability is sucked into those regions, a kind of quantum drawing and quartering. Rubbish leads to rubbish. The method introduced here does not assume that the state is stationary; it develops a condition that must be satisfied if the state is to be stationary. No drawing and quartering need by inflicted on the particle’s probability. A non-eigen energy value can correspond to a perfectly valid normalizable mixed state. However, that state will not be stationary.

11 iEt1iEt2 (,)xt a12 () xe  b() xe O.K., but not stationary.

9/17/2010 SP425 Notes –Time Independent Schrödinger Equation Ch2-47 The classical (phase integral) or WKB1 approximation: The WKB approximation is a generalization of the phase condition for eigenstates to the cases of non-constant potentials. It is the topic of chapter 8.

NOTE: The next sample calculation uses a different notation and barrier description than does the corresponding presentation in Griffiths. Study the development here and use it to guide you as you complete the details of the Griffiths example. Exploit correspondences. Under what conditions should the probability of transmission through the barriers be the same, etc.

SC10.) The barrier potential problem: A free particle with energy E is to be incident from the left on a rectangular barrier of height VB and width d. The potential is  0(0)x   represented as:Vx() VB (0 x d ) [SE(ti).41] and the wavefunction as:   0(x  d )

 Aeikx Be ikx  iKx iKx ()xCeDe .  ikx  Fe Notation Alert: The notation and structure of the barrier vary from reference to reference. For example, Griffiths has a barrier depth of Vo whereas the barrier of this

example has a height above the background zero of +VB. The barrier in this example runs between 0 and d. For comparison with Griffiths, one should replace d by 2a. 2mE The propagation constant in the region I (x < 0) and in region III (x > d) is k =  .

2(mEV B ) The propagation constant in the region II (0 < x < d) is K =  .

1 The WKB approximation as applied in quantum mechanics is named after Wentzel, Kramer and Brillouin. The mathematical underpinnings of the method are due to Liouville, Rayleigh and Jeffreys.

9/17/2010 SP425 Notes –Time Independent Schrödinger Equation Ch2-48 Solve for 9/17/2010 The notation Solve for Matching at Matching at 2 Equate theexpressionsfor2 me thatthenewsymbol iscloselyrelatedto

C C E: Particle Energy and and x x F = =0: = d D D F e : : : SP425 Notes–Time Indepe ikd 2 hasbeenintroducedasashorthand.Iused K C= 2 K D= ( ( K +k K -k K C (actuallyfor2 2 K ( (actually for2 K D= K C= K +k ( ) ) ( C e k K -k C e A A and 2 ( +(

+( V : Barrier Energy

A -B B iKd iKd A +B=CD ) ( ) ( ( ( +De K +k K –k K +k -De K -k K D ndent Schrödinger Equation F e F e ) = ikd ikd . K C K C ) ) ) ) ) K ) B F -iKd e A -iKd B = e A ( . +( -iKd and 2 = ( and 2 +( iKd ) =Fe C -D =kFe =( =( ( K +k K +k K -k K –k K D K D K +k K -k ikd ) ) ) ) ) ) )

ikd

F F B B ) e e ) F iKd F -iKd e e 

iKd -iKd ()

xCe De outside thebarrier inside thebarrier

[SE(ti).42] F F K  =

k      = = eBe Ae F e 2( mEV K iKx iKx

k ikx ikx Fe 2 ikd  mE 

toremind   ikx

Ch2 B )

-49  

Solve for F and B in terms of A using the addition-subtraction method. Replace F with F eikd.

Mega-Results: Fk2 Keikd  [SE(ti).43] A 2cos[](kK Kd i K22 k )sin[ Kd]

BiKkKd()sin[]22  [SE(ti).44] A 2kK cos[ Kd ] i ( K22 k )sin[ Kd ]

Exercise: Show that these equations are equivalent to [2.167] and [2.168] on page 82 of Griffiths.

Exercise: What is the condition on Kd that leads to a zero amplitude for the reflected wave. If nothing is reflected, what must be the probability for transmission?  F Exercise: Show that /A has magnitude one for Kd = m . What is the transmission probability for Kd = m ? 

F 2 B 2 Exercise: Show that | /A| + | /A| has magnitude one for the case of the barrier

[SE(ti).41] studied for which k1 = k3.

Exercise: Suppose that the potential were modified such that the kinetic energy of the particle in region III was one quarter of its value in region I. What relation would you F 2 B 2 expect to hold in place of: | /A| + | /A| = 1? 

9/17/2010 SP425 Notes –Time Independent Schrödinger Equation Ch2-50 Equation [SE(ti).43] predicts 100% transmission in the cases that Kd = m. That is: whenever the barrier width is an integral number of half wavelengths of the wave inside the barrier. [This behavior is called the Ramsauer-Townsend Effect. An optical analog can be found in the optical Fabry-Perot interferometer and in multilayer coatings. A half-lambda thick layer optical layer is sometimes called an absentee layer in the optical coating business.]

Hours of drudgery have revealed that the inverse of the transmission probability is more easily interpreted the transmission T itself.

2 22 22 1 A 2k K cos[ Kd ] i ( K k )sin[ Kd ] 2 k K cos[ Kd ]  i ( K k )sin[ Kd ] T  ikd ikd F 22kK e kK e

2 22 2 2k K cos[ Kd ] ( K k )sin[ Kd ] ()Kk222 TK1 cos22 [d ] sin [Kd ] 2kK 2 4kK22 The expression involves cos2[Kd] and sin2[Kd]  use the number one trig identity.

22222 12 2()4Kk kK 2 TKdKdcos [ ] sin [ ]22 sin [Kd ] 4kK

2mE 2(mEV B ) Recalling the energetics of the problem, k =  and K =  . 2 12([EV E )  4 EEV (  ) TK1sBBin[d] 4(EE VB ) 22 12([2EVBB ) 4 EEV (  ) VB22(mEV B ) TK1sin[]1sind  d 4(EE VBB ) 4(EE V )

Exercise: This result appears to differ by a sign from the result in Griffiths on page 82. The results actually agree. Explain.

Exercise: The cos2+sin2 identity was used. What is the analogous identity satisfied by sinh and cosh? K would be imaginary if VB were greater than E. What is sin[i]?

Fabry-Perot behavior for Transmission plotted versus barrier thickness for various ratios of VB to E.

9/17/2010 SP425 Notes –Time Independent Schrödinger Equation Ch2-51

VB = 0.98 E VB = 0.5 E e=1; vb = .98; Plot[ 1/(1 + (vb^2/(4 e (e - e=1; vb = .5; Plot[ 1/(1 + (vb^2/(4 e vb)) (Sin[(e -vb) d])^2)), {d,0,400}] (e - vb)(Sin[(e -vb) d])^2)), {d,0,20}]

To add: Show that the amplitude inside the barrier is large in the case of VB = 0.98 E and high transmission. That is the wave is resonant in the barrier. Consider the special case of resonant transmission: This occurs for sin[Kd] = 0. It follows that:

Fk2 Keikd 1 for sin[Kd ] 0 AkKKdiKkKd2cos[](22 )sin[]

BiKkKd()sin[]22 0forsin[Kd ] 0 AkKKdiKk2 cos[ ] (22 )sin[ Kd ] What is true about the amplitudes of the wavefunction inside the barrier for the condition of resonant transmission? 2K C = (K + k) A + (K – k) B  2K C = (K + k) A 2K D = (K - k) A + (K + k) B 2K D = (K - k) A

2mE 2(mEV B ) k =  and K =  .

AK k A E A E  CD1;1 22K  2  E VEBBV

The amplitudes inside the barrier can by very large in the limit that |E – VB| << E. In this case, one observe the shape transmission spikes with ‘near zero’ transmission off C D the resonance condition. For the case that VB = 0.98 E, /A  8.07and /A  6.07. One

9/17/2010 SP425 Notes –Time Independent Schrödinger Equation Ch2-52 must square these to find the probability density inside the barrier. In this case of sharp resonance, the total probability density inside the barrier is about 100 times that in the incident beam. The strongly enhanced probability in side the barrier in the case of sharp resonance is rarely mentioned even though it is a crucial point in explaining some VB = 0.98 E phenomena. C must have a greater magnitude than D because the net probability current in side the barrier must be in the +x direction. We can even find a rather indirect connection to thermal neutron induced nuclear fission of 235U.

Thermal (low kinetic energy) neutrons are more effective than faster neutrons at triggering the fission of 235U. One factor is just the time that the neutron ‘spends in the nucleus’. This factor can by thought of as v-1, the inverse of the neutron’s speed. An additional factor is the existence of a virtual bound state with E just above 0 for the potential well that the nucleus presents to the neutron. A low energy neutron almost satisfies the condition to be resonant in the well and so it builds an enhanced probability density in the well. Eventually, it would leak out of the well, but it ‘spends more time inside the well’ before leaking away due to the resonance making it more probable that the neutron will perturb the nucleus in to a state from which it will fission. (Reliability: less than average) The resonance condition is that after bouncing around for a cycle, the net phase change of the wave is an integer multiple of 2. The wave partially reflects at each position of rapid change of the potential The waves interfere constructively and thus show a resonance by building a larger amplitude if the net phase change for a cycle of bouncing is about m(2).

AAPLICATIONS: Quantum Wells (http://www.utdallas.edu/~frensley/technical/hetphys/node11.html)

If one makes a heterostructure with sufficiently thin layers, quantum interference effects begin to appear prominently in the motion of the electrons. The simplest structure in which these may be observed is a quantum well, which simply consists of a thin layer of a narrower-gap semiconductor

9/17/2010 SP425 Notes –Time Independent Schrödinger Equation Ch2-53 between thicker layers of a wider-gap material [37]. The band profile then shows a ``rectangular well,'' as illustrated in Fig. 11.

Figure 11: Energy-band profile of a structure containing three quantum wells, showing the confined states in each well. The structure consists of GaAs wells of thickness 11, 8, and 5 nm in Al Ga As barrier layers. The gaps in the lines indicating the confined state energies show the locations of nodes of the corresponding wavefunctions.

Modulating retro-reflector From Wikipedia, the free encyclopedia Jump to: navigation, search According to the U.S. Navy's Naval Research Laboratory, a modulating retro-reflector (MRR) system combines an optical retro-reflector and an electro-optic shutter to allow two-way optical communications. [1]

Free space optical communication technology has emerged in recent years as an attractive alternative to the conventional Radio Frequency (RF) systems. This emergence is due in large part to the increasing maturity of lasers and compact optical systems that enable exploitation of the inherent advantages (over RF) of the much shorter wavelengths characteristic of optical and near-infrared carriers[1]:

Modulating Retro-reflector Technology Overview.[1]Larger bandwidth Low probability of intercept Immunity from interference or jamming Frequency spectrum allocation issue relief Smaller, lighter, lower power

9/17/2010 SP425 Notes –Time Independent Schrödinger Equation Ch2-54

Technology An MRR couples an optical retro-reflector with a Multiple Quantum Well (MQW) modulator to direct optical signals to receiver providing communications device.[2][3]

[edit] Origin of the concept of quantum wells In 1972, Charles H. Henry, a physicist and newly-appointed Head of the Semiconductor Electronics Research Department at Bell Laboratories, had a keen interest in the subject of integrated optics, the fabrication of optical circuits in which the light travels in waveguides.

In late 1972, while pondering the problems associated with waveguides, he had a sudden insight, a realization that a double hetero-structure is a waveguide for electron waves, not just for light waves. On further reflection, he saw that there is a complete analogy between the confinement of light by a slab waveguide and the confinement of electrons by the potential well that is formed from the difference in band gaps in a double hetero-structure.

Henry realized that there should be discrete modes (levels) in the potential well, and a simple estimate showed that if the active layer of the hetero-structure is as thin as several tens of nanometers, the electron levels would be split apart by tens of milli- electron volts, which should be observable. This structure is now called a quantum well.

9/17/2010 SP425 Notes –Time Independent Schrödinger Equation Ch2-55 Henry then calculated how this quantization would alter the optical absorption edge of the semiconductor. His conclusion was that instead of the optical absorption increasing smoothly, the absorption edge of a thin hetero-structure would appear as a series of steps.

http://en.wikipedia.org/wiki/Quantum_well_laser

[edit] Invention of the quantum well laser After this experiment showed the reality of the predicted quantum well energy levels, Henry tried to think of an application. He realized that the quantum well structure would alter the density of states of the semiconductor, and result in an improved semiconductor laser requiring fewer electrons and electron holes to reach laser threshold. Also, he realized that the laser wavelength could be changed merely by changing the thickness of the thin quantum well layers, whereas in the conventional laser a change in wavelength requires a change in layer composition. Such a laser, he reasoned, would have superior performance characteristics compared to the standard double hetero-structure lasers being made at that time.

Dingle and Henry received a patent on this new type of semiconductor laser comprising a pair of wide bandgap layers having an active region sandwiched between them, in which "the active layers are thin enough (e.g., about 1 to 50 nanometers) to separate the quantum levels of electrons confined therein. These lasers exhibit wavelength tunability by changing the thickness of the active layers. Also described is the possibility of threshold reductions resulting from modification of the density of electron states." The patent was issued on September 21, 1976, entitled "Quantum Effects in Hetero-structure Lasers," U.S. Patent No. 3,982,297 [2].

Quantum well lasers require fewer electrons and holes to reach threshold than conventional double hetero-structure lasers. A well-designed quantum well laser can have an exceedingly low threshold current.

Moreover, since quantum efficiency (photons-out per electrons-in) is largely limited by optical absorption by the electrons and holes, very high quantum efficiencies can be achieved with the quantum well laser.

To compensate for the reduction in active layer thickness, a small number of identical quantum wells are often used. This structure is called a multi-quantum well laser.

Appendix

9/17/2010 SP425 Notes –Time Independent Schrödinger Equation Ch2-56 Problems 1.) The following function has the Fourier series, 32 1 f ()tm33 sin[2 1]t  m0 (2m  1)

1

0.8

0.6

0.4

0.2  1(tttfort 1)222   0  2 ft() -2 -1.5 -1 -0.5 0.5 1 1.5 2   22  1(tttfort  1)2    2  0 -0.2 -0.4

-0.6

-0.8 -1 Parseval’s theorem states that:

T 2 2  222 ft() dt T c11 a  b for   ffm0 22fm T  mm11  2

     f ()tcffmfq0  acosmq t  bsin t . mq11

-3 -3 Use Parseval’s theorem for cfo = 0, afm = 0 and bfk = 32  (2 k + 1) and T = 2 to  1 evaluate the sum  6 which was needed in Ex. 2.2 in chapter 2 of Griffiths. m0 (2m  1) 6  Answer: /960.

2.) It is known that a finite square well in one dimension always has at least one bound state. The negative Dirac delta potential has exactly one bound state. Consider a finite well of depth |V | and width a. Use particle in a box formula E  n22  2 to estimate the o n 2ma2 number of bound states in the well. A delta function potential can be represented as a square well in the limit that a  0 while |Vo| a is held constant. a.) Show that the estimate for the number of bound states approaches zero (or perhaps one) in the limit a  0.

9/17/2010 SP425 Notes –Time Independent Schrödinger Equation Ch2-57 b.) Argue that a square well will always have at least one bound state (in one dimension). 

11 amxipmxˆˆcc 22mmcc   x 

† 11 amxipmxˆˆcc 22mmcc   x  [aaˆˆ,]† 1 Use the fundamental commutation relation [xˆ,]pˆ  i and the definitions above to

prove that [aaˆˆ,† ] 1.

4.) Use the commutation relation [aaˆˆ,† ] 1 and the form of the hamiltonian

ˆ  ˆˆ†  Haac ½ to show that if m is and eigenfunction with eigenvalue Em then

axˆ m () is an eigenfunction with eigenvalue Em c (or it is zero). ˆ  ˆˆ† Eigenvalue relation: Hmc()xaaxE½  mmm ()   ()x 

5.) Find o and raise it to find 1 and 2 using [SE(ti).21]. Show that they are normalized. Find expressions for xˆ and pˆˆin terms ofa and aˆ† .Find expression for x

n and p n based on the known results:

† axˆˆnnn() n 111 ()and x axn () n () x [SE(ti).21]

6.) An operator is has the value or equivalent to its action on and arbitrary function.

Consider the operator [x, px] which is the commutator of x and px.   [,xˆˆpxppxixixxx ] ˆˆ ˆˆ x  x x

 Let it act on a function f(x). [,xpˆˆx ] fx ()  ixxx i xfx () ?

9/17/2010 SP425 Notes –Time Independent Schrödinger Equation Ch2-58 a.) What is the value of [,xˆˆpx ]? Answer: [,xˆˆpix ] 

b.) What is the value of [,xˆˆp y ]? Use g(x,y) as the arbitrary function.

8.) The lowest two states of the QHO have the forms:

2 2 3  x /2 i(½c )t  x /2 it()2c 0 (,)xt Ae e and 1(,)xt Bxe e . Find real values for the normalization constants A and B. Consider the mixed state (,)x txt1 [ (,)  (,)]xt. Compute the probability density, x and p  for this 2 01 x mixed state.

9. An operator is has the value or equivalent to its action on an arbitrary function. a.) Consider the operator [x, px] which is the commutator of x and px.   [,xˆˆpxppxixixxx ] ˆˆ ˆˆ x  x x

 Let it act on a function f(x). [,xpˆˆx ] fx ()  ixxx i xfx () ?

What is the value of [,xˆˆpx ]? Answer: [,xˆˆpix ] 

2 2 b.) Consider the operator [x, x2 ] which is the commutator of x and x2 .

2 Let it act on a function f(x). [x, x2 ] f(x) = ? c.) Consider the operator [x, Hˆ ] which is the commutator of x and Hˆ where ˆ 2  ˆ H 2m x2 Vx(). Let it act on a function f(x). [x, H ] f(x) = ?

ˆ ˆ 10. A Hermitian operator Q has a complete set of eigenfunctions. Qqnn n. ˆ a.) Show that the expectation value of Q in the eigenstate m is qm. b.) Write down the defining condition for the operator Qˆ to be Hermitian.

9/17/2010 SP425 Notes –Time Independent Schrödinger Equation Ch2-59 c.) Show that the eigenvalues of Qˆ are real.

 ˆ d.) Show that the expectation value of Q in a mixed state  (,)x tax  mm (,)t is m1

 ˆ 2 Qa  mmq. m1 Conclude that Hermitian operators have real eigenvalues and expectation values. These properties make them candidates to represent observables.

 ˆ k e.) Show that the expectation value of (Q ) in a mixed state  (,)x tax  mm (,)t is m1

 ˆ 2 k Qaq  mm(). m1

11.) Assume that an eigenfunction has been found. ˆ  ˆˆ† Hmm()xaaxE0 ½  () m m ()x

Consider the new wavefunction âm(x); show that, if it exists, it is an eigenstate of the problem with eigenvalue Em - o.

12.) Use the operator relations axˆˆ††() n 1 () x annn 1  1 nn1 axˆˆnn() n1 () x annn1 x  1 1 [aaˆˆ† ] pˆˆ  i [aa†  ˆ] 2   2

to evaluate x n(x) and pˆ n(x).

Answer: x()xx1  nn()1  ()x ; nn 2211n

pˆ()xi  nn() x  1 () x nn 2211n

9/17/2010 SP425 Notes –Time Independent Schrödinger Equation Ch2-60

13.) a.) Use the operator relations

Answer: x()xx1  nn()1  ()x ; nn 2211n

pˆ()xi  nn() x  1 () x nn 2211n to evaluate the expectation value of x and pˆ in the mixed state a 0(x) + b 1(x) where a and b are complex numbers. b.) What condition must be place on a and b to ensure that a 0(x) + b 1(x) is normalized? 2 2 c.) Compute the expectation value of x and pˆ in the mixed state a 0(x) + b 1(x)

14.) Repeat the finite square well problem to find the analog of [SE(ti).38] for the odd bound states of the finite well. Continue to find the analog of [SE(ti).39].

15.) Show that the eigen-conditions for both the even and the odd bound states of a finite well can be extracted from [SE(ti).40]: ka tan[ka – m(½ )] =a.

16.) [Keep number fixed, reference shallow well]A shallow square well (one with

2ma2 potential depth V0  2 , a fraction of the ground state kinetic energy in an infinite well of width 2a) has one bound state with a binding energy that is quadratic in Vo. That is the state is bound by an energy proportional to the well depth times the well depth measured in infinite square well level separations. In particular,

V E 0 V 2 0 2ma2 Assume that the binding energy is much less than the well depth (or that the kinetic energy in the well is almost Vo. Use the Schrödinger equation to estimate the ratio of

9/17/2010 SP425 Notes –Time Independent Schrödinger Equation Ch2-61 the second derivative of  to the value of  at x = 0. Assuming that the change is very small, estimate ratio of the slope of the wavefunction to the wavefunction at x = a. Show that value is the decay constant  in the forbidden region. Express  in terms of

V the binding energy of the state. Conclude that E 0 V for a shallow well. binding 2 0 2ma2 17.) Repeat the process that led to equations [SE(ti).43] and [SE(ti).44] making the changes required for the case that VB > E. Discuss your results. What limit should lead to 100% reflection? 18.) Solve these relations for xˆ and pˆ . Use the commutation relation for the raising and lowering operators to deduce the relation for xˆ and pˆ .

11  amxipmxˆˆcc 22mmcc   x 

† 11 amxipmxˆˆcc 22mmcc   x  [aaˆˆ,]† 1

mc Express the result using  =  . What are the dimensions of ? Check the dimensions of for xˆ and pˆ .

19.) The eigenfunctions of the QSHO are known to be a complete set so a time t = 0, a

 solution can be represented as (,0)x  cmm ()x where the m(z) are the standard m0 ortho-normal eigen-set. The general form for a time dependent wave function

 im(½)0 t is (,)xt cmm () xe . Find the expectation value of the coordinate x in a m0 general state of the QSHO. Compute that expectation value of position as:

 [(,)]x tx (,) xtdx using series representations for the wave function and its 

9/17/2010 SP425 Notes –Time Independent Schrödinger Equation Ch2-62  complex conjugate. Also use the orthogonality relation,  ()xxdx ()  . Use:  mn mn

x  1 1 []aaˆˆ† and axˆ†()n 1 ()and x axnˆ() () x. Redefine the 2 nn11nn summation index in one sum as needed to make the summation ranges identical and show that the sums are the complex conjugates of one another. The result can be written as:

 1*Ceit00 Ce it 1|| Ceii eit0  || Ce e it  0  Acos[ t ] x 0

 i ½  -1 where CCe|| (2)  cckkk1 1 , A = 2  |C| k 0 That is: the expectation value of position has the same time dependence as does the position of a classical oscillator with the same m and k. Note that the expectation value of the classical observable x is real-valued.

34.) Study equations [SE(ti).43] and [SE(ti).44] making the changes required for the

case K = i that occurs for VB > E. Use the definitions of sine and cosine to evaluate sin[i] and cos[i] for real values .

18.) A particle is in a square 2D infinite well: (0  x  a; 0  y  a). The eigenfunctions

2 are: (,)xt 2 sinkx sinny eiknt with eigenvalues  =  kn22 . An kn a  aa   kn 2ma2   ensemble of identical infinite well systems is prepared according to the recipe:

(x,0) = 0.2 (x,0) +0.6 (x,0) + 0.3 (x,0) + 0.5 (x,0) + 0.5 (x,0) + 0.1 (x,0) a.) Find  b.) List the possible results for precise energy determinations made on each member of the ensemble. Give the relative probabilities for measuring each possible value. c.) An energy measurement returns the value 522 . Immediately after the 2ma2 measurement what are the probability densities at (½ a, ½ a) and at (¼ a, ¼ a)?

9/17/2010 SP425 Notes –Time Independent Schrödinger Equation Ch2-63 d.) An energy measurement on a new system in the ensemble returns the value 422 . ma2 1 Immediately after the measurement what are the probability densities at (½ a, /3 a) and 1 at ( /3 a, ½ a)?

19.) An ensemble of identical QHOs is prepared according to the recipe  ()x  1  ()xx 2 () ()x 6  123 What is the expectation value for the energy? What index value is assigned to the QHO ground state?

20.) An ensemble of hydrogen atom systems is prepared according to the recipe   ()rirrirri1 2 () ()  ()  ()  ()r 22 100 200 21, 1 210 211  a.) An energy measurement returns the value -3.4 eV. Immediately after the energy measurement, what are the possible values and relative probabilities of a measurement of the square of angular momentum? b.) Beginning with a fresh member of the ensemble, an energy measurement returns the value -3.4 eV. It is followed immediately by a measurement of Lz which returns 1.

Immediately after that measurement, what are the possible values and relative probabilities of a measurement of the square of angular momentum? c.) Beginning with a fresh member of the ensemble, a measurement of Lz which returns 0 . Immediately after that measurement, what are the possible values and relative probabilities of a measurement of the energy?

d.) Beginning with a fresh member of the ensemble, a measurement of Lz which returns 0 . Immediately after that measurement, what are the possible values and relative probabilities of a measurement of the square of the angular momentum?

9/17/2010 SP425 Notes –Time Independent Schrödinger Equation Ch2-64

21.) Verify that the number operator nˆ (=â†â) is Hermitian. First use the explicit form,

11 nˆ = mxccmx . Add a second proof using the 2mmcc x  2 x  † definition of Hermitian conjugate and the result that â+ (= â ) is the Hermitian

 conjugate of â (= â). Qqˆˆ† (,t) (,) qtd  (,) qtQqtd (,) . Note that - ii i i this defining property is a prescription for moving one operator at a time from the ket to the bra.

22.) The figure below includes a representation of a wavefunction for a finite well state. Which is it, the blue or the magenta plotted line? Assign n = 1 to the ground state. What is the n value of the finite well state illustrated?

The non-finite square well state plot represents a state appropriate for what problem? What n would be assigned to it in that case?

23.) The barrier penetration problem illustrates several important wave behaviors. Use the equations [SE(ti).44] and [SE(ti).42] to find that there is 100% transmission in the C D 2 case that Kd = m. What are /A and /A in this case? In particular, estimate the |C|

9/17/2010 SP425 Notes –Time Independent Schrödinger Equation Ch2-65 2 inside the barrier as a multiple of |A| when Kd = m and VB = 0.96 E. The condition reflects a resonance of the particle-wave inside the barrier.

C E D E 2 2 Wild Guesses: A ½½ ; A ½½ ; |C| = 9|A| EVB EVB

A  BFG 24.) Solve the equations for F and G. A(12)iBiF (12) G

ab A F Find the matrix such that:  . What equations follow if G = 0? cd B G

AG  FB 25.) Solve the equations for F and B. A(1 2iGFBi ) (1 2 )

ab A F Find the matrix such that:  . What equations follow if G = 0? cdG B The matrix related the incoming to outgoing amplitudes. It is called the scattering matrix of the S matrix.

C D 26.) For the barrier penetration problem, find expressions for /A and /A that are

k 2 analogous to [SE(ti).43]. The incident probability flux is /m |A| . Find an expression

for the flux inside the barrier when the resonance condition sin[Kd] = 0 is satisfied. Comment on the result.

Problem Development:

9/17/2010 SP425 Notes –Time Independent Schrödinger Equation Ch2-66 Exercise: The wavelength of a free particle with kinetic energy equal to the well depth

  h V would be well 2mV . Define the dimensionless parameter  = well/L where L is the full width of the well. Develop expressions for the number of even and odd bound states in terms of .

Exercise: The wavelength of a free particle with kinetic energy equal to the well depth

  h V would be well 2mV . Define the dimensionless parameter  = well/L where L is the full width of the well. Imagine that V is fixed and L is varied. Show that the uncertainty product in the ground state is a minimum for (not checked!)

References: 1. David J. Griffiths, Introduction to Quantum Mechanics, 2nd Edition, Pearson Prentice Hall (2005). 2. Richard Fitzpatrick, Quantum Mechanics Note Set, University of Texas. 3. Robert Leighton, Principles of Modern Physics, McGraw-Hill (1959).

Delta potential model: If the well-depth V  n222  , the well has n + 1 bound states. 0 2(2)ma2

2 boundstates  2(2)maVa1 8(mVa0 ) 1 220 22

9/17/2010 SP425 Notes –Time Independent Schrödinger Equation Ch2-67 For the long tall rectangle model of the delta function, Vo a remains constant in the limit that a goes to zero. We see that the number of bound states goes to one in this limit.

9/17/2010 SP425 Notes –Time Independent Schrödinger Equation Ch2-68