QM1 - Tutorial 6 Finite Potential Well
Yaakov Yudkin 30 November 2017
Contents
1 Scattering VS Bound States 1 1.1 Scattering ...... 1 1.2 Bound States ...... 1
2 Example: Half In nite Potential Well 2
3 Example: In nite Square Well with Delta Barrier 5
1 Scattering VS Bound States
Finite Well Potential Consider a nite potential well described by ( −V 0 < x < L V (x) = 0 0 else
The energy E of the particle can be either positive or negative but is must be larger than −V0.
1.1 Scattering
Energy If E > 0 this is a scattering problem. Since the particle is free the energy spectrum is continuous. Moreover, the energy is given as an initial condition. The task is then to nd the transmission and re ection coe cients using the probability currents.
Classical Picture Think of a classical particle in the potential V (x) described above. It can be anywhere between x = −∞ and x = +∞. This is why the particle is free.
Solution of Schrodinger's Equation In order to solve for the scattering coe cients we rst need to solve the Schrodinger equation. This is done by splitting up space and solving each region individually. In this case we get Aeikx + Be−ikx x < 0 ψ(x) = Ceiqx + De−iqx 0 < x < L F eikx x > L where √ 2mE p2m(E + V ) k = and q = 0 ~ ~ Using the boundary conditions we can then nd the scattering coe cients.
1.2 Bound States
Energy In the lecture we will discuss the case E < 0. The particle is inside the potential well. We thus expect discrete energy levels. The task is to nd them.
1 Classical Picture For E < 0 a classical particle cannot be anywhere. It is con ned to the region of the well, i.e. between x = 0 and x = L. Hence, this situation is called a bound state. The particle is bound to a certain region. In quantum mechanics the particle is still con ned to the region of the well but there is a nite probability to nd the particle inside the wall.
Solution of Schrodinger's Equation As mentioned above the goal is to nd the energy levels. For this we must solve the eigen value equation familiar to us as the time independent Schrodinger equation. Since the energy E is negative it will prove very useful to write E = − |E|. The Schrodinger equation is
2 d2 − ~ + V (x) ψ (x) = − |E| ψ (x) 2m dx2
We divide space into three regions and solve for ψ (x) in each of them. We get Aekx x < 0 ψ(x) = Beiqx + Ce−iqx 0 < x < L De−kx x > L where p2m |E| p2m (V − |E|) k = and q = 0 ~ ~ From the boundary conditions the eigen energies can be found. In general an exact solution is not possible. Instead we will obtain a transcendental equation which can be solved graphically.
2 Example: Half In nite Potential Well
Statement of the Problem Consider the well described by ∞ x < 0 V (x) = 0 0 < x < L V0 x > L
A particle of mass m is subject to this potential and found in a bound state. Find a transcendental equation for the energy eigen values and solve it graphically. Find the minimal depth V0 for a bound state to exist. Also nd the eigen functions.
Relevant Energy Interval Think classically. The energy must be higher then the bottom of the well but low enough so it cant leave. This is the case for the interval 0 < E < V0.
Split Space and Schrodinger We divide space into three. For the region x < 0 the wave function vanishes, i.e. ψ(x < 0) = 0. This is simple. The other two regions are 0 < x < L and x > L. The Schrodinger equation in those regions is ( 2 2 ~ ∂ − 2m ∂x2 ψ(x) = Eψ(x) 0 < x < L 2 2 ~ ∂ − 2m ∂x2 ψ(x) + V0ψ(x) = Eψ(x) x > L We rearrange them to the form ( ∂2 2mE ∂x2 ψ(x) = − 2 ψ(x) 0 < x < L 2 2 ~ ~ ∂ 2m(E−V0) 2 ψ(x) = − ψ(x) x > L 2m ∂x ~ In the rst equation we replace k2 = 2mE/~2. In the second equation we rst put in the minus sign. 2m (E − V ) 2m (V − E) − 0 = 0 = q2 ~ ~ 2 This is important. Since E < V0 this is a positive quantity. We can now replace it by q without having to worry about imaginary numbers. The equations become
( ∂2 2 ∂x2 ψ(x) = −k ψ(x) 0 < x < L 2 2 ~ ∂ 2 2m ∂x2 ψ(x) = q ψ(x) x > L
2 We solve them and add the region x < 0 for which the wave function vanishes. The total wave function is 0 x < 0 ψ(x) = Aeikx + Be−ikx 0 < x < L Ceqx + De−qx x > L
Boundary Conditions Next we must use boundary conditions. Since there are four unknown constants and the unknown energy we will need ve equations. One of them will obviously be the normalization condition, but lets leave that to the end. In order to solve the problem we need four boundary conditions. They are:
1. ψ(0) = 0 2. ψ(x → ∞) = 0
3. ψ(L−) = ψ(L+) 4. ψ0(L−) = ψ0(L+) Condition (1) gives A + B = 0 The solution in region 0 < x < L can thus be rephrased
Aeikx + Be−ikx = A eikx − e−ikx = 2A sin (kx) → A sin(kx)
From condition (2) we get C = 0 The total wave function is thus 0 x < 0 ψ(x) = A sin(kx) 0 < x < L De−qx x > L
Transcendental Equation for the Energy Now we use the boundary conditions (3) and (4). They give
A sin(kL) = De−qL
kA cos(kL) = −qDe−qL By dividing the second by the rst we can get rid of both A and D at once. One obtains q cot (kL) = − k Plugging the expression for k and q back in we get √ ! r 2mE V cot L = − 0 − 1 ~ E
The only unknown in this equation is the energy E. We cannot solve for E analytically (aka. transcendental equation).
Change of Variables An easy way to get a idea of the solution is to solve graphically. For simplicity we de ne the dimensionless quantities √ √ 2mE 2mV0 θ = L and θ0 = L ~ ~ The condition 0 < E < V0 becomes 0 < θ < θ0. The transcendental equation can now be written as s θ 2 cot θ = − 0 − 1 θ
We see that for θ > θ0 the rhs is imaginary. This re ects the condition 0 < θ < θ0 for which a bound state is possible.
3 Figure 1: Graphically solving the equation for the energy levels.
Graphical Solution For the graphical solution follow a few easy steps: 1. plot the lhs (here cot θ) as a function of θ
2. plot the rhs as a function of θ for dierent choices of θ0 (play around with it) 3. the points at which the two plots intersect are a solution
Minimal Depth Since the lhs is positive until π/2 there will be no solution to the equation if θ0 < π/2. This implies that if √ 2 2 2mV0 π ~ π L < → V0 < ~ 2 8mL2 there will be no bound state at all. The potential is to shallow for the particle to be con ned inside it.
Eigen Functions Next we want to nd the eigen functions belonging to the energies. Since we have found them we can treat k and q as known. Going back to condition (3) we have A sin(kL) = De−qL → D = A sin(kL)eqL and thus the total wave function is 0 x < 0 ψ(x) = A sin(kx) 0 < x < L A sin(kL)e−q(x−L) x > L
The remaining constant (A) is found by imposing normalization. Watch out: the total wave function must be normalized and not the individual parts. The condition is Z ∞ |ψ (x)|2 dx = 1 −∞ Here it becomes Z L Z ∞ |ψ (0 < x < L)|2 dx + |ψ (x > L)|2 dx = 1 0 L We can solve the integral
Z L Z ∞ 1 =|A|2 sin2(kx)dx + |A|2 sin2(kL)e2qL e−2qxdx 0 L L sin(2kL) e−2qL =|A|2 − + |A|2 sin2(kL)e2qL 2 4k 2q L sin(2kL) sin2(kL) =|A|2 − + 2 4k 2q
4 and nd the normalization constant
−1/2 L sin(2kL) sin2(kL) A = − + 2 4k 2q 3 Example: In nite Square Well with Delta Barrier
Statement of the Problem Find the bound states for a particle of mass m in the potential ( βδ(x) |x| < L V (x) = ∞ else
Compare your results to the in nite potential well without a delta function. Discuss the limit of vanishing and in nite delta. Hint: It might be easier to use sinusoidal functions over exponential.
Schrodinger Equation and Wave Function The wave function must vanish for |x| > L. Inside the well there are two distinct regions. The time independent Schrodinger equation is
2 d2 − ~ ψ(x) = Eψ(x) 2m dx2 in both regions and the solution is ( √ A cos (kx) + B sin (kx) −L < x < 0 2mE ψ(x) = where k = C cos (kx) + D sin (kx) 0 < x < L ~
Parity Since the potential is symmetric around x = 0 the wave functions must have a de nite parity, i.e. ψ(x) is either even or odd. We can treat both cases individually. We will rst treat the even case and then the odd case.
Even Wave Function The wave function must satisfy
ψ(x) = ψ(−x)
So if we replace x by −x in ψ(0 < x < L) it must be equal to ψ (−L < x < 0). This implies
C cos (kx) − D sin (kx) = A cos (kx) + B sin (kx) → A = C and B = −D and so the even wave function is ( A cos (kx) + B sin (kx) −L < x < 0 ψeven(x) = A cos (kx) − B sin (kx) 0 < x < L
Boundary Condition: Continuity of Wave Function We now use the boundary conditions at x = ±L. Since the potential is in nite there the wave function must vanish. The equations at x = L and at x = −L are identical. We get A cos (k : L) − B sin (kL) = 0 → B = A cot (kL) which means that the wave function is ( A cos (kx) + A cot (kL) sin (kx) −L < x < 0 ψeven(x) = A cos (kx) − A cot (kL) sin (kx) 0 < x < L
As can be seen the wave function ψ(x) takes the value A at x = 0 no matter what side we come from. This means that ψ(x) is already continuous at x = 0.
5 Boundary Condition: Discontinuity of Derivative The next condition we have to use is the discontinuity of the derivative. For this we take the Schrodinger equation
2 d2 − ~ ψ(x) + βδ(x)ψ(x) = Eψ(x) 2m dx2 and integrate from − to . By taking the limit → 0 we obtain the condition
2 ~ dψ dψ − = βψ(0) 2m dx 0+ dx 0− We use this condition. 2 ~ (−kA cot (kL) − kA cot (kL)) = αA 2m Canceling A and cleaning up a little we get
mα 2 cot (kL) = − → tan (kL) = − ~ kL ~2k mLα which is a transcendental equation for kL.
Even Parity Eigen Energies The transcendental equation can be solved graphically. The solutions kL = zn can be put into a list {zn}. We then nd the energies √ 2 2 2mE even ~ zn L = zn → En = ~ 2mL2 Odd Wave Function Starting from ( A cos (kx) + B sin (kx) −L < x < 0 ψ(x) = C cos (kx) + D sin (kx) 0 < x < L we impose the condition ψ(x) = −ψ(−x) This means that replacing x → −x in ψ (0 < x < L) results in −ψ (−L < x < 0) (note the minus sign). We have
C cos (kx) − D sin (kx) = −A cos (kx) − B sin (kx) → A = −C and B = D and so the wave function can be written ( A cos (kx) + B sin (kx) −L < x < 0 ψodd(x) = −A cos (kx) + B sin (kx) 0 < x < L
Boundary Conditions In this case it is easiest to rst use the continuity condition at x = 0. Since
ψ(0+) = ψ(0−) we get A = −A → A = 0 and so the odd wave function is ( B sin (kx) −L < x < 0 ψodd(x) = → ψodd(x) = B sin (kx) , |x| < L B sin (kx) 0 < x < L
The derivative at x = 0 is thus continuous. This is so albeit the delta function because ψ(0) = 0. If this were not so the solution would not be odd. The last condition we have to use is the vanishing of the wave function at x = ±L. From it we get the quantization
B sin (kL) = 0 → kL = nπ , n = 1, 2,...
6 Odd Parity Eigen Energies From the quantization of kL the energies are obtained √ 2 2 2 2mE odd ~ π n L = nπ → En = ~ 2mL2 Interpretation of the Odd Parity Lets look at the energies of the odd wave functions rst. We found
2 2 2 2 2 2 odd ~ π n ~ π (2n) En = = 2mL2 2m (2L)2
These are the even-numbered eigen energies of an in nite potential well of length 2L. They were not altered by the delta function because their wave functions vanish at the location of the delta. The solution is independent of β. The wave functions obtained here are identical to the odd ones (odd function but even-numbered) obtained when considering the in nite potential well in the interval [−L, L].
Interpretation of the Even Parity The even wave functions ψeven(x) are altered by the delta function. The integration constants A and B will depend on the height of the potential β. We take a look at the limiting cases β → 0 and β → ∞. For a vanishing delta function (β → 0) we reproduce the in nite well of length 2L. We can see this directly from the transcendental equation for the energies
2 tan (kL) = − ~ kL mLβ since for β → 0 we must have π tan (kL) = ∞ → kL = (2n + 1) 2 The energies in this case are √ 2 2mE π ~2π2 (2n + 1) L = (2n + 1) → En = ~ 2 2m (2L)2 We have found the missing are odd-numbered eigen energies of the in nite potential well. For an in nitely high delta (β → ∞) we get
tan (kL) = 0 → kL = nπ which is the same condition we found for the odd wave functions. We see that taking the delta to ∞ splits the well of length 2L into two independent wells of length L.
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