Monthly Archives: January 2017

Angular momentum – Poisson bracket to commutator

Shankar, R. (1994), Principles of Quantum Mechanics, Plenum Press. Section 7.4, Exercise 7.4.8.

The classical angular momentum components are

\displaystyle   \ell_{x} \displaystyle  = \displaystyle  yp_{z}-zp_{y}\ \ \ \ \ (1)
\displaystyle  \ell_{y} \displaystyle  = \displaystyle  zp_{x}-xp_{z}\ \ \ \ \ (2)
\displaystyle  \ell_{z} \displaystyle  = \displaystyle  xp_{y}-yp_{x} \ \ \ \ \ (3)

In the position basis, we can replace each coordinate by its quantum operator {x\rightarrow X}, {y\rightarrow Y} and {z\rightarrow Z}, and each momentum component by the derivative {p_{i}\rightarrow-i\hbar\partial/\partial q_{i}}, where {q_{i}} is the {i}th coordinate. This gives

\displaystyle   L_{x} \displaystyle  = \displaystyle  -i\hbar\left(y\frac{\partial}{\partial z}-z\frac{\partial}{\partial y}\right)\ \ \ \ \ (4)
\displaystyle  L_{y} \displaystyle  = \displaystyle  -i\hbar\left(z\frac{\partial}{\partial x}-x\frac{\partial}{\partial z}\right)\ \ \ \ \ (5)
\displaystyle  L_{z} \displaystyle  = \displaystyle  -i\hbar\left(x\frac{\partial}{\partial y}-y\frac{\partial}{\partial x}\right) \ \ \ \ \ (6)

Because coordinates always commute with momentum components of other coordinates ({x} commutes with {p_{y}} and {p_{z}}, etc), there is no ordering ambiguity in making the transition from classical to quantum mechanics. That is, we could place the coordinate on either side of the momentum in each term for all components {L_{i}}.

Classically, we can calculate the Poisson brackets for the angular momentum components. For example

\displaystyle   \left\{ \ell_{x},\ell_{y}\right\} \displaystyle  = \displaystyle  \sum_{i}\left(\frac{\partial\ell_{x}}{\partial q_{i}}\frac{\partial\ell_{y}}{\partial p_{i}}-\frac{\partial\ell_{x}}{\partial p_{i}}\frac{\partial\ell_{y}}{\partial q_{i}}\right)\ \ \ \ \ (7)
\displaystyle  \displaystyle  = \displaystyle  -p_{y}\left(-x\right)-yp_{z}\ \ \ \ \ (8)
\displaystyle  \displaystyle  = \displaystyle  xp_{y}-yp_{z}\ \ \ \ \ (9)
\displaystyle  \displaystyle  = \displaystyle  \ell_{z} \ \ \ \ \ (10)

According to the rule for converting classical Poisson brackets to quantum commutators, we should get (since there is no ordering ambiguity)

\displaystyle  \left[L_{x},L_{y}\right]=i\hbar L_{z} \ \ \ \ \ (11)

As we’ve seen earlier, this is verified by direct calculation using the position-momentum commutator

\displaystyle  \left[q_{i},p_{j}\right]=i\hbar\delta_{ij} \ \ \ \ \ (12)

Poisson brackets to commutators: classical to quantum

Shankar, R. (1994), Principles of Quantum Mechanics, Plenum Press. Section 7.4, Exercise 7.4.7.

The postulates of quantum mechanics that we described earlier included specifications for the matrix elements of position {X} and momentum {P} in position space:

\displaystyle \left\langle x\left|X\right|x^{\prime}\right\rangle \displaystyle = \displaystyle x\delta\left(x-x^{\prime}\right)\ \ \ \ \ (1)
\displaystyle \left\langle x\left|P\right|x^{\prime}\right\rangle \displaystyle = \displaystyle -i\hbar\delta^{\prime}\left(x-x^{\prime}\right) \ \ \ \ \ (2)

A more fundamental form of this postulate is to specify the commutation relation between {X} and {P}, which is independent of the basis and is

\displaystyle \left[X,P\right]=i\hbar \ \ \ \ \ (3)

 

This allows the construction of explicit forms of the operators in other bases, such as the momentum basis, where

\displaystyle X \displaystyle = \displaystyle i\hbar\frac{d}{dp}\ \ \ \ \ (4)
\displaystyle P \displaystyle = \displaystyle p \ \ \ \ \ (5)

We can verify this by calculating the commutator by applying it to a function {f\left(p\right)}:

\displaystyle \left[X,P\right]f \displaystyle = \displaystyle i\hbar\frac{d}{dp}\left(pf\left(p\right)\right)-i\hbar p\frac{d}{dp}f\left(p\right)\ \ \ \ \ (6)
\displaystyle \displaystyle = \displaystyle i\hbar f\left(p\right)+i\hbar p\frac{d}{dp}f\left(p\right)-i\hbar p\frac{d}{dp}f\left(p\right)\ \ \ \ \ (7)
\displaystyle \displaystyle = \displaystyle i\hbar f\left(p\right) \ \ \ \ \ (8)

Thus 3 is satisfied in the momentum basis as well.

The standard recipe for converting a classical system to a quantum one is to first calculate the Poisson bracket for two physical quantities in the classical system, which gives

\displaystyle \left\{ \omega,\lambda\right\} =\sum_{i}\left(\frac{\partial\omega}{\partial q_{i}}\frac{\partial\lambda}{\partial p_{i}}-\frac{\partial\omega}{\partial p_{i}}\frac{\partial\lambda}{\partial q_{i}}\right) \ \ \ \ \ (9)

where {q_{i}} and {p_{i}} are the canonical coordinates and momenta. To convert to a quantum commutator, we replace the classical quantities by their quantum operator equivalents and the Poisson bracket by {i\hbar} times the corresponding commutator. That is

\displaystyle \left[\Omega,\Lambda\right]=i\hbar\left\{ \omega,\lambda\right\} \ \ \ \ \ (10)

For the case of {X} and {P}, we have, in classical mechanics in one dimension

\displaystyle \left\{ x,p\right\} =\frac{\partial x}{\partial x}\frac{\partial p}{\partial p}-\frac{\partial x}{\partial p}\frac{\partial p}{\partial x}=1 \ \ \ \ \ (11)

 

so the quantum commutator is given by 3.

For other quantities, we can use the theorems on the Poisson brackets to reduce them:

\displaystyle \left\{ \omega,\lambda\right\} \displaystyle = \displaystyle -\left\{ \lambda,\omega\right\} \ \ \ \ \ (12)
\displaystyle \left\{ \omega,\lambda+\sigma\right\} \displaystyle = \displaystyle \left\{ \omega,\lambda\right\} +\left\{ \omega,\sigma\right\} \ \ \ \ \ (13)
\displaystyle \left\{ \omega,\lambda\sigma\right\} \displaystyle = \displaystyle \left\{ \omega,\lambda\right\} \sigma+\left\{ \omega,\sigma\right\} \lambda \ \ \ \ \ (14)

Quantum commutators obey similar rules

\displaystyle \left[\Omega,\Lambda\right] \displaystyle = \displaystyle -\left[\Lambda,\Omega\right]\ \ \ \ \ (15)
\displaystyle \left[\Omega,\Lambda+\Gamma\right] \displaystyle = \displaystyle \left[\Omega,\Lambda\right]+\left[\Omega,\Gamma\right]\ \ \ \ \ (16)
\displaystyle \left[\Omega\Lambda,\Gamma\right] \displaystyle = \displaystyle \Omega\left[\Lambda,\Gamma\right]+\left[\Omega,\Gamma\right]\Lambda \ \ \ \ \ (17)

The main difference between Poisson brackets and commutators is that, for the latter, the order of the operators in the last equation can make a difference. That is, in 14 we could also have written

\displaystyle \left\{ \omega,\lambda\sigma\right\} =\sigma\left\{ \omega,\lambda\right\} +\lambda\left\{ \omega,\sigma\right\} \ \ \ \ \ (18)

since all three quantities are numerical (not operators), so multiplication commutes. In 17 it is not true in general that, for example

\displaystyle \Omega\left[\Lambda,\Gamma\right]+\left[\Omega,\Gamma\right]\Lambda=\left[\Lambda,\Gamma\right]\Omega+\left[\Omega,\Gamma\right]\Lambda \ \ \ \ \ (19)

The conversion from classical to quantum mechanics can then be achieved in general by replacing

\displaystyle \left\{ \omega\left(x,p\right),\lambda\left(x,p\right)\right\} =\gamma\left(x,p\right) \ \ \ \ \ (20)

by

\displaystyle \left[\Omega\left(X,P\right),\Lambda\left(X,P\right)\right]=i\hbar\Gamma\left(X,P\right) \ \ \ \ \ (21)

where each of the operators in the last equation is obtained by replacing {x} in the first equation by {X} and {p} by {P}. We do need to be careful with the ordering of the operators in the quantum version, however.

As an example, suppose we have

\displaystyle \Omega \displaystyle = \displaystyle X\ \ \ \ \ (22)
\displaystyle \Lambda \displaystyle = \displaystyle X^{2}+P^{2} \ \ \ \ \ (23)

In the classical version, we calculate the Poisson bracket

\displaystyle \left\{ \omega,\lambda\right\} \displaystyle = \displaystyle \left\{ x,x^{2}+p^{2}\right\} \ \ \ \ \ (24)
\displaystyle \displaystyle = \displaystyle \left\{ x,x^{2}\right\} +\left\{ x,p^{2}\right\} \ \ \ \ \ (25)
\displaystyle \displaystyle = \displaystyle 0+2\left\{ x,p\right\} p\ \ \ \ \ (26)
\displaystyle \displaystyle = \displaystyle 2p \ \ \ \ \ (27)

Thus, by our rule above, the quantum version should be

\displaystyle \left[\Omega,\Lambda\right]=2i\hbar P \ \ \ \ \ (28)

We can verify this using 17

\displaystyle \left[X,X^{2}+P^{2}\right] \displaystyle = \displaystyle \left[X,X^{2}\right]+\left[X,P^{2}\right]\ \ \ \ \ (29)
\displaystyle \displaystyle = \displaystyle 0-\left[P^{2},X\right]\ \ \ \ \ (30)
\displaystyle \displaystyle = \displaystyle -P\left[P,X\right]-\left[P,X\right]P\ \ \ \ \ (31)
\displaystyle \displaystyle = \displaystyle -P\left(-i\hbar\right)-\left(-i\hbar\right)P\ \ \ \ \ (32)
\displaystyle \displaystyle = \displaystyle 2i\hbar P \ \ \ \ \ (33)

In this case, there is no ordering ambiguity in the quantum version, since {\left[X,P\right]=i\hbar} is just a number.

For a second example, suppose we have

\displaystyle \Omega \displaystyle = \displaystyle X^{2}\ \ \ \ \ (34)
\displaystyle \Lambda \displaystyle = \displaystyle P^{2} \ \ \ \ \ (35)

The classical version gives us, using the relations 14, 11 and 27

\displaystyle \left\{ x^{2},p^{2}\right\} \displaystyle = \displaystyle -\left\{ p^{2},x^{2}\right\} \ \ \ \ \ (36)
\displaystyle \displaystyle = \displaystyle -2\left\{ p^{2},x\right\} x\ \ \ \ \ (37)
\displaystyle \displaystyle = \displaystyle 2\left\{ x,p^{2}\right\} x\ \ \ \ \ (38)
\displaystyle \displaystyle = \displaystyle 4px \ \ \ \ \ (39)

In the classical case, this result is the same as {4xp}, but because {X} and {P} don’t commute in the quantum form, we need to be careful about the ordering.

We can do the calculation:

\displaystyle \left[X^{2},P^{2}\right]=X\left[X,P^{2}\right]+\left[X,P^{2}\right]X \ \ \ \ \ (40)

From 33 we have

\displaystyle \left[X,P^{2}\right]=2i\hbar P \ \ \ \ \ (41)

so we get

\displaystyle \left[X^{2},P^{2}\right]=2i\hbar\left(XP+PX\right) \ \ \ \ \ (42)

Thus if the Poisson bracket involves a product of {p} and {x}, this should be replaced by

\displaystyle xp\mbox{ or }px\rightarrow\frac{1}{2}\left(XP+PX\right) \ \ \ \ \ (43)

in the quantum version.

Changing the position basis with a unitary transformation

Shankar, R. (1994), Principles of Quantum Mechanics, Plenum Press. Section 7.4, Exercise 7.4.9.

The standard representation of the position and momentum operators in the position basis is

\displaystyle   X \displaystyle  \rightarrow \displaystyle  x\ \ \ \ \ (1)
\displaystyle  P \displaystyle  \rightarrow \displaystyle  -i\hbar\frac{d}{dx} \ \ \ \ \ (2)

It turns out it’s possible to modify this definition by adding some arbitrary function of position {f\left(x\right)} to {P} so we have

\displaystyle   X^{\prime} \displaystyle  \rightarrow \displaystyle  x\ \ \ \ \ (3)
\displaystyle  P^{\prime} \displaystyle  \rightarrow \displaystyle  -i\hbar\frac{d}{dx}+f\left(x\right) \ \ \ \ \ (4)

Since any function of {x} commutes with {X}, the commutation relations remain unchanged, so we have

\displaystyle  \left[X^{\prime},P^{\prime}\right]=i\hbar \ \ \ \ \ (5)

Another way of interpreting this change in operators is by using the unitary transformation of the {X} basis, in the form

\displaystyle  \left|x\right\rangle \rightarrow\left|\tilde{x}\right\rangle =e^{ig\left(X\right)/\hbar}\left|x\right\rangle =e^{ig\left(x\right)/\hbar}\left|x\right\rangle  \ \ \ \ \ (6)

where

\displaystyle  g\left(x\right)\equiv\int^{x}f\left(x^{\prime}\right)dx^{\prime} \ \ \ \ \ (7)

The last equality in 6 comes from the fact that operating on {\left|x\right\rangle } with any function of the {X} operator (provided the function can be expanded in a power series) results in multiplying {\left|x\right\rangle } by the same function, but with the operator {X} replaced by the numeric position value.

To verify this works, we can calcuate the matrix elements of the old {X} and {P} operators in the new basis. We have

\displaystyle  \left\langle \tilde{x}\left|X\right|\tilde{x}^{\prime}\right\rangle =\left\langle x\left|e^{-ig\left(x\right)/\hbar}Xe^{ig\left(x^{\prime}\right)/\hbar}\right|x^{\prime}\right\rangle  \ \ \ \ \ (8)

At this stage, since the two exponentials are numerical functions and not operators, we can take them outside the bracket to

\displaystyle   \left\langle \tilde{x}\left|X\right|\tilde{x}^{\prime}\right\rangle \displaystyle  = \displaystyle  e^{-ig\left(x\right)/\hbar}e^{ig\left(x^{\prime}\right)/\hbar}\left\langle x\left|X\right|x^{\prime}\right\rangle \ \ \ \ \ (9)
\displaystyle  \displaystyle  = \displaystyle  e^{-ig\left(x\right)/\hbar}e^{ig\left(x^{\prime}\right)/\hbar}x^{\prime}\delta\left(x-x^{\prime}\right)\ \ \ \ \ (10)
\displaystyle  \displaystyle  = \displaystyle  x\delta\left(x-x^{\prime}\right) \ \ \ \ \ (11)

The exponentials cancel in the last line since the delta function is non-zero only when {x=x^{\prime}}.

The above result can also be obtained by inserting a couple of identity operators into 8:

\displaystyle   \left\langle x\left|e^{-ig\left(x\right)/\hbar}Xe^{ig\left(x^{\prime}\right)/\hbar}\right|x^{\prime}\right\rangle \displaystyle  = \displaystyle  \int\int\left\langle x\left|e^{-ig\left(x\right)/\hbar}\right|y\right\rangle \left\langle y\left|X\right|z\right\rangle \left\langle z\left|e^{ig\left(x^{\prime}\right)/\hbar}\right|x^{\prime}\right\rangle dy\;dz\ \ \ \ \ (12)
\displaystyle  \displaystyle  = \displaystyle  \int\int\left\langle x\left|e^{-ig\left(x\right)/\hbar}\right|y\right\rangle z\delta\left(y-z\right)\left\langle z\left|e^{ig\left(x^{\prime}\right)/\hbar}\right|x^{\prime}\right\rangle dy\;dz\ \ \ \ \ (13)
\displaystyle  \displaystyle  = \displaystyle  \int\left\langle x\left|e^{-ig\left(x\right)/\hbar}\right|z\right\rangle z\left\langle z\left|e^{ig\left(x^{\prime}\right)/\hbar}\right|x^{\prime}\right\rangle dz\ \ \ \ \ (14)
\displaystyle  \displaystyle  = \displaystyle  \int e^{i\left[g\left(x^{\prime}\right)-g\left(x\right)\right]/\hbar}\left\langle x\left|z\right.\right\rangle z\left\langle z\left|x^{\prime}\right.\right\rangle dz\ \ \ \ \ (15)
\displaystyle  \displaystyle  = \displaystyle  \int e^{i\left[g\left(x^{\prime}\right)-g\left(x\right)\right]/\hbar}\delta\left(x-z\right)z\delta\left(z-x^{\prime}\right)dz\ \ \ \ \ (16)
\displaystyle  \displaystyle  = \displaystyle  e^{i\left[g\left(x^{\prime}\right)-g\left(x\right)\right]/\hbar}x^{\prime}\delta\left(x-x^{\prime}\right)\ \ \ \ \ (17)
\displaystyle  \displaystyle  = \displaystyle  x\delta\left(x-x^{\prime}\right) \ \ \ \ \ (18)

The momentum operator works as follows. Using the original definition 2 on the modified basis we have

\displaystyle   \left\langle \tilde{x}\left|P\right|\tilde{x}^{\prime}\right\rangle \displaystyle  = \displaystyle  -i\hbar\left\langle x\left|e^{-ig\left(x\right)/\hbar}\frac{d}{dx^{\prime}}e^{ig\left(x^{\prime}\right)/\hbar}\right|x^{\prime}\right\rangle \ \ \ \ \ (19)
\displaystyle  \displaystyle  = \displaystyle  -i\hbar\left\langle x\left|e^{-ig\left(x\right)/\hbar}\frac{i}{\hbar}e^{ig\left(x^{\prime}\right)/\hbar}\frac{dg\left(x^{\prime}\right)}{dx^{\prime}}\right|x^{\prime}\right\rangle -\ \ \ \ \ (20)
\displaystyle  \displaystyle  \displaystyle  i\hbar\left\langle x\left|e^{-ig\left(x\right)/\hbar}e^{ig\left(x^{\prime}\right)/\hbar}\frac{d}{dx^{\prime}}\right|x^{\prime}\right\rangle \ \ \ \ \ (21)

From 7 we have

\displaystyle  \frac{dg\left(x\right)}{dx}=\frac{d}{dx}\int^{x}f\left(x^{\prime}\right)dx^{\prime}=f\left(x\right) \ \ \ \ \ (22)

This gives

\displaystyle   \left\langle \tilde{x}\left|P\right|\tilde{x}^{\prime}\right\rangle \displaystyle  = \displaystyle  \left\langle x\left|e^{i\left[g\left(x^{\prime}\right)-g\left(x\right)\right]/\hbar}\left[f\left(x^{\prime}\right)-i\hbar\frac{d}{dx^{\prime}}\right]\right|x^{\prime}\right\rangle \ \ \ \ \ (23)
\displaystyle  \displaystyle  = \displaystyle  e^{i\left[g\left(x^{\prime}\right)-g\left(x\right)\right]/\hbar}\left[f\left(x^{\prime}\right)-i\hbar\frac{d}{dx^{\prime}}\right]\left\langle x\left|x^{\prime}\right.\right\rangle \ \ \ \ \ (24)
\displaystyle  \displaystyle  = \displaystyle  e^{i\left[g\left(x^{\prime}\right)-g\left(x\right)\right]/\hbar}\left[f\left(x^{\prime}\right)-i\hbar\frac{d}{dx^{\prime}}\right]\delta\left(x-x^{\prime}\right)\ \ \ \ \ (25)
\displaystyle  \displaystyle  = \displaystyle  \left[f\left(x\right)-i\hbar\frac{d}{dx}\right]\delta\left(x-x^{\prime}\right) \ \ \ \ \ (26)

This shows that by a unitary change of {X} basis 6, we transform the position and momentum operators (well, just the momentum operator, really) according to 3. We’ve multiplied the original {\left|x\right\rangle } states by a phase factor which depends on some function {f\left(x\right)}. This doesn’t change the matrix elements of {X}, but it does add {f\left(x\right)} to the matrix elements of {P}. The commonly used definition of {P} is thus with {f\left(x\right)=0}.

Harmonic oscillator – raising and lowering operators as functions of time

Shankar, R. (1994), Principles of Quantum Mechanics, Plenum Press. Section 7.4, Exercise 7.4.6.

We’ll consider here the problem of finding the averages of the raising and lowering operators (from the harmonic oscillator) as functions of time, that is, we want to find {\left\langle a\left(t\right)\right\rangle } and {\left\langle a^{\dagger}\left(t\right)\right\rangle }. At first glance we might think they are both zero, since they are defined in terms of position and momentum as

\displaystyle   a^{\dagger} \displaystyle  = \displaystyle  \frac{1}{\sqrt{2\hbar m\omega}}\left[-iP+m\omega X\right]\ \ \ \ \ (1)
\displaystyle  a \displaystyle  = \displaystyle  \frac{1}{\sqrt{2\hbar m\omega}}\left[iP+m\omega X\right] \ \ \ \ \ (2)

and the averages of {P} and {X} in any of the energy eigenstates of the harmonic oscillator are all zero. However, suppose we have a mixed state {\left|\psi\right\rangle } which can be written as a sum over the eigenstates as

\displaystyle   \psi\left(t\right) \displaystyle  = \displaystyle  \sum_{n=0}^{\infty}c_{n}e^{-iE_{n}t/\hbar}\left|n\right\rangle \ \ \ \ \ (3)
\displaystyle  \displaystyle  = \displaystyle  \sum_{n=0}^{\infty}c_{n}e^{-i\left(2n+1\right)\omega t/2}\left|n\right\rangle \ \ \ \ \ (4)

where in the second line we used the energies of the oscillator as

\displaystyle  E_{n}=\hbar\omega\left(n+\frac{1}{2}\right) \ \ \ \ \ (5)

We now have

\displaystyle   \left\langle a\left(t\right)\right\rangle \displaystyle  = \displaystyle  \left\langle \psi\left|a\right|\psi\right\rangle \ \ \ \ \ (6)
\displaystyle  \displaystyle  = \displaystyle  \sum_{n=0}^{\infty}\sum_{m=0}^{\infty}c_{m}^*e^{i\left(2m+1\right)\omega t/2}c_{n}e^{-i\left(2n+1\right)\omega t/2}\left\langle m\left|a\right|n\right\rangle \ \ \ \ \ (7)
\displaystyle  \displaystyle  = \displaystyle  \sum_{n=0}^{\infty}\sum_{m=0}^{\infty}c_{m}^*c_{n}e^{i\left(m-n\right)\omega t}\left\langle m\left|a\right|n\right\rangle \ \ \ \ \ (8)

We can now use the formula

\displaystyle  a\left|n\right\rangle =\sqrt{n}\left|n-1\right\rangle \ \ \ \ \ (9)

This gives

\displaystyle   \left\langle a\left(t\right)\right\rangle \displaystyle  = \displaystyle  \sum_{n=0}^{\infty}\sum_{m=0}^{\infty}c_{m}^*c_{n}e^{i\left(m-n\right)\omega t}\sqrt{n}\left\langle m\left|n-1\right.\right\rangle \ \ \ \ \ (10)
\displaystyle  \displaystyle  = \displaystyle  \sum_{n=0}^{\infty}\sum_{m=0}^{\infty}c_{m}^*c_{n}e^{i\left(m-n\right)\omega t}\sqrt{n}\delta_{m,n-1}\ \ \ \ \ (11)
\displaystyle  \displaystyle  = \displaystyle  e^{-i\omega t}\sum_{n=0}^{\infty}c_{n-1}^*c_{n}\sqrt{n}\ \ \ \ \ (12)
\displaystyle  \displaystyle  = \displaystyle  e^{-i\omega t}\left\langle a\left(0\right)\right\rangle \ \ \ \ \ (13)

Note that if {\left|\psi\right\rangle } is an eigenstate, then only one of the coefficients {c_{n}} is non-zero, so {\left\langle a\left(0\right)\right\rangle =0} as we’d expect.

The derivation for {\left\langle a^{\dagger}\left(t\right)\right\rangle } is similar:

\displaystyle   \left\langle a^{\dagger}\left(t\right)\right\rangle \displaystyle  = \displaystyle  \left\langle \psi\left|a^{\dagger}\right|\psi\right\rangle \ \ \ \ \ (14)
\displaystyle  \displaystyle  = \displaystyle  \sum_{n=0}^{\infty}\sum_{m=0}^{\infty}c_{m}^*e^{i\left(2m+1\right)\omega t/2}c_{n}e^{-i\left(2n+1\right)\omega t/2}\left\langle m\left|a^{\dagger}\right|n\right\rangle \ \ \ \ \ (15)
\displaystyle  \displaystyle  = \displaystyle  \sum_{n=0}^{\infty}\sum_{m=0}^{\infty}c_{m}^*c_{n}e^{i\left(m-n\right)\omega t}\left\langle m\left|a^{\dagger}\right|n\right\rangle \ \ \ \ \ (16)

We can now use the formula

\displaystyle  a^{\dagger}\left|n\right\rangle =\sqrt{n+1}\left|n+1\right\rangle \ \ \ \ \ (17)

This gives

\displaystyle   \left\langle a^{\dagger}\left(t\right)\right\rangle \displaystyle  = \displaystyle  \sum_{n=0}^{\infty}\sum_{m=0}^{\infty}c_{m}^*c_{n}e^{i\left(m-n\right)\omega t}\sqrt{n+1}\left\langle m\left|n+1\right.\right\rangle \ \ \ \ \ (18)
\displaystyle  \displaystyle  = \displaystyle  \sum_{n=0}^{\infty}\sum_{m=0}^{\infty}c_{m}^*c_{n}e^{i\left(m-n\right)\omega t}\sqrt{n+1}\delta_{m,n+1}\ \ \ \ \ (19)
\displaystyle  \displaystyle  = \displaystyle  e^{i\omega t}\sum_{n=0}^{\infty}c_{n+1}^*c_{n}\sqrt{n+1}\ \ \ \ \ (20)
\displaystyle  \displaystyle  = \displaystyle  e^{i\omega t}\left\langle a^{\dagger}\left(0\right)\right\rangle \ \ \ \ \ (21)

Harmonic oscillator – mixed initial state and Ehrenfest’s theorem

Shankar, R. (1994), Principles of Quantum Mechanics, Plenum Press. Section 7.4, Exercise 7.4.5.

We’ve already done an example of a harmonic oscillator in a mixed initial state, but it’s useful to do this other example from Shankar so we can see how the modified Ehrenfest’s theorem fits in. In this case, we start with a particle in the mixed initial state

\displaystyle \left|\psi\left(0\right)\right\rangle =\frac{1}{\sqrt{2}}\left[\left|0\right\rangle +\left|1\right\rangle \right] \ \ \ \ \ (1)

The time-dependent solution is therefore

\displaystyle \left|\psi\left(t\right)\right\rangle \displaystyle = \displaystyle \frac{1}{\sqrt{2}}\left[e^{-iE_{0}\hbar t}\left|0\right\rangle +e^{-iE_{1}t}\left|1\right\rangle \right]\ \ \ \ \ (2)
\displaystyle \displaystyle = \displaystyle \frac{1}{\sqrt{2}}\left[e^{-i\omega t/2}\left|0\right\rangle +e^{-3i\omega t/2}\left|1\right\rangle \right] \ \ \ \ \ (3)

since the first two energies are {E_{0}=\hbar\omega/2} and {E_{1}=3\hbar\omega/2}.

The position and momentum operators can be written in terms of the raising and lowering operators

\displaystyle X \displaystyle = \displaystyle \sqrt{\frac{\hbar}{2m\omega}}\left(a^{\dagger}+a\right)\ \ \ \ \ (4)
\displaystyle P \displaystyle = \displaystyle i\sqrt{\frac{\hbar m\omega}{2}}\left(a^{\dagger}-a\right) \ \ \ \ \ (5)

To find the mean position and momentum, we can use these equations:

\displaystyle \left\langle X\left(0\right)\right\rangle \displaystyle = \displaystyle \left\langle \psi\left(0\right)\left|X\right|\psi\left(0\right)\right\rangle \ \ \ \ \ (6)
\displaystyle \displaystyle = \displaystyle \frac{1}{2}\sqrt{\frac{\hbar}{2m\omega}}\left[\left\langle 0\right|+\left\langle 1\right|\right]\left(a^{\dagger}+a\right)\left[\left|0\right\rangle +\left|1\right\rangle \right] \ \ \ \ \ (7)

To work out the last line, remember that the stationary states are orthogonal so that {\left\langle 0\left|1\right.\right\rangle =0}, and that

\displaystyle a^{\dagger}\left|n\right\rangle \displaystyle = \displaystyle \sqrt{n+1}\left|n+1\right\rangle \ \ \ \ \ (8)
\displaystyle a\left|n\right\rangle \displaystyle = \displaystyle \sqrt{n}\left|n-1\right\rangle \ \ \ \ \ (9)

We therefore get

\displaystyle \left\langle X\left(0\right)\right\rangle =\frac{1}{2}\sqrt{\frac{\hbar}{2m\omega}}\left(1+1\right)=\sqrt{\frac{\hbar}{2m\omega}} \ \ \ \ \ (10)

Doing a similar analysis for the momentum, we have

\displaystyle \left\langle P\left(0\right)\right\rangle \displaystyle = \displaystyle \left\langle \psi\left(0\right)\left|P\right|\psi\left(0\right)\right\rangle \ \ \ \ \ (11)
\displaystyle \displaystyle = \displaystyle \frac{i}{2}\sqrt{\frac{\hbar m\omega}{2}}\left[\left\langle 0\right|+\left\langle 1\right|\right]\left(a^{\dagger}-a\right)\left[\left|0\right\rangle +\left|1\right\rangle \right]\ \ \ \ \ (12)
\displaystyle \displaystyle = \displaystyle \sqrt{\frac{\hbar m\omega}{2}}\frac{1}{2i}\left[\left\langle 0\right|+\left\langle 1\right|\right]\left(a-a^{\dagger}\right)\left[\left|0\right\rangle +\left|1\right\rangle \right]\ \ \ \ \ (13)
\displaystyle \displaystyle = \displaystyle \sqrt{\frac{\hbar m\omega}{2}}\frac{1}{2i}\left(1-1\right)\ \ \ \ \ (14)
\displaystyle \displaystyle = \displaystyle 0 \ \ \ \ \ (15)

We can expand these equations to give the averages of position and momentum at all times by plugging in 3:

\displaystyle \left\langle X\left(t\right)\right\rangle \displaystyle = \displaystyle \left\langle \psi\left(t\right)\left|X\right|\psi\left(t\right)\right\rangle \ \ \ \ \ (16)
\displaystyle \displaystyle = \displaystyle \frac{1}{2}\sqrt{\frac{\hbar}{2m\omega}}\left[\left\langle 0\right|e^{i\omega t/2}+\left\langle 1\right|e^{3i\omega t/2}\right]\left(a^{\dagger}+a\right)\left[e^{-i\omega t/2}\left|0\right\rangle +e^{-3i\omega t/2}\left|1\right\rangle \right]\ \ \ \ \ (17)
\displaystyle \displaystyle = \displaystyle \frac{1}{2}\sqrt{\frac{\hbar}{2m\omega}}\left(e^{-i\omega t}+e^{i\omega t}\right)\ \ \ \ \ (18)
\displaystyle \displaystyle = \displaystyle \sqrt{\frac{\hbar}{2m\omega}}\cos\omega t \ \ \ \ \ (19)
\displaystyle \left\langle P\left(t\right)\right\rangle \displaystyle = \displaystyle \left\langle \psi\left(t\right)\left|P\right|\psi\left(t\right)\right\rangle \ \ \ \ \ (20)
\displaystyle \displaystyle = \displaystyle \frac{i}{2}\sqrt{\frac{\hbar m\omega}{2}}\left[\left\langle 0\right|e^{i\omega t/2}+\left\langle 1\right|e^{3i\omega t/2}\right]\left(a^{\dagger}-a\right)\left[e^{-i\omega t/2}\left|0\right\rangle +e^{-3i\omega t/2}\left|1\right\rangle \right]\ \ \ \ \ (21)
\displaystyle \displaystyle = \displaystyle \sqrt{\frac{\hbar m\omega}{2}}\frac{1}{2i}\left[\left\langle 0\right|e^{i\omega t/2}+\left\langle 1\right|e^{3i\omega t/2}\right]\left(a-a^{\dagger}\right)\left[e^{-i\omega t/2}\left|0\right\rangle +e^{-3i\omega t/2}\left|1\right\rangle \right]\ \ \ \ \ (22)
\displaystyle \displaystyle = \displaystyle \sqrt{\frac{\hbar m\omega}{2}}\frac{1}{2i}\left(e^{-i\omega t}-e^{i\omega t}\right)\ \ \ \ \ (23)
\displaystyle \displaystyle = \displaystyle -\sqrt{\frac{\hbar m\omega}{2}}\sin\omega t \ \ \ \ \ (24)

Although we can calculate {\left\langle \dot{X}\left(t\right)\right\rangle } and {\left\langle \dot{P}\left(t\right)\right\rangle } directly by taking the time derivative, we can also do it by using Ehrenfest’s theorem in the form

\displaystyle \frac{d\left\langle \Omega\right\rangle }{dt}=-\frac{i}{\hbar}\left\langle \left[\Omega,H\right]\right\rangle \ \ \ \ \ (25)

for some operator {\Omega}.

Since the energy of the oscillator in state {\left|n\right\rangle } is {\left(n+\frac{1}{2}\right)\hbar\omega}, we can write the hamiltonian as

\displaystyle H=\hbar\omega\left(a^{\dagger}a+\frac{1}{2}\right) \ \ \ \ \ (26)

We also have the commutator

\displaystyle \left[a,a^{\dagger}\right]=1 \ \ \ \ \ (27)

To use this for {X} and {P} we need the commutators {\left[a,H\right]} and {\left[a^{\dagger},H\right]}, which amounts to finding

\displaystyle \left[a,a^{\dagger}a\right] \displaystyle = \displaystyle aa^{\dagger}a-a^{\dagger}aa\ \ \ \ \ (28)
\displaystyle \displaystyle = \displaystyle \left(1+a^{\dagger}a\right)a-a^{\dagger}aa\ \ \ \ \ (29)
\displaystyle \displaystyle = \displaystyle a\ \ \ \ \ (30)
\displaystyle \left[a^{\dagger},a^{\dagger}a\right] \displaystyle = \displaystyle a^{\dagger}a^{\dagger}a-a^{\dagger}aa^{\dagger}\ \ \ \ \ (31)
\displaystyle \displaystyle = \displaystyle a^{\dagger}a^{\dagger}a-a^{\dagger}\left(1+a^{\dagger}a^{\dagger}a\right)\ \ \ \ \ (32)
\displaystyle \displaystyle = \displaystyle -a^{\dagger} \ \ \ \ \ (33)

Therefore we have

\displaystyle \left[a,H\right] \displaystyle = \displaystyle \hbar\omega a\ \ \ \ \ (34)
\displaystyle \left[a^{\dagger},H\right] \displaystyle = \displaystyle -\hbar\omega a^{\dagger} \ \ \ \ \ (35)

Finally we get

\displaystyle \left\langle \dot{X}\left(t\right)\right\rangle \displaystyle = \displaystyle -\frac{i}{\hbar}\left\langle \left[X,H\right]\right\rangle \ \ \ \ \ (36)
\displaystyle \displaystyle = \displaystyle -\frac{i}{\hbar}\left\langle \left[a+a^{\dagger},H\right]\right\rangle \ \ \ \ \ (37)
\displaystyle \displaystyle = \displaystyle -\frac{i}{\hbar}\hbar\omega\sqrt{\frac{\hbar}{2m\omega}}\left\langle a-a^{\dagger}\right\rangle \ \ \ \ \ (38)
\displaystyle \displaystyle = \displaystyle i\omega\sqrt{\frac{\hbar}{2m\omega}}\sqrt{\frac{2}{\hbar m\omega}}\frac{1}{i}\left\langle P\left(t\right)\right\rangle \ \ \ \ \ (39)
\displaystyle \displaystyle = \displaystyle -\omega\sqrt{\frac{\hbar}{2m\omega}}\sin\omega t \ \ \ \ \ (40)

where we used 5 in the fourth line and 24 in the last line. The last line is indeed the time derivative of 19, so fortunately Ehrenfest’s theorem gives the correct answer.

For the momentum, we have

\displaystyle \left\langle \dot{P}\left(t\right)\right\rangle \displaystyle = \displaystyle -\frac{i}{\hbar}\left\langle \left[P,H\right]\right\rangle \ \ \ \ \ (41)
\displaystyle \displaystyle = \displaystyle -\frac{i}{\hbar}\left\langle \left[a^{\dagger}-a,H\right]\right\rangle \ \ \ \ \ (42)
\displaystyle \displaystyle = \displaystyle -\frac{i}{\hbar}\hbar\omega\sqrt{\frac{\hbar m\omega}{2}}i\left\langle -a^{\dagger}-a\right\rangle \ \ \ \ \ (43)
\displaystyle \displaystyle = \displaystyle \omega\sqrt{\frac{\hbar m\omega}{2}}\sqrt{\frac{\hbar}{2m\omega}}\sqrt{\frac{2m\omega}{\hbar}}\left\langle -X\left(t\right)\right\rangle \ \ \ \ \ (44)
\displaystyle \displaystyle = \displaystyle -\omega\sqrt{\frac{\hbar m\omega}{2}}\cos\omega t \ \ \ \ \ (45)

which is the correct derivative of 24.

Virial theorem in classical mechanics; application to harmonic oscillator

Shankar, R. (1994), Principles of Quantum Mechanics, Plenum Press. Section 7.4, Exercise 7.4.3.

We’ve seen the virial theorem in quantum mechanics, but this theorem was originally devised in classical mechanics. For a single particle, we consider the quantity

\displaystyle  G=\mathbf{r}\cdot\mathbf{p} \ \ \ \ \ (1)

that is, the product of position and momentum. Taking the time derivative, we have

\displaystyle   \frac{dG}{dt} \displaystyle  = \displaystyle  \mathbf{p}\cdot\frac{d\mathbf{r}}{dt}+\mathbf{r}\cdot\frac{d\mathbf{p}}{dt}\ \ \ \ \ (2)
\displaystyle  \displaystyle  = \displaystyle  mv^{2}+\mathbf{r}\cdot\mathbf{F} \ \ \ \ \ (3)

where {\mathbf{v}=d\mathbf{r}/dt} is the velocity of the particle, {\mathbf{p}=m\mathbf{v}} and {\mathbf{F}=d\mathbf{p}/dt} is the force acting on the particle. If the force is a central force (that is, it depends only on the particle’s distance from some centre point {\mathbf{r}=0}, then the force can be written as the negative gradient of a potential {V} that depends only on {\mathbf{r}}. In the case where {V} depends only on a power of {r}, we have

\displaystyle   V \displaystyle  = \displaystyle  ar^{k}\ \ \ \ \ (4)
\displaystyle  \mathbf{F} \displaystyle  = \displaystyle  -\frac{dV}{dr}=-kar^{k-1}\hat{\mathbf{r}} \ \ \ \ \ (5)

In that case, we have

\displaystyle   \frac{dG}{dt} \displaystyle  = \displaystyle  mv^{2}-\mathbf{r}\cdot kar^{k-1}\hat{\mathbf{r}}\ \ \ \ \ (6)
\displaystyle  \displaystyle  = \displaystyle  2T-kar^{k}\ \ \ \ \ (7)
\displaystyle  \displaystyle  = \displaystyle  2T-kV \ \ \ \ \ (8)

where {T} is the kinetic energy {T=\frac{1}{2}mv^{2}}. If the particle is moving in a circular orbit then its average position and average momentum (averaged over one orbit) do not change with time, so {\frac{dG}{dt}=0} and we get

\displaystyle   2\left\langle T\right\rangle -k\left\langle V\right\rangle \displaystyle  = \displaystyle  0\ \ \ \ \ (9)
\displaystyle  \left\langle T\right\rangle \displaystyle  = \displaystyle  \frac{k}{2}\left\langle V\right\rangle \ \ \ \ \ (10)

Another way of seeing this is that, in a circular orbit at constant orbital speed, the only force acting is the centripetal force holding the particle in its orbit, which is

\displaystyle  F_{cen}=-\frac{mv^{2}}{r} \ \ \ \ \ (11)

where the minus sign indicates that the force acts in the opposite direction to the outward pointing radius vector.

This force is provided by the gradient of the potential, so we have

\displaystyle  F_{cen}=-\frac{dV}{dr}=-kar^{k-1} \ \ \ \ \ (12)

We therefore have

\displaystyle   \frac{mv^{2}}{r} \displaystyle  = \displaystyle  kar^{k-1}\ \ \ \ \ (13)
\displaystyle  mv^{2}=2T \displaystyle  = \displaystyle  kar^{k}=kV\ \ \ \ \ (14)
\displaystyle  \left\langle T\right\rangle \displaystyle  = \displaystyle  \frac{k}{2}\left\langle V\right\rangle \ \ \ \ \ (15)

For the case of a harmonic oscillator, {V=\frac{1}{2}m\omega^{2}x^{2}} so the exponent is {k=2} and we have {T=V}. We can verify this by calculating the mean kinetic and potential energies explicitly, using earlier results. In the oscillator state {\left|n\right\rangle } we have

\displaystyle   \left\langle x^{2}\right\rangle \displaystyle  = \displaystyle  \frac{\hbar}{m\omega}\left(n+\frac{1}{2}\right)\ \ \ \ \ (16)
\displaystyle  \left\langle p^{2}\right\rangle \displaystyle  = \displaystyle  \hbar m\omega\left(n+\frac{1}{2}\right) \ \ \ \ \ (17)

The energies are

\displaystyle   \left\langle T\right\rangle \displaystyle  = \displaystyle  \frac{\left\langle p^{2}\right\rangle }{2m}=\frac{\hbar\omega}{2}\left(n+\frac{1}{2}\right)\ \ \ \ \ (18)
\displaystyle  \left\langle V\right\rangle \displaystyle  = \displaystyle  \frac{1}{2}m\omega^{2}\left\langle x^{2}\right\rangle =\frac{\hbar\omega}{2}\left(n+\frac{1}{2}\right) \ \ \ \ \ (19)

Therefore {\left\langle T\right\rangle =\left\langle V\right\rangle } as required.

Harmonic oscillator – eigenfunctions in momentum space

Shankar, R. (1994), Principles of Quantum Mechanics, Plenum Press. Section 7.3, Exercise 7.3.7.

We’ve seen how to solve the Schrödinger equation for the harmonic oscillator in the position basis, where the independent variable is {x}. It’s actually fairly easy to adapt this solution to find the wave functions in momentum space. (We’ve also found these functions by using the Fourier transform of the position functions, but the present post shows an easier way.)

The Schrödinger equation for the stationary states of the harmonic oscillator is, in operator form:

\displaystyle  \frac{P^{2}}{2m}\psi+\frac{1}{2}m\omega^{2}X^{2}\psi=E\psi \ \ \ \ \ (1)

To work in momentum space, we use the results

\displaystyle   P \displaystyle  = \displaystyle  p\ \ \ \ \ (2)
\displaystyle  X \displaystyle  = \displaystyle  i\hbar\frac{\partial}{\partial p} \ \ \ \ \ (3)

This gives

\displaystyle  \frac{p^{2}}{2m}\psi-\frac{1}{2}\hbar^{2}m\omega^{2}\frac{d^{2}\psi}{dp^{2}}=E\psi \ \ \ \ \ (4)

Dividing through by {\left(m\omega\right)^{2}} we get

\displaystyle  -\frac{\hbar^{2}}{2m}\psi^{\prime\prime}+\frac{p^{2}}{2m^{3}\omega^{2}}=\frac{E}{\left(m\omega\right)^{2}}\psi \ \ \ \ \ (5)

where a prime on {\psi} indicates a derivative with respect to {p}.

This is similar to the Schrödinger equation in position space:

\displaystyle  -\frac{\hbar^{2}}{2m}\psi^{\prime\prime}+\frac{1}{2}m\omega^{2}x^{2}\psi=E\psi \ \ \ \ \ (6)

(where a prime here indicates a derivative with respect to {x}). When we solved the position space equation, we introduced a dimensionless variable

\displaystyle  y\equiv\sqrt{\frac{m\omega}{\hbar}}x \ \ \ \ \ (7)

Using this technique to solve 5, we try a definition for {y} of

\displaystyle  y\equiv\frac{p}{\sqrt{\hbar m\omega}} \ \ \ \ \ (8)

(You can check the units of {\sqrt{\hbar m\omega}} to see they are the units of momentum, so {y} is indeed dimensionless here.) Making this substitution, we get

\displaystyle   \frac{d^{2}\psi}{dp^{2}} \displaystyle  = \displaystyle  \frac{1}{\hbar m\omega}\frac{d^{2}\psi}{dy^{2}}\ \ \ \ \ (9)
\displaystyle  \frac{p^{2}}{2m^{3}\omega^{2}} \displaystyle  = \displaystyle  \frac{\hbar y^{2}}{2m^{2}\omega} \ \ \ \ \ (10)

Thus 5 becomes

\displaystyle   -\frac{\hbar^{2}}{2m}\frac{1}{\hbar m\omega}\frac{d^{2}\psi}{dy^{2}}+\frac{\hbar y^{2}}{2m^{2}\omega}\psi \displaystyle  = \displaystyle  \frac{E}{\left(m\omega\right)^{2}}\psi\ \ \ \ \ (11)
\displaystyle  \frac{\hbar^{2}}{2m}\left[-\frac{d^{2}\psi}{dy^{2}}+y^{2}\psi\right] \displaystyle  = \displaystyle  \frac{\hbar E}{\omega}\psi \ \ \ \ \ (12)

We can now use the same dimensionless parameter we used in the earlier derivation:

\displaystyle  \varepsilon\equiv\frac{E}{\hbar\omega} \ \ \ \ \ (13)

This results in the differential equation

\displaystyle  \psi^{\prime\prime}+\left(2\varepsilon-y^{2}\right)\psi=0 \ \ \ \ \ (14)

where a prime now indicates a derivative with respect to {y}. This is exactly the same differential equation that we got for the position basis, except that the independent variable {y} is now defined in terms of {p} by 8 instead of {x}. We can solve it in the same way, which results in the same quantization condition on the allowable energies of {E_{n}=\left(n+\frac{1}{2}\right)\hbar\omega}. The eigenfunctions look the same when expressed in terms of {y}:

\displaystyle  \psi_{n}(y)=A\frac{1}{\sqrt{2^{n}n!}}H_{n}(y)e^{-y^{2}/2} \ \ \ \ \ (15)

where {A} is a normalization constant with the value in the position basis of

\displaystyle  A=\left(\frac{m\omega}{\pi\hbar}\right)^{1/4} \ \ \ \ \ (16)

and {H_{n}} is a Hermite polynomial. We can get the eigenfunctions in momentum space by replacing {y} by 8. We can see that this amounts to replacing {x\rightarrow p} and {m\omega\rightarrow\frac{1}{m\omega}}, so we get

\displaystyle  \psi_{n}(p)=\frac{1}{\left(\pi\hbar m\omega\right)^{1/4}}\frac{1}{\sqrt{2^{n}n!}}H_{n}\left(\frac{p}{\sqrt{\hbar m\omega}}\right)e^{-p^{2}/2\hbar m\omega} \ \ \ \ \ (17)

In particular, the ground state is

\displaystyle  \psi_{0}\left(p\right)=\frac{1}{\left(\pi\hbar m\omega\right)^{1/4}}e^{-p^{2}/2\hbar m\omega} \ \ \ \ \ (18)

Harmonic oscillator – zero-point energy from uncertainty principle

Shankar, R. (1994), Principles of Quantum Mechanics, Plenum Press. Section 7.3.

There is a nice result derived in Shankar’s section 7.3 in which he shows that we can actually derive the ground state energy and wave function for the harmonic oscillator from the uncertainty principle. Classically, the energy of a harmonic oscillator is

\displaystyle H=\frac{p^{2}}{2m}+\frac{1}{2}m\omega^{2}x^{2} \ \ \ \ \ (1)

where both {p} and {x} are continuous variables that can, in principle, take on any values. Thus classically it is possible for an oscillator to have {x=p=0} giving a ground state with zero energy. In quantum mechanics, because {X} and {P} don’t commute, the position and momentum cannot both have precise values, which means that the ground state must have an energy greater than zero. This so-called zero-point energy is (as found by Solving Schrödinger’s equation)

\displaystyle E_{0}=\frac{\hbar\omega}{2} \ \ \ \ \ (2)

To derive this without needing to solve Schrödinger’s equation, we first recall that a state in which the position-momentum uncertainty is a minimum must be a gaussian of form

\displaystyle \Psi\left(x\right)=Ae^{-a(x-\langle x\rangle)^{2}/2\hbar}e^{i\langle p\rangle x/\hbar} \ \ \ \ \ (3)

where {a} is a positive real constant, {A} is the normalization constant, {\left\langle x\right\rangle } is the mean position and {\left\langle p\right\rangle } is the mean momentum. For a harmonic oscillator centred at {x=0}, we have that both {\left\langle x\right\rangle =\left\langle p\right\rangle =0}, so we know that the ground state wave function has the form

\displaystyle \psi\left(x\right)=Ae^{-ax^{2}/2\hbar} \ \ \ \ \ (4)

 

To normalize this we require (assuming {A} is real)

\displaystyle \int_{-\infty}^{\infty}\psi^{2}\left(x\right)dx=1 \ \ \ \ \ (5)

Using the standard result for a gaussian integral (see Appendix 2 in Shankar or use Google)

\displaystyle \int_{-\infty}^{\infty}\psi^{2}\left(x\right)dx \displaystyle = \displaystyle A^{2}\int_{-\infty}^{\infty}e^{-ax^{2}/\hbar}dx\ \ \ \ \ (6)
\displaystyle \displaystyle = \displaystyle A^{2}\sqrt{\frac{\pi\hbar}{a}} \ \ \ \ \ (7)

Therefore

\displaystyle A=\left(\frac{a}{\pi\hbar}\right)^{1/4} \ \ \ \ \ (8)

We need to find {a} such that {\Delta X\Delta P} is minimized. The harmonic oscillator hamiltonian is

\displaystyle H=\frac{P^{2}}{2m}+\frac{1}{2}m\omega^{2}X^{2} \ \ \ \ \ (9)

 

Since {\left\langle X\right\rangle =\left\langle P\right\rangle =0}, the uncertainties become

\displaystyle \left(\Delta X\right)^{2} \displaystyle = \displaystyle \left\langle X^{2}\right\rangle -\left\langle X\right\rangle ^{2}=\left\langle X^{2}\right\rangle \ \ \ \ \ (10)
\displaystyle \left(\Delta P\right)^{2} \displaystyle = \displaystyle \left\langle P^{2}\right\rangle -\left\langle P\right\rangle ^{2}=\left\langle P^{2}\right\rangle \ \ \ \ \ (11)

Averaging 9 we get

\displaystyle \left\langle H\right\rangle \displaystyle = \displaystyle \frac{\left\langle P^{2}\right\rangle }{2m}+\frac{1}{2}m\omega^{2}\left\langle X^{2}\right\rangle \ \ \ \ \ (12)
\displaystyle \displaystyle = \displaystyle \frac{\left(\Delta P\right)^{2}}{2m}+\frac{1}{2}m\omega^{2}\left(\Delta X\right)^{2} \ \ \ \ \ (13)

At minimum uncertainty

\displaystyle \Delta X\Delta P=\frac{\hbar}{2} \ \ \ \ \ (14)

so we have

\displaystyle \Delta P \displaystyle = \displaystyle \frac{\hbar}{2\Delta X}\ \ \ \ \ (15)
\displaystyle \left\langle H\right\rangle \displaystyle = \displaystyle \frac{\hbar^{2}}{8m\left(\Delta X\right)^{2}}+\frac{1}{2}m\omega^{2}\left(\Delta X\right)^{2} \ \ \ \ \ (16)

The minimum energy can now be found by finding the value of {\left(\Delta X\right)^{2}} that minimizes this function. Treating {\left(\Delta X\right)^{2}} (not just {\Delta X}) as the independent variable, we have

\displaystyle \frac{\partial\left\langle H\right\rangle }{\partial\left(\Delta X\right)^{2}} \displaystyle = \displaystyle -\frac{\hbar^{2}}{8m\left[\left(\Delta X\right)^{2}\right]^{2}}+\frac{1}{2}m\omega^{2}\ \ \ \ \ (17)
\displaystyle \displaystyle = \displaystyle -\frac{\hbar^{2}}{8m\left(\Delta X\right)^{4}}+\frac{1}{2}m\omega^{2}=0\ \ \ \ \ (18)
\displaystyle \left(\Delta X\right)^{2} \displaystyle = \displaystyle \frac{\hbar}{2m\omega} \ \ \ \ \ (19)

This gives a minimum value for the mean energy of

\displaystyle \left\langle H\right\rangle _{min}=\frac{\hbar\omega}{2} \ \ \ \ \ (20)

To complete the derivation, we need to find the gaussian 4 that gives the correct value 19 for {\left(\Delta X\right)^{2}}. That is, we need to find {a} such that

\displaystyle \left(\Delta X\right)^{2}=\left\langle X^{2}\right\rangle =\frac{\hbar}{2m\omega} \ \ \ \ \ (21)

This requires doing another gaussian integral:

\displaystyle \left\langle X^{2}\right\rangle \displaystyle = \displaystyle \int_{-\infty}^{\infty}x^{2}\psi^{2}\left(x\right)dx\ \ \ \ \ (22)
\displaystyle \displaystyle = \displaystyle \sqrt{\frac{a}{\pi\hbar}}\int_{-\infty}^{\infty}x^{2}e^{-ax^{2}/\hbar}dx\ \ \ \ \ (23)
\displaystyle \displaystyle = \displaystyle \sqrt{\frac{a}{\pi\hbar}}\sqrt{\frac{\pi\hbar}{a}}\frac{h}{2a}\ \ \ \ \ (24)
\displaystyle \displaystyle = \displaystyle \frac{\hbar}{2a} \ \ \ \ \ (25)

We therefore get

\displaystyle \frac{\hbar}{2a} \displaystyle = \displaystyle \frac{\hbar}{2m\omega}\ \ \ \ \ (26)
\displaystyle a \displaystyle = \displaystyle m\omega \ \ \ \ \ (27)

which gives a normalized minimum energy wave function

\displaystyle \psi_{min}\left(x\right)=\left(\frac{m\omega}{\pi\hbar}\right)^{1/4}e^{-m\omega x^{2}/2\hbar} \ \ \ \ \ (28)

 

This is the lowest possible value for the energy, but is it actually the ground state energy? What we have shown so far is that

\displaystyle \left\langle \psi_{min}\left|H\right|\psi_{min}\right\rangle \le\left\langle \psi_{0}\left|H\right|\psi_{0}\right\rangle =E_{0} \ \ \ \ \ (29)

 

where {\left|\psi_{0}\right\rangle } is the ground state energy. However, we can invoke the variational principle which states that if {\psi} is any normalized function, then the ground state energy {E_{0}} of any hamiltonian {H} satisfies

\displaystyle E_{0}\le\left\langle \psi\left|H\right|\psi\right\rangle \ \ \ \ \ (30)

Using {\psi=\psi_{min}} we therefore have

\displaystyle E_{0}\le\left\langle \psi_{min}\left|H\right|\psi_{min}\right\rangle \ \ \ \ \ (31)

 

Combining 29 and 31 we have

\displaystyle \left\langle \psi_{min}\left|H\right|\psi_{min}\right\rangle \le E_{0}\le\left\langle \psi_{min}\left|H\right|\psi_{min}\right\rangle \ \ \ \ \ (32)

which means that

\displaystyle E_{0}=\left\langle \psi_{min}\left|H\right|\psi_{min}\right\rangle \ \ \ \ \ (33)

and therefore that {\left|\psi_{0}\right\rangle =\left|\psi_{min}\right\rangle }, that is, 28 is actually the ground state wave function.

Although this clever little derivation gives us the ground state energy and wave function, it doesn’t say anything about the higher energy states, or tell us that they are all equally spaced with a spacing of {\hbar\omega}. Nevertheless, it’s a pleasant exercise.

Harmonic oscillator – mean position and momentum

Shankar, R. (1994), Principles of Quantum Mechanics, Plenum Press. Section 7.3, Exercise 7.3.5.

The energy eigenfunctions of the harmonic oscillator are

\displaystyle \psi_{n}(y)=\left(\frac{m\omega}{\pi\hbar}\right)^{1/4}\frac{1}{\sqrt{2^{n}n!}}H_{n}(y)e^{-y^{2}/2} \ \ \ \ \ (1)

where

\displaystyle y\equiv\sqrt{\frac{m\omega}{\hbar}}x \ \ \ \ \ (2)

 

and {H_{n}} is the Hermite polynomial of order {n}. Since an even (odd) Hermite polynomial is an even (odd) function, {\psi_{n}} is even (odd) if {n} is even (odd), so we can use this fact to show that

\displaystyle \left\langle n\left|X\right|n\right\rangle =\int_{-\infty}^{\infty}x\psi_{n}^{2}\left(x\right)\;dx=0 \ \ \ \ \ (3)

This follows because the square of either an even or odd function gives an even function, and {x} itself is odd, so the integrand is the product of an odd and even function, which is odd. The integral over any interval symmetric about {x=0} of an odd function is odd. Thus the mean position {\left\langle X\right\rangle } of a particle in any of the harmonic oscillator’s energy eigenstates is zero.

For the momentum {P}, we have

\displaystyle \left\langle n\left|P\right|n\right\rangle =-i\hbar\int_{-\infty}^{\infty}\psi_{n}\frac{d\psi_{n}}{dx}\;dx \ \ \ \ \ (4)

As we showed earlier

\displaystyle -i\hbar\frac{d\psi_{n}}{dx}=i\sqrt{\hbar m\omega}\left(\frac{m\omega}{\pi\hbar}\right)^{1/4}\frac{1}{\sqrt{2^{n}n!}}e^{-y^{2}/2}\left[nH_{n-1}-\frac{1}{2}H_{n+1}\right] \ \ \ \ \ (5)

If {n} is even (odd), then {nH_{n-1}-\frac{1}{2}H_{n+1}} is odd (even), so the product {\psi_{n}\frac{d\psi_{n}}{dx}} is always the product of one odd and one even function, making it odd. Thus

\displaystyle \left\langle n\left|P\right|n\right\rangle =0 \ \ \ \ \ (6)

Thus the mean momentum {\left\langle P\right\rangle =0} in all energy eigenstates.

This means that the uncertainties in position and momentum are determined entirely by the mean square values:

\displaystyle \left(\Delta X\right)^{2} \displaystyle = \displaystyle \left\langle X^{2}\right\rangle -\left\langle X\right\rangle ^{2}=\left\langle X^{2}\right\rangle \ \ \ \ \ (7)
\displaystyle \left(\Delta P\right)^{2} \displaystyle = \displaystyle \left\langle P^{2}\right\rangle \ \ \ \ \ (8)

We can work out these values for a couple of specific states. For {n=1} we have

\displaystyle \left\langle 1\left|X^{2}\right|1\right\rangle \displaystyle = \displaystyle \int_{-\infty}^{\infty}x^{2}\psi_{1}^{2}\left(x\right)\;dx\ \ \ \ \ (9)
\displaystyle \displaystyle = \displaystyle \frac{1}{2}\sqrt{\frac{m\omega}{\pi\hbar}}\int_{-\infty}^{\infty}x^{2}H_{1}^{2}\left(y\right)e^{-y^{2}}dx\ \ \ \ \ (10)
\displaystyle \displaystyle = \displaystyle \frac{1}{2}\frac{\hbar}{\sqrt{\pi}m\omega}\int_{-\infty}^{\infty}y^{2}H_{1}^{2}\left(y\right)e^{-y^{2}}dy\ \ \ \ \ (11)
\displaystyle \displaystyle = \displaystyle \frac{2\hbar}{\sqrt{\pi}m\omega}\int_{-\infty}^{\infty}y^{4}e^{-y^{2}}dy\ \ \ \ \ (12)
\displaystyle \displaystyle = \displaystyle \frac{2\hbar}{\sqrt{\pi}m\omega}\frac{3\sqrt{\pi}}{4}\ \ \ \ \ (13)
\displaystyle \displaystyle = \displaystyle \frac{3\hbar}{2m\omega} \ \ \ \ \ (14)

We’ve used

\displaystyle H_{1}\left(y\right)=2y \ \ \ \ \ (15)

and formula just before A.2.3 from the appendix in Shankar, which gives

\displaystyle I_{4}\left(\alpha\right) \displaystyle = \displaystyle \int_{-\infty}^{\infty}x^{4}e^{-\alpha x^{2}}dx\ \ \ \ \ (16)
\displaystyle \displaystyle = \displaystyle \frac{\partial^{2}}{\partial\alpha^{2}}\int_{-\infty}^{\infty}e^{-\alpha x^{2}}dx\ \ \ \ \ (17)
\displaystyle \displaystyle = \displaystyle \frac{\partial^{2}}{\partial\alpha^{2}}I_{0}\left(\alpha\right) \ \ \ \ \ (18)

From formula A.2.2

\displaystyle I_{4}\left(\alpha\right) \displaystyle = \displaystyle \frac{\partial^{2}}{\partial\alpha^{2}}\sqrt{\frac{\pi}{\alpha}}\ \ \ \ \ (19)
\displaystyle \displaystyle = \displaystyle \frac{3\sqrt{\pi}}{4\alpha^{5/2}} \ \ \ \ \ (20)

Setting {\alpha=1} gives

\displaystyle \int_{-\infty}^{\infty}y^{4}e^{-y^{2}}dy=\frac{3\sqrt{\pi}}{4} \ \ \ \ \ (21)

For {P}, we have

\displaystyle \left\langle 1\left|P^{2}\right|1\right\rangle =-\hbar^{2}\int_{-\infty}^{\infty}\psi_{1}\frac{d^{2}}{dx^{2}}\psi_{1}\;dx \ \ \ \ \ (22)

 

The derivative is

\displaystyle \frac{d^{2}}{dx^{2}}\psi_{1} \displaystyle = \displaystyle \frac{d^{2}\psi_{1}}{dy^{2}}\left(\frac{dy}{dx}\right)^{2}\ \ \ \ \ (23)
\displaystyle \displaystyle = \displaystyle \frac{1}{\sqrt{2}}\left(\frac{m\omega}{\pi\hbar}\right)^{1/4}\frac{m\omega}{\hbar}\frac{d^{2}}{dy^{2}}\left[H_{1}(y)e^{-y^{2}/2}\right]\ \ \ \ \ (24)
\displaystyle \displaystyle = \displaystyle \frac{1}{\sqrt{2}}\left(\frac{m\omega}{\pi\hbar}\right)^{1/4}\frac{m\omega}{\hbar}\frac{d^{2}}{dy^{2}}\left[2ye^{-y^{2}/2}\right]\ \ \ \ \ (25)
\displaystyle \displaystyle = \displaystyle \left(\frac{m\omega}{4\pi\hbar}\right)^{1/4}\frac{m\omega}{\hbar}e^{-y^{2}/2}\left[2y^{3}-6y\right] \ \ \ \ \ (26)

We can now evaluate 22:

\displaystyle -\hbar^{2}\int_{-\infty}^{\infty}\psi_{1}\frac{d^{2}}{dx^{2}}\psi_{1}\;dx \displaystyle = \displaystyle -\hbar^{2}\sqrt{\frac{m\omega}{4\pi\hbar}}\frac{m\omega}{\hbar}\sqrt{\frac{\hbar}{m\omega}}\int_{-\infty}^{\infty}2ye^{-y^{2}}2\left[y^{3}-3y\right]dy\ \ \ \ \ (27)
\displaystyle \displaystyle = \displaystyle -\frac{2m\omega\hbar}{\sqrt{\pi}}\int_{-\infty}^{\infty}e^{-y^{2}}\left[y^{4}-3y^{2}\right]dy\ \ \ \ \ (28)
\displaystyle \displaystyle = \displaystyle -\frac{2m\omega\hbar}{\sqrt{\pi}}\left[\frac{3\sqrt{\pi}}{4}-\frac{3\sqrt{\pi}}{2}\right]\ \ \ \ \ (29)
\displaystyle \displaystyle = \displaystyle \frac{3}{2}m\omega\hbar \ \ \ \ \ (30)

Thus for the {n=1} state

\displaystyle \Delta X \displaystyle = \displaystyle \sqrt{\frac{3\hbar}{2m\omega}}\ \ \ \ \ (31)
\displaystyle \Delta P \displaystyle = \displaystyle \sqrt{\frac{3}{2}m\omega\hbar}\ \ \ \ \ (32)
\displaystyle \Delta X\Delta P \displaystyle = \displaystyle \frac{3}{2}\hbar>\frac{\hbar}{2} \ \ \ \ \ (33)

For the {n=0} (ground) state, we can use {H_{0}=1} to get

\displaystyle \left\langle X^{2}\right\rangle \displaystyle = \displaystyle \int_{-\infty}^{\infty}x^{2}\psi_{0}^{2}\left(x\right)\;dx\ \ \ \ \ (34)
\displaystyle \displaystyle = \displaystyle \sqrt{\frac{m\omega}{\pi\hbar}}\int_{-\infty}^{\infty}x^{2}H_{0}^{2}\left(y\right)e^{-y^{2}}dx\ \ \ \ \ (35)
\displaystyle \displaystyle = \displaystyle \frac{\hbar}{\sqrt{\pi}m\omega}\int_{-\infty}^{\infty}y^{2}e^{-y^{2}}dy\ \ \ \ \ (36)
\displaystyle \displaystyle = \displaystyle \frac{\hbar}{2m\omega} \ \ \ \ \ (37)

For {P}:

\displaystyle \frac{d^{2}}{dx^{2}}\psi_{0} \displaystyle = \displaystyle \frac{d^{2}\psi_{0}}{dy^{2}}\left(\frac{dy}{dx}\right)^{2}\ \ \ \ \ (38)
\displaystyle \displaystyle = \displaystyle \left(\frac{m\omega}{\pi\hbar}\right)^{1/4}\frac{m\omega}{\hbar}e^{-y^{2}/2}\left(y^{2}-1\right)\ \ \ \ \ (39)
\displaystyle \left\langle P^{2}\right\rangle =-\hbar^{2}\int_{-\infty}^{\infty}\psi_{0}\frac{d^{2}}{dx^{2}}\psi_{0}\;dx \displaystyle = \displaystyle -\hbar^{2}\sqrt{\frac{m\omega}{\pi\hbar}}\frac{m\omega}{\hbar}\sqrt{\frac{\hbar}{m\omega}}\int_{-\infty}^{\infty}e^{-y^{2}}\left(y^{2}-1\right)dy\ \ \ \ \ (40)
\displaystyle \displaystyle = \displaystyle -\frac{m\omega\hbar}{\sqrt{\pi}}\left[\frac{\sqrt{\pi}}{2}-\sqrt{\pi}\right]\ \ \ \ \ (41)
\displaystyle \displaystyle = \displaystyle \frac{1}{2}m\omega\hbar \ \ \ \ \ (42)

The uncertainty principle in this case gives

\displaystyle \Delta X\Delta P=\frac{\hbar}{2} \ \ \ \ \ (43)

so it saturates the condition {\Delta X\Delta P\ge\frac{\hbar}{2}}.

Harmonic oscillator: matrix elements using Hermite polynomials

Shankar, R. (1994), Principles of Quantum Mechanics, Plenum Press. Section 7.3, Exercise 7.3.4.

Earlier, we found the matrix elements of {X} and {P} of the harmonic oscillator using the raising and lowering operators. We can also find these matrix elements using the recursion relations and orthogonality of Hermite polynomials. The required relations are also given as Shankar’s equations 7.3.24 – 7.3.26.

\displaystyle   H_{n}^{\prime}\left(y\right) \displaystyle  = \displaystyle  2nH_{n-1}\left(y\right)\ \ \ \ \ (1)
\displaystyle  H_{n+1}\left(y\right) \displaystyle  = \displaystyle  2yH_{n}\left(y\right)-2nH_{n-1}\left(y\right)\ \ \ \ \ (2)
\displaystyle  \int_{-\infty}^{\infty}H_{n}\left(y\right)H_{n^{\prime}}\left(y\right)e^{-y^{2}}dy \displaystyle  = \displaystyle  \sqrt{\pi}2^{n}n!\delta_{nn^{\prime}} \ \ \ \ \ (3)

The energy eigenfunctions of the harmonic oscillator are

\displaystyle  \psi_{n}(y)=\left(\frac{m\omega}{\pi\hbar}\right)^{1/4}\frac{1}{\sqrt{2^{n}n!}}H_{n}(y)e^{-y^{2}/2} \ \ \ \ \ (4)

where

\displaystyle  y\equiv\sqrt{\frac{m\omega}{\hbar}}x \ \ \ \ \ (5)

The matrix elements of {x} are therefore

\displaystyle  \left\langle n^{\prime}\left|X\right|n\right\rangle =\int_{-\infty}^{\infty}\psi_{n^{\prime}}\left(x\right)x\psi_{n}\left(x\right)dx \ \ \ \ \ (6)

where we’ve used the fact that {\psi_{n^{\prime}}\left(x\right)} is real, so its complex conjugate is the same as the original. Converting to the variable {y} using 5 we have

\displaystyle   \left\langle n^{\prime}\left|X\right|n\right\rangle \displaystyle  = \displaystyle  \frac{\hbar}{m\omega}\int_{-\infty}^{\infty}\psi_{n^{\prime}}\left(y\right)y\psi_{n}\left(y\right)dy\ \ \ \ \ (7)
\displaystyle  \displaystyle  = \displaystyle  \frac{\hbar}{m\omega}\sqrt{\frac{m\omega}{\pi\hbar}}\frac{1}{\sqrt{2^{n^{\prime}}2^{n}n^{\prime}!n!}}\int_{-\infty}^{\infty}H_{n}\left(y\right)yH_{n^{\prime}}\left(y\right)e^{-y^{2}}dy\ \ \ \ \ (8)
\displaystyle  \displaystyle  = \displaystyle  \sqrt{\frac{\hbar}{\pi m\omega}}\frac{1}{\sqrt{2^{n^{\prime}}2^{n}n^{\prime}!n!}}\int_{-\infty}^{\infty}\frac{1}{2}\left[H_{n+1}+2nH_{n-1}\right]H_{n^{\prime}}e^{-y^{2}}dy\ \ \ \ \ (9)
\displaystyle  \displaystyle  = \displaystyle  \sqrt{\frac{\hbar}{m\omega}}\frac{2^{n^{\prime}}n^{\prime}!}{2\sqrt{2^{n^{\prime}}2^{n}n^{\prime}!n!}}\left[\delta_{n^{\prime},n+1}+2n\delta_{n^{\prime},n-1}\right]\ \ \ \ \ (10)
\displaystyle  \displaystyle  = \displaystyle  \sqrt{\frac{\hbar}{m\omega}}\left[\frac{2^{n+1}\left(n+1\right)!}{2\sqrt{2^{2n+1}\left(n+1\right)!n!}}\delta_{n^{\prime},n+1}+\frac{2^{n}n\left(n-1\right)!}{2\sqrt{2^{2n-1}\left(n-1\right)!n!}}\delta_{n^{\prime},n-1}\right]\ \ \ \ \ (11)
\displaystyle  \displaystyle  = \displaystyle  \sqrt{\frac{\hbar}{2m\omega}}\left[\frac{\left(n+1\right)n!}{\sqrt{\left(n+1\right)\left(n!\right)^{2}}}\delta_{n^{\prime},n+1}+\frac{n\left(n-1\right)!}{\sqrt{n\left[\left(n-1\right)!\right]^{2}}}\delta_{n^{\prime},n-1}\right]\ \ \ \ \ (12)
\displaystyle  \displaystyle  = \displaystyle  \sqrt{\frac{\hbar}{2m\omega}}\left[\sqrt{n+1}\delta_{n^{\prime},n+1}+\sqrt{n}\delta_{n^{\prime},n-1}\right] \ \ \ \ \ (13)

We used 2 to get the third line and 3 to do the integrals.

For the matrix elements of {P} we use

\displaystyle   P \displaystyle  = \displaystyle  -i\hbar\frac{d}{dx}=-i\sqrt{\hbar m\omega}\frac{d}{dy}\ \ \ \ \ (14)
\displaystyle  P\psi_{n}\left(y\right) \displaystyle  = \displaystyle  -i\sqrt{\hbar m\omega}\left(\frac{m\omega}{\pi\hbar}\right)^{1/4}\frac{1}{\sqrt{2^{n}n!}}e^{-y^{2}/2}\left[H_{n}^{\prime}\left(y\right)-yH_{n}\left(y\right)\right]\ \ \ \ \ (15)
\displaystyle  \displaystyle  = \displaystyle  -i\sqrt{\hbar m\omega}\left(\frac{m\omega}{\pi\hbar}\right)^{1/4}\frac{1}{\sqrt{2^{n}n!}}e^{-y^{2}/2}\left[2nH_{n-1}-\frac{1}{2}\left(H_{n+1}+2nH_{n-1}\right)\right]\ \ \ \ \ (16)
\displaystyle  \displaystyle  = \displaystyle  -i\sqrt{\hbar m\omega}\left(\frac{m\omega}{\pi\hbar}\right)^{1/4}\frac{1}{\sqrt{2^{n}n!}}e^{-y^{2}/2}\left[nH_{n-1}-\frac{1}{2}H_{n+1}\right] \ \ \ \ \ (17)

We used 2 to get the third line.

We get, using {dx=\sqrt{\hbar/m\omega}dy} from 5

\displaystyle   \left\langle n^{\prime}\left|P\right|n\right\rangle \displaystyle  = \displaystyle  -i\sqrt{\hbar m\omega}\sqrt{\frac{m\omega}{\pi\hbar}}\sqrt{\frac{\hbar}{m\omega}}\frac{1}{\sqrt{2^{n^{\prime}}n^{\prime}!2^{n}n!}}\int_{-\infty}^{\infty}H_{n^{\prime}}\left[nH_{n-1}-\frac{1}{2}H_{n+1}\right]e^{-y^{2}}dy\ \ \ \ \ (18)
\displaystyle  \displaystyle  = \displaystyle  -i\sqrt{\hbar m\omega}\left[\frac{2^{n^{\prime}}n^{\prime}!n}{\sqrt{2^{n^{\prime}}n^{\prime}!2^{n}n!}}\delta_{n^{\prime},n-1}-\frac{2^{n^{\prime}-1}n^{\prime}!}{\sqrt{2^{n^{\prime}}n^{\prime}!2^{n}n!}}\delta_{n^{\prime},n+1}\right]\ \ \ \ \ (19)
\displaystyle  \displaystyle  = \displaystyle  -i\sqrt{\hbar m\omega}\left[\frac{2^{n-1}\left(n-1\right)!n}{\sqrt{2^{n-1}\left(n-1\right)!2^{n}n!}}\delta_{n^{\prime},n-1}-\frac{2^{n}\left(n+1\right)!}{\sqrt{2^{n+1}\left(n+1\right)!2^{n}n!}}\delta_{n^{\prime},n+1}\right]\ \ \ \ \ (20)
\displaystyle  \displaystyle  = \displaystyle  i\sqrt{\frac{\hbar m\omega}{2}}\left[\sqrt{n+1}\delta_{n^{\prime},n+1}-\sqrt{n}\delta_{n^{\prime},n-1}\right] \ \ \ \ \ (21)