Class 5: Quantum Harmonic Oscillator – Ladder Operators

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Class 5: Quantum harmonic oscillator – Ladder operators

Ladder operators The time independent Schrödinger equation for the quantum harmonic oscillator can be written as

1 ()pmx2+ 222ω ψ = E ψ , (5.1) 2m

where the momentum operator p is

d p= − i ℏ . (5.2) dx

If p were a number, we could factorize

p2+ m 222ω x =−+( ipmxipmx ω)( + ω ). (5.3)

However, we need to remember that p is an operator. In fact, because

d  d  ()()−+ipmxipmxω + ωψ =−+ℏ mx ω  ℏ + mx ωψ  dx  dx  d  d ψ  =−+ℏmxω  ℏ + mx ω ψ  dx  dx  d dψ  d ψ =−ℏ ℏ +mxωψ  + mx ω ℏ + mx2 ωψ 2 2 dxdx  dx (5.4) d2ψ d d ψ =−ℏ2 − ℏ()mxωψ + mx ω ℏ + mx2 ωψ 2 2 dx2 dx dx d 2ψ =−ℏ2 −mωψ ℏ + m 222 ω x ψ dx 2 =p2ψ + mx 222 ωψ − m ωψℏ

we see that there is an extra term, such that

p2+ m 222ω x =−+( ipmxipmx ω)( + ωω) + m ℏ (5.5)

Even so it is useful to consider the two dimensionless operators

1 a+ =() − ipmx + ω , (5.6) 2ℏmω and

1 a− =() + ipmx + ω . (5.7) 2ℏmω

The reason for labeling the operators with subscripts + and – will become clear later. Since the order of operation is important, we have

1 1 aa− + =() ++ ipmxω() −+ ipmx ω 2ℏmω 2 ℏ m ω (5.8) 1 2 222 =p + mω x − im ω () xp − px  , 2ℏmω

and

1 1 aa+ − =() −+ ipmxω() ++ ipmx ω 2ℏmω 2 ℏ m ω (5.9) 1 2 222 =p + mω x + im ω () xp − px  . 2ℏmω

Each expression contains the combination xp− px which is called the commutator of x and p.

The standard notation for the commutator of two operators A and B is

[ A, B] = AB − BA . (5.10)

Since

dψ d []xp,ψ=− ixℏ + i ℏ() x ψ = i ℏ ψ , (5.11) dx dx we can write

[x, p] = i ℏ . (5.12)

Also, because the Hamiltonian is

p2 mω 2 x 2 1 H=+ =() pmx2 + 222ω , (5.13) 2m 2 2 m

we have

1 1  H 1 a a= H +ℏω =+ . (5.14) − + ℏω2  ℏ ω 2

Similarly

1 1  H 1 a a= H +ℏω =− . (5.15) + − ℏω2  ℏ ω 2

These last two equations can be combined to show that

1 ℏ H=ω () aaaa−+ + +− . (5.16) 2

We also see that

[aa−+,] = aa −+ − aa +− = 1. (5.17)

The Schrödinger equation can be written as

1  E a− a + −  ψ = ψ , (5.18) 2  ℏω or as

1  E a+ a − +  ψ = ψ . (5.19) 2  ℏω

Suppose we have a solution ψ of the Schrödinger equation corresponding to energy E. Then a+ψ is also a solution to the Schrödinger equation but for energy E + ℏω. To see why, consider

ℏ 1  Ha()+ψ= ω aa +− +  () a + ψ 2  ℏ 1  =ωaaa+−+ ψ + a + ψ  (5.20) 2  ℏ 1  =ωa+ a − a + ψ + ψ  . 2 

Now using equation (5.14), this gives

ℏ1 1 1  ℏ Ha()+ψω= aH +  ψψψ ++=  aE+ () ψωψ + ℏω 2 2  (5.21) ℏ =()E + ω a + ψ , which proves our contention.

Similarly, if ψ is a solution of the Schrödinger equation corresponding to energy E, then a−ψ is also a ℏ solution to the Schrödinger equation but for energy E − ω. The operators a+ and a− are called ladder

ℏ operators , because the raising operator a+ moves up the energy ladder by a step of ω and the lowering ℏ operator a− moves down the energy ladder by a step of ω.

Since the minimum value of the potential energy is zero and occurs at a single value of x, the lowest energy for the QHO must be greater than zero. Let the wave function for the minimum energy be ψ 0 ( x). Since there is no energy level below this minimum value, we must have

a−ψ 0 = 0. (5.22)

When combined with the normalization condition, this equation allows us to find ψ 0 ( x), since it gives

(+ip + mω x ) ψ 0 = 0. (5.23)

The normalized solution is found to be

1 4 mω x 2 − mω  2ℏ ψ 0 (x )=   e . (5.24) π ℏ 

The energy of this state is easily found by applying the Hamiltonian operator given by equation (5.15)

ℏ1  1 ℏ Hψω0= a+ a − +  ψ 0 = ωψ 0 . (5.25) 2  2

We see that the lowest energy is

1 E = ℏω, (5.26) 0 2

and by applying the raising operator, the energy of the nth state is

1  ℏ En = n +  ω, (5.27) 2 

Normalization of the wave functions Although the ladder operators can be used to create a new wave function from a given normalized wave function, the new wave function is not normalized. To determine the normalization constant, we need to explore some more properties of the ladder operators. First consider

∞ ∗ ∫ f() ag+ dx , (5.28) −∞ where f and g are functions of x. Putting in the explicit form of the operator, we have

∞ ∞   ∗ 1 ℏ ∗dg ∗ fagdx()+ = − f + mxfgdxω ∫ℏ ∫   −∞2 mω −∞ dx  ∗ 1 ∞ d() f g df ∗  = −ℏ ++ ℏ g mxfgdxω ∗  (5.29) ℏ ∫ dx dx 2 mω −∞   ∞ ∗ 1∗ ∞ 1 df ∗  =−+ℏf g   ℏ g + mω xf g  dx . ℏ  −∞ ℏ ∫ 2mω 2 m ω −∞ dx 

Provided f∗ g → 0 as x → ∞ , the first term on the last line is zero. Also the remaining integral is

1 ∞ df ∗  ∞ ℏg+ mxfgdxω ∗ = gaf ∗ dx . (5.30) ℏ ∫  ∫ ()− 2 mω −∞ dx  −∞

Hence we have shown that

∞ ∞ ∗ ∗ ∫f() agdx+= ∫ gaf() − dx . (5.31) −∞ −∞

In a similar manner, we can show that

∞ ∞ ∗ ∗ ∫f() agdx−= ∫ gaf() + dx . (5.32) −∞ −∞

(The operators a− and a+ are said to be Hermitian conjugates .)

Suppose ψ n represents a normalized wave functions for any value of n. Let g =ψ n and f= a +ψ n in equation (5.31). Then

∞ ∞ ∞ ∗ ∗   ∗ (5.33) ∫()()()a++ψψnn a dx= ∫ ψ n aa −+ ψ n  dx = ∫ ψψ nn() aa −+ dx . −∞ −∞ −∞

Using equation (5.14) and equation (5.27), we have

* * a− a + ψn=( n + 1) ψ n . (5.34)

On using this in equation (5.33), we get

∞ ∞ ∗ ∗ ∫()()()a+ψψnn a + dx= ∫ ψ n n + 1 ψ n dx . (5.35) −∞ −∞

Since the wave functions are assumed normalized, we see that

a+ψn= n + 1 ψ n + 1 . (5.36)

In a similar way, by starting with equation (5.32), we find

a−ψn= n ψ n − 1. (5.37)

Since

1 12 1 3 ⋯ ψψn==a+− n1() a +− ψ n 2 =() a +− ψ n 3 = , n nn()−1 nnn()() − 1 − 2 we see that

1 n ψn = ()a+ ψ 0. (5.38) n!

Problem 2.12 Find x,, p x2 , p 2 , and T for the nth stationary state of the harmonic oscillator. Check that the Uncertainty Principle is satisfied.

From equations (5.6) and (5.7), we find

2mω a+ a = x , (5.39) − + ℏ and

2 aa− = ip . (5.40) − + ℏmω

Hence

∞ℏ ∞ x=∫ψψnn xdx = ∫ ψ n() aadx− += + ψ n 0, −∞2mω −∞ where use has been made of the orthogonality of the wave functions, and the lowering and raising properties of the ladder operators.

Similarly we see that p = 0.

Also

∞ℏ ∞ ℏ ∞ 2 2 2 2 2 x=ψψn xdx n = ψψ n() aa−+ += n dx ψ n () aaaaaa −−++−+ +++ ψ n dx ∫2mω ∫ 2 m ω ∫ −∞ −∞ −∞ E 1  ℏ =n =n +  , mω2 2  m ω where we have used equation (5.16). Similarly

∞ℏ ∞ ℏ ∞ 2 2 mω2 m ω 2 2 p=ψψn pdx n =− ψψ n() aa− − + n dx =− ψ n () aaaaaa− −−+ −+ +_ + ψ n dx ∫2 ∫ 2 ∫ −∞ −∞ −∞ ℏmω∞ ℏ m ω 2 E  1  =ψaaaa + ψ dx =n =+ nℏ m ω . ∫ n()− + + _ n ℏ   2−∞ 2ω  2 

Also,

2 p 1 1  ℏ 1 T= =+ n  ω = E n . 2m 22  2

To check whether the Uncertainty Principle holds, consider

ℏ 2 2 1 1ℏ  1 ℏ σx σ p =xp =+ n ⋅+ nmnω =+  . 2mω 2  2

We see that the principle holds, with equality for n = 0.

Orthonormality of the wave functions The stationary state wave functions for the harmonic oscillator are orthonormal, i.e.

∞ ∫ ψmn ψdx = δ mn . (5.41) −∞

The orthogonality of the wave function can be shown by considering

∞ ∞ ∞ ∫ψψmnHdx= ∫ ψψ mnn EdxE = nmn ∫ ψψ dx . (5.42) −∞ −∞ −∞

Now

∞ ∞ ℏ22 ℏ 2∞ 2 ∞ d  d ψ n ∫∫ψψψmnmHdx=−+2 Vdx  ψ n =− ∫∫ ψ m2 dx + ψψ mn Vdx (5.43) −∞ −∞ 2mdx  2 m−∞ dx −∞

By integration by parts, we have

∞ d2ψ d ψ  ∞ ∞ dd ψψ ψndx= ψ n − mn dx ∫mdx2  m dx  ∫ dxdx −∞ −∞ −∞ (5.44) ∞ ∞ ∞ 2 dψn  d ψ m  d ψ m =ψm  − ψ n  + ∫ 2 ψ n dx . dx −∞ dx  −∞ −∞ dx

Applying the boundary conditions, the first two terms on the right hand side are both equal to zero. Putting the result in equation (5.43), we have

∞ℏ2 ∞ 2 ∞∞ ∞ d ψ m ∫ψψmnHdx=− ∫∫∫ ψ n2 dx + ψψ nm Vdx = ψψ nmmnm HdxE = ∫ ψψ dx . (5.45) −∞2m −∞ dx −∞ −∞ −∞

Hence

∞ ∞ Em∫ψψ nm dxE= n ∫ ψψ nm dx . (5.46) −∞ −∞

Because the energy values are not equal if m≠ n , we see that the wave functions must be orthogonal.

This is a quite general result, which does not depend on the particular potential energy function. Two wave functions that have different energies are orthogonal.