Tag Archives: harmonic oscillator

Harmonic oscillator in a magnetic field

Shankar, R. (1994), Principles of Quantum Mechanics, Plenum Press. Chapter 12, Exercise 12.3.8.

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As another example of the harmonic oscillator, we’ll look at a charged particle moving in a magnetic field. The field {\mathbf{B}} is given in terms of the magnetic vector potential

\displaystyle \mathbf{A}=\frac{B}{2}\left(-y\hat{\mathbf{x}}+x\hat{\mathbf{y}}\right) \ \ \ \ \ (1)

 

The field is

\displaystyle \mathbf{B} \displaystyle = \displaystyle \nabla\times\mathbf{A}\ \ \ \ \ (2)
\displaystyle \displaystyle = \displaystyle \left(\frac{\partial A_{y}}{\partial x}-\frac{\partial A_{x}}{\partial y}\right)\hat{\mathbf{z}}\ \ \ \ \ (3)
\displaystyle \displaystyle = \displaystyle B\hat{\mathbf{z}} \ \ \ \ \ (4)

If the particle is confined to the {xy} plane and the magnetic field provides the only force, the force is given by the Lorentz force law

\displaystyle \mathbf{F}=q\mathbf{v}\times\mathbf{B} \ \ \ \ \ (5)

Since {\mathbf{F}} is always perpendicular to the direction of motion {\mathbf{v}}, the magnetic force does no work, so the kinetic energy and hence the speed {v} of the particle is constant. Classically, the particle is thus confined to move in a circle with {\mathbf{F}} providing the centripetal force, so we have

\displaystyle qvB \displaystyle = \displaystyle \frac{\mu v^{2}}{\rho}\ \ \ \ \ (6)
\displaystyle v \displaystyle = \displaystyle \frac{qB\rho}{\mu} \ \ \ \ \ (7)

where {q} is the charge, {\mu} is the mass and {\rho} is the radius of the circle. The period of the orbit is

\displaystyle T=\frac{2\pi\rho}{v}=\frac{2\pi\mu}{qB} \ \ \ \ \ (8)

which gives an angular frequency of

\displaystyle \omega_{0}=\frac{2\pi}{T}=\frac{qB}{\mu} \ \ \ \ \ (9)

This is the result in SI units; Shankar uses Gaussian units, in which the magnetic field picks up a factor of {\frac{1}{c}}, so in Shankar’s notation, this is

\displaystyle \omega_{0}=\frac{qB}{\mu c} \ \ \ \ \ (10)

As the rest of the problem relies on Gaussian units, we’ll stick to them from now on.

Classically, the Hamiltonian for the electromagnetic force is

\displaystyle H=\frac{\left|\mathbf{p}-q\mathbf{A}/c\right|^{2}}{2\mu}+q\phi \ \ \ \ \ (11)

where {\phi} is the electric potential, which is zero here. Thus using 1, we have for the quantum version in which {\mathbf{p}} and the position vector are replaced by operators

\displaystyle H=\frac{\left(P_{x}+qYB/2c\right)^{2}}{2\mu}+\frac{\left(P_{y}-qXB/2c\right)^{2}}{2\mu} \ \ \ \ \ (12)

 

We can perform a canonical transformation by defining

\displaystyle Q \displaystyle \equiv \displaystyle \frac{1}{qB}\left(cP_{x}+\frac{qYB}{2}\right)\ \ \ \ \ (13)
\displaystyle P \displaystyle \equiv \displaystyle P_{y}-\frac{qXB}{2c} \ \ \ \ \ (14)

We can verify that these coordinates are canonical by checking their commutator:

\displaystyle \left[Q,P\right] \displaystyle = \displaystyle \frac{1}{qB}\left[cP_{x}+\frac{qYB}{2},P_{y}-\frac{qXB}{2c}\right]\ \ \ \ \ (15)
\displaystyle \displaystyle = \displaystyle \frac{1}{qB}\left(-\frac{qB}{2}\left[P_{x},X\right]+\frac{qB}{2}\left[Y,P_{y}\right]\right)\ \ \ \ \ (16)
\displaystyle \displaystyle = \displaystyle \frac{i\hbar}{2}+\frac{i\hbar}{2}\ \ \ \ \ (17)
\displaystyle \displaystyle = \displaystyle i\hbar \ \ \ \ \ (18)

Thus {Q} and {P} have the correct commutator for a pair of position and momentum variables.

Rewriting 12 in terms of {Q} and {P}, we have

\displaystyle H \displaystyle = \displaystyle \frac{q^{2}B^{2}}{2\mu c^{2}}Q^{2}+\frac{P^{2}}{2\mu}\ \ \ \ \ (19)
\displaystyle \displaystyle = \displaystyle \frac{P^{2}}{2\mu}+\frac{\mu}{2}\omega_{0}^{2}Q^{2} \ \ \ \ \ (20)

Thus {H} has the same form as that for a one-dimensional harmonic oscillator with frequency {\omega_{0}}, so the energy levels of this system must be

\displaystyle E=\left(n+\frac{1}{2}\right)\hbar\omega_{0} \ \ \ \ \ (21)

 

We can expand 12 in terms of the original position and momentum variables to get

\displaystyle H \displaystyle = \displaystyle \frac{P_{x}^{2}+P_{y}^{2}}{2\mu}+\frac{1}{2}\mu\left(\frac{qB}{2\mu c}\right)^{2}\left(X^{2}+Y^{2}\right)+\frac{qB}{2\mu c}\left(P_{x}Y-P_{y}X\right)\ \ \ \ \ (22)
\displaystyle \displaystyle = \displaystyle \frac{P_{x}^{2}+P_{y}^{2}}{2\mu}+\frac{1}{2}\mu\left(\frac{\omega_{0}}{2}\right)^{2}\left(X^{2}+Y^{2}\right)-\frac{\omega_{0}}{2}\left(XP_{y}-YP_{x}\right)\ \ \ \ \ (23)
\displaystyle \displaystyle = \displaystyle H\left(\frac{\omega_{0}}{2},\mu\right)-\frac{\omega_{0}}{2}L_{z} \ \ \ \ \ (24)

where {H\left(\frac{\omega_{0}}{2},\mu\right)} is the hamiltonian for a 2-dim harmonic oscillator with frequency {\omega_{0}/2}. As we saw when solving that system, the Hamiltonian for the isotropic oscillator commutes with {L_{z}} since the potential is radially symmetric, thus the eigenfunctions of {H} are also eigenfunctions of {L_{z}}. In terms of the present problem, this means that the eigenfunctions of {H\left(\frac{\omega_{0}}{2},\mu\right)} are also eigenfunctions of {L_{z}} and thus also eigenfunctions of {H}. In our solution of the 2-dim isotropic oscillator, we found that the energy levels are given by

\displaystyle E=\hbar\omega\left(2k+\left|m\right|+1\right) \ \ \ \ \ (25)

 

where {k=0,1,2,\ldots} and {m} is the angular momentum (in units of {\hbar}). Thus for the oscillator with Hamiltonian {H\left(\frac{\omega_{0}}{2},\mu\right)}, the energy levels are

\displaystyle E \displaystyle = \displaystyle \frac{1}{2}\hbar\omega_{0}\left(2k+\left|m\right|+1\right)\ \ \ \ \ (26)
\displaystyle \displaystyle = \displaystyle \hbar\omega_{0}\left(k+\frac{1}{2}\left|m\right|+\frac{1}{2}\right) \ \ \ \ \ (27)

The energy levels of the original {H} are therefore, from 24

\displaystyle E \displaystyle = \displaystyle \hbar\omega_{0}\left(k+\frac{1}{2}\left|m\right|+\frac{1}{2}\right)-\frac{\omega_{0}}{2}m\hbar\ \ \ \ \ (28)
\displaystyle \displaystyle = \displaystyle \hbar\omega_{0}\left(k+\frac{1}{2}\left|m\right|-\frac{1}{2}m+\frac{1}{2}\right) \ \ \ \ \ (29)

[Shankar says the {k} can be ‘any integer’, but from our original derivation of 25, we found that {k} is a non-negative integer.] Equation 29 gives the same energies as 21, since if {m>0}, we get {E=\hbar\omega_{0}\left(k+\frac{1}{2}\right)}, while if {m<0} we have {E=\hbar\omega_{0}\left(k+\left|m\right|+\frac{1}{2}\right)}. Both {k+\frac{1}{2}} and {k+\left|m\right|+\frac{1}{2}} give the same sequence of values as {n+\frac{1}{2}}. [I’m not quite sure the two methods are equivalent, though, since 21, being the solution of a one-dimensional system is non-degenerate, while 29, being a two-dimensional system does have degenerate energy levels.]

Harmonic oscillator in 2 dimensions: comparison with rectangular coordinates

Shankar, R. (1994), Principles of Quantum Mechanics, Plenum Press. Chapter 12, Exercise 12.3.7 (8) – (10).

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In this post, we’ll continue with the solution of the 2-d isotropic harmonic oscillator. In the previous post, we found that the radial equation {R} can be written as

\displaystyle R\left(y\right)=y^{\left|m\right|}e^{-y^{2}/2}U\left(y\right) \ \ \ \ \ (1)

 

where the dimensionless variables are given by

\displaystyle y \displaystyle \equiv \displaystyle \sqrt{\frac{\mu\omega}{\hbar}}\rho\ \ \ \ \ (2)
\displaystyle \varepsilon \displaystyle \equiv \displaystyle \frac{E}{\hbar\omega} \ \ \ \ \ (3)

and {U} has a solution as a power series

\displaystyle U\left(y\right)=\sum_{r=0}^{\infty}C_{r}y^{r} \ \ \ \ \ (4)

 

The coefficients {C_{r}} satisfy the recursion relation

\displaystyle C_{r+2}=\frac{2\left(r+\left|m\right|+1-\varepsilon\right)}{\left(r+2\right)^{2}}C_{r} \ \ \ \ \ (5)

Only {C_{r}} for even {r} are non-zero.

In order for {U} to remain finite for large {y}, the series must terminate, which gives the allowable values for the energy as

\displaystyle E=\hbar\omega\left(n+1\right) \ \ \ \ \ (6)

with

\displaystyle n\equiv2k+\left|m\right| \ \ \ \ \ (7)

 

and {k=0,1,2,\ldots}.

We can now compare the solution obtained in polar coordinates with our earlier solution in terms of rectangular coordinates. First, what are the possible values for {m} for a given energy {E=\hbar\omega\left(n+1\right)}? From the relation 7, we can look at even and odd {n} separately. For even {n}, {k} can take values {0,1,\ldots,\frac{n}{2}-1,\frac{n}{2}}. The first {\frac{n}{2}} of these values for {k} (that is, for {k=0,1,\ldots,\frac{n}{2}-1}) each allow two values of {m} such that {\left|m\right|=n-2k}, namely {m=\pm\left(n-2k\right)}. If {k=\frac{n}{2}}, then we must have {m=0}. Thus for even {n} the total number of combinations is {2\times\frac{n}{2}+1=n+1}.

For odd {n}, {k} can take on values {0,1,\ldots,\frac{n-1}{2}}, giving a total of {\frac{n+1}{2}} possible values. (If this isn’t obvious, write it out for the first few values of odd {n} to see the pattern.) For each of these values of {k}, {m} can take on the two values {m=\pm\left(n-2k\right)}, thus there are again {2\times\frac{n+1}{2}=n+1} different combinations. Thus a state with energy {E=\hbar\omega\left(n+1\right)} has a degeneracy {n+1}.

We can construct the actual eigenfunctions for a couple of values of {n} by plugging in the appropriate formulas. For {n=0} there is only one function, which we find by setting {k=m=0}. From 4, we have

\displaystyle U_{0}\left(y\right)=C_{0} \ \ \ \ \ (8)

and from 1 we have

\displaystyle R_{0}\left(y\right)=C_{0}e^{-y^{2}/2} \ \ \ \ \ (9)

or, in terms of the original variables

\displaystyle R_{0}\left(\rho\right)=C_{0}e^{-\mu\omega\rho^{2}/2\hbar} \ \ \ \ \ (10)

The complete solution is given by

\displaystyle \psi_{m}\left(\rho,\phi\right) \displaystyle = \displaystyle R\left(\rho\right)\Phi_{m}\left(\phi\right)\ \ \ \ \ (11)
\displaystyle \displaystyle = \displaystyle \frac{1}{\sqrt{2\pi}}R\left(\rho\right)e^{im\phi} \ \ \ \ \ (12)

so for {m=0} we have

\displaystyle \psi_{0}\left(\rho,\phi\right)=\frac{C_{0}}{\sqrt{2\pi}}e^{-\mu\omega\rho^{2}/2\hbar} \ \ \ \ \ (13)

The constant {C_{0}} can be found by normalizing:

\displaystyle 1 \displaystyle = \displaystyle \int_{0}^{\infty}\int_{0}^{2\pi}\left|\psi_{0}\right|^{2}\rho d\phi d\rho\ \ \ \ \ (14)
\displaystyle \displaystyle = \displaystyle \left|C_{0}\right|^{2}\int_{0}^{\infty}e^{-\mu\omega\rho^{2}/\hbar}\rho d\rho\ \ \ \ \ (15)
\displaystyle \displaystyle = \displaystyle \left|C_{0}\right|^{2}\frac{\hbar}{2\mu\omega}\ \ \ \ \ (16)
\displaystyle C_{0} \displaystyle = \displaystyle \sqrt{\frac{2\mu\omega}{\hbar}}\ \ \ \ \ (17)
\displaystyle \psi_{0}\left(\rho,\phi\right) \displaystyle = \displaystyle \sqrt{\frac{\mu\omega}{\pi\hbar}}e^{-\mu\omega\rho^{2}/2\hbar} \ \ \ \ \ (18)

This agrees with the earlier result in rectangular coordinates (eqn 26 in this post). This must be the case, since the {n=0} state is non-degenerate.

For {n=1}, we have {k=0} and {m=\pm1} so we have two solutions:

\displaystyle \psi_{1} \displaystyle = \displaystyle \frac{C_{0}}{\sqrt{2\pi}}\sqrt{\frac{\mu\omega}{\hbar}}\rho e^{-\mu\omega\rho^{2}/2\hbar}e^{i\phi}\ \ \ \ \ (19)
\displaystyle \psi_{-1} \displaystyle = \displaystyle \frac{C_{0}}{\sqrt{2\pi}}\sqrt{\frac{\mu\omega}{\hbar}}\rho e^{-\mu\omega\rho^{2}/2\hbar}e^{-i\phi} \ \ \ \ \ (20)

Again, we normalize

\displaystyle 1 \displaystyle = \displaystyle \int_{0}^{\infty}\int_{0}^{2\pi}\left|\psi_{\pm1}\right|^{2}\rho d\phi d\rho\ \ \ \ \ (21)
\displaystyle \displaystyle = \displaystyle \frac{\mu\omega}{\hbar}\left|C_{0}\right|^{2}\int_{0}^{\infty}e^{-\mu\omega\rho^{2}/\hbar}\rho^{3}d\rho\ \ \ \ \ (22)
\displaystyle C_{0} \displaystyle = \displaystyle \sqrt{\frac{2\mu\omega}{\hbar}}\ \ \ \ \ (23)
\displaystyle \psi_{\pm1} \displaystyle = \displaystyle \frac{\mu\omega}{\hbar\sqrt{\pi}}\rho e^{-\mu\omega\rho^{2}/2\hbar}e^{\pm i\phi} \ \ \ \ \ (24)

These solutions are linear combinations of the corresponding solutions in rectangular coordinates (converted to polar for comparison):

\displaystyle \psi_{10} \displaystyle = \displaystyle \frac{\sqrt{2}\mu\omega}{\hbar\sqrt{\pi}}e^{-\mu\omega\rho^{2}/2\hbar}\rho\cos\phi\ \ \ \ \ (25)
\displaystyle \psi_{01} \displaystyle = \displaystyle \frac{\sqrt{2}\mu\omega}{\hbar\sqrt{\pi}}e^{-\mu\omega\rho^{2}/2\hbar}\rho\sin\phi \ \ \ \ \ (26)

The combinations are

\displaystyle \psi_{+1} \displaystyle = \displaystyle \frac{1}{\sqrt{2}}\left(\psi_{10}+i\psi_{01}\right)\ \ \ \ \ (27)
\displaystyle \psi_{-1} \displaystyle = \displaystyle \frac{1}{\sqrt{2}}\left(\psi_{10}-i\psi_{01}\right) \ \ \ \ \ (28)

The parity of the states is found from their behaviour under the transformation (in rectangular coordinates) {x\rightarrow-x} and {y\rightarrow-y}. In polar coordinates this is equivalent to the transformation {\phi\rightarrow\phi+\pi} and from 18 and 24 we see that

\displaystyle \psi_{0}\left(\rho,\phi+\pi\right) \displaystyle = \displaystyle \psi_{0}\left(\rho,\phi\right)\ \ \ \ \ (29)
\displaystyle \psi_{\pm1}\left(\rho,\phi+\pi\right) \displaystyle = \displaystyle \psi_{\pm1}\left(\rho,\phi\right)e^{\pm\pi}\ \ \ \ \ (30)
\displaystyle \displaystyle = \displaystyle -\psi_{\pm1}\left(\rho,\phi\right) \ \ \ \ \ (31)

Thus the parity of {n=0} is even, and that of {n=1} is odd. In general, since the {\phi} dependence enters only through the term {e^{im\phi}=e^{in\phi}e^{-2ik\phi}}, we see that adding {\pi} to {\phi} leaves the {e^{-2ik\phi}} term unchanged and multiplies the {e^{in\phi}} term by {e^{in\pi\phi}=\left(-1\right)^{n}}, so the parity of state {n} is {\left(-1\right)^{n}}.

Two-dimensional harmonic oscillator – Part 2: Series solution

Shankar, R. (1994), Principles of Quantum Mechanics, Plenum Press. Chapter 12, Exercise 12.3.7 (6) – (7).

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In this post, we’ll continue with the solution of the 2-d isotropic harmonic oscillator. In the last post, we started with the ODE for the radial function in the form

\displaystyle -\frac{\hbar^{2}}{2\mu}\left(\frac{d^{2}}{d\rho^{2}}+\frac{1}{\rho}\frac{d}{d\rho}-\frac{m^{2}}{\rho^{2}}\right)R+\frac{1}{2}\mu\omega^{2}\rho^{2}R=ER \ \ \ \ \ (1)

 

We introduced dimensionless variables

\displaystyle y \displaystyle \equiv \displaystyle \sqrt{\frac{\mu\omega}{\hbar}}\rho\ \ \ \ \ (2)
\displaystyle \varepsilon \displaystyle \equiv \displaystyle \frac{E}{\hbar\omega} \ \ \ \ \ (3)

and found that {R} could be written as

\displaystyle R\left(y\right)=y^{\left|m\right|}e^{-y^{2}/2}U\left(y\right) \ \ \ \ \ (4)

 

with {U} given by the solution of the ODE

\displaystyle U^{\prime\prime}+\left(\frac{2\left|m\right|+1}{y}-2y\right)U^{\prime}+\left(2\varepsilon-2\left|m\right|-2\right)U=0 \ \ \ \ \ (5)

 

We can solve this by using a power series of the form

\displaystyle U\left(y\right)=\sum_{r=0}^{\infty}C_{r}y^{r} \ \ \ \ \ (6)

where the coefficients {C_{r}} are constants.

The derivatives are

\displaystyle U^{\prime} \displaystyle = \displaystyle \sum_{r=0}^{\infty}C_{r}ry^{r-1}\ \ \ \ \ (7)
\displaystyle \displaystyle = \displaystyle 0+C_{1}+2C_{2}y+3C_{3}+y^{2}+\ldots\ \ \ \ \ (8)
\displaystyle \displaystyle = \displaystyle \sum_{r=0}^{\infty}C_{r+1}\left(r+1\right)y^{r}\ \ \ \ \ (9)
\displaystyle U^{\prime\prime} \displaystyle = \displaystyle \sum_{r=0}^{\infty}C_{r+1}r\left(r+1\right)y^{r-1}\ \ \ \ \ (10)
\displaystyle \displaystyle = \displaystyle 0+\left(1\right)\left(2\right)C_{2}+\left(2\right)\left(3\right)C_{3}y+\ldots\ \ \ \ \ (11)
\displaystyle \displaystyle = \displaystyle \sum_{r=0}^{\infty}C_{r+2}\left(r+1\right)\left(r+2\right)y^{r} \ \ \ \ \ (12)

Plugging these into 5 we have (we’ll drop the absolute value signs on {\left|m\right|} to make the notation simpler; we can restore them at the end):

\displaystyle \sum_{r=0}^{\infty}C_{r+2}\left(r+1\right)\left(r+2\right)y^{r}+\left(2m+1\right)\sum_{r=0}^{\infty}C_{r}ry^{r-2}- \displaystyle \displaystyle \ldots\ \ \ \ \ (13)
\displaystyle 2\sum_{r=0}^{\infty}C_{r}ry^{r}+2\left(\varepsilon-m-1\right)\sum_{r=0}^{\infty}C_{r}y^{r} \displaystyle = \displaystyle 0 \ \ \ \ \ (14)

The second sum in the first line is

\displaystyle \sum_{r=0}^{\infty}C_{r}ry^{r-2} \displaystyle = \displaystyle 0+C_{1}y^{-1}+2C_{2}+3C_{3}y+\ldots\ \ \ \ \ (15)
\displaystyle \displaystyle = \displaystyle \sum_{r=-1}^{\infty}C_{r+2}\left(r+2\right)y^{r} \ \ \ \ \ (16)

The sum thus becomes

\displaystyle \left(2m+1\right)C_{1}y^{-1}+\sum_{r=0}^{\infty}y^{r}C_{r+2}\left(r+2\right)^{2}+2\sum_{r=0}^{\infty}y^{r}C_{r}\left[-r+\varepsilon-m-1\right] \displaystyle = \displaystyle 0 \ \ \ \ \ (17)

A basic theorem about power series is that if the sum of the series equals zero for all {y}, then the coefficient of each power must be zero. This shows that {C_{1}=0} since otherwise the series would blow up as {y\rightarrow0}. This results in a recursion relation for the {C_{r}}:

\displaystyle C_{r+2}=\frac{2\left(r+m+1-\varepsilon\right)}{\left(r+2\right)^{2}}C_{r} \ \ \ \ \ (18)

 

Since {C_{1}=0}, all {C_{r}=0} for odd {r}. For large {r} we have

\displaystyle \frac{C_{r+2}}{C_{r}}\rightarrow\frac{2}{r} \ \ \ \ \ (19)

 

If the series is allowed to be infinite, this leads to a divergent series as we can see from the following (based on Shankar’s section 7.3). Suppose we look at {y^{m}e^{y^{2}}}, which clearly goes to infinity at large {y} (remember, {m} is positive). In series form this is

\displaystyle y^{m}e^{y^{2}}=\sum_{k=0}^{\infty}\frac{y^{2k+m}}{k!} \ \ \ \ \ (20)

The coefficient {C_{n}} of {y^{n}} , with {n=2k+m} in this series is

\displaystyle C_{n}=\frac{1}{\left[\left(n-m\right)/2\right]!} \ \ \ \ \ (21)

Similarly,

\displaystyle C_{n+2}=\frac{1}{\left[\left(n+2-m\right)/2\right]!} \ \ \ \ \ (22)

The ratio is

\displaystyle \frac{C_{n+2}}{C_{n}} \displaystyle = \displaystyle \frac{\left[\left(n-m\right)/2\right]!}{\left[\left(n+2-m\right)/2\right]!}\ \ \ \ \ (23)
\displaystyle \displaystyle = \displaystyle \frac{1}{\left(n-m\right)/2+1}\ \ \ \ \ (24)
\displaystyle \displaystyle \rightarrow \displaystyle \frac{2}{n} \ \ \ \ \ (25)

In other words, the coefficients of our series solution have the same behaviour 19 for large {r} as those in the series for {y^{m}e^{y^{2}}}. Referring back to 4, we see that this gives an overall behaviour for the radial function {R} of

\displaystyle R\rightarrow y^{m}e^{-y^{2}/2}y^{m}e^{y^{2}}=y^{2m}e^{y^{2}/2} \ \ \ \ \ (26)

Thus if we allow the series for {U} to be infinite, the overall solution diverges, which is not acceptable. We therefore require that the series terminates at some finite value of {r}, and from 18 we see that this happens if

\displaystyle \varepsilon=r+m+1 \ \ \ \ \ (27)

for some {r}. From the definition 3 this gives us the allowed values for the energy:

\displaystyle E \displaystyle = \displaystyle \hbar\omega\left(r+\left|m\right|+1\right)\ \ \ \ \ (28)
\displaystyle \displaystyle = \displaystyle \hbar\omega\left(2k+\left|m\right|+1\right) \ \ \ \ \ (29)

where the last line follows because {r} must be even. If

\displaystyle n\equiv2k+\left|m\right| \ \ \ \ \ (30)

then the allowed energies are

\displaystyle E=\hbar\omega\left(n+1\right) \ \ \ \ \ (31)

Two-dimensional harmonic oscillator – Part 1

Shankar, R. (1994), Principles of Quantum Mechanics, Plenum Press. Chapter 12, Exercise 12.3.7 (1) – (5).

[If some equations are too small to read easily, use your browser’s magnifying option (Ctrl + on Chrome, probably something similar on other browsers).]

In this problem, we’ll look at solving the 2-dimensional isotropic harmonic oscillator. The solution is fairly lengthy, so we’ll split it into two posts, with this being the first. The method of solution is similar to that used in the one-dimensional harmonic oscillator, so you may wish to refer back to that before proceeding.

The Hamiltonian is, in rectangular coordinates:

\displaystyle  H=\frac{P_{x}^{2}+P_{y}^{2}}{2\mu}+\frac{1}{2}\mu\omega^{2}\left(X^{2}+Y^{2}\right) \ \ \ \ \ (1)

The potential term is radially symmetric (it doesn’t depend on the polar angle {\phi}) so we have a problem of the form considered earlier. We saw there that for such potentials {\left[H,L_{z}\right]=0}. [If you don’t believe this, you can grind through the calculations using the commutation relations for {L_{z}} with the rectangular momenta and coordinates, but I won’t go through that here.]

As a result, {L_{z}} and {H} have simultaneous eigenfunctions of form

\displaystyle  \psi\left(\rho,\phi\right)=R\left(\rho\right)\Phi_{m}\left(\phi\right) \ \ \ \ \ (2)

where

\displaystyle  \Phi_{m}\left(\phi\right)=\frac{1}{\sqrt{2\pi}}e^{im\phi} \ \ \ \ \ (3)

The radial function satisfies the ODE

\displaystyle  -\frac{\hbar^{2}}{2\mu}\left(\frac{d^{2}R}{d\rho^{2}}+\frac{1}{\rho}\frac{dR}{d\rho}-\frac{m^{2}}{\rho^{2}}R\right)+V\left(\rho\right)R=ER \ \ \ \ \ (4)

where in this case

\displaystyle  V\left(\rho\right)=\frac{1}{2}\mu\omega^{2}\rho^{2} \ \ \ \ \ (5)

Thus the equation we must solve is

\displaystyle  -\frac{\hbar^{2}}{2\mu}\left(\frac{d^{2}}{d\rho^{2}}+\frac{1}{\rho}\frac{d}{d\rho}-\frac{m^{2}}{\rho^{2}}\right)R+\frac{1}{2}\mu\omega^{2}\rho^{2}R=ER \ \ \ \ \ (6)

To get a feel for the solution, we examine the behaviour in two limiting cases: {\rho\rightarrow0} and {\rho\rightarrow\infty}. It’s actually easier if we introduce dimensionless variables now, rather than in Shankar’s step 4, so we define

\displaystyle   y \displaystyle  \equiv \displaystyle  \sqrt{\frac{\mu\omega}{\hbar}}\rho\ \ \ \ \ (7)
\displaystyle  \varepsilon \displaystyle  \equiv \displaystyle  \frac{E}{\hbar\omega} \ \ \ \ \ (8)

This transforms 6 to

\displaystyle   -\frac{\hbar^{2}}{2\mu}\left(\frac{\mu\omega}{\hbar}\right)\left(\frac{d^{2}}{dy^{2}}+\frac{1}{y}\frac{d}{dy}-\frac{m^{2}}{y^{2}}\right)R+\frac{1}{2}\hbar\omega y^{2}R \displaystyle  = \displaystyle  ER\ \ \ \ \ (9)
\displaystyle  -\left(\frac{d^{2}}{dy^{2}}+\frac{1}{y}\frac{d}{dy}-\frac{m^{2}}{y^{2}}+2\varepsilon\right)R+y^{2}R \displaystyle  = \displaystyle  0 \ \ \ \ \ (10)

We can now look at {y\rightarrow0}, and we neglect the terms {2\varepsilon R} and {y^{2}R} to get

\displaystyle  \left(\frac{d^{2}}{dy^{2}}+\frac{1}{y}\frac{d}{dy}-\frac{m^{2}}{y^{2}}\right)R=0 \ \ \ \ \ (11)

If we try a solution of form

\displaystyle  R=y^{\left|m\right|} \ \ \ \ \ (12)

we have

\displaystyle  \left|m\right|\left(\left|m\right|-1\right)y^{\left|m\right|-2}+\left|m\right|y^{\left|m\right|-2}-m^{2}y^{\left|m\right|-2}=0 \ \ \ \ \ (13)

Thus 12 is indeed a solution in this limiting case.

For {y\rightarrow\infty}, we can ignore the terms {\frac{1}{y}\frac{d}{dy}}, {\frac{m^{2}}{y^{2}}R} and {2\varepsilon R} to get

\displaystyle  -\frac{d^{2}}{dy^{2}}R+y^{2}R=0 \ \ \ \ \ (14)

or

\displaystyle  R^{\prime\prime}=y^{2}R \ \ \ \ \ (15)

We try a solution of form

\displaystyle  R=y^{a}e^{-y^{2}/2} \ \ \ \ \ (16)

where {a} is some constant. We find

\displaystyle   R^{\prime} \displaystyle  = \displaystyle  \left(ay^{a-1}-y^{a+1}\right)e^{-y^{2}/2}\ \ \ \ \ (17)
\displaystyle  R^{\prime\prime} \displaystyle  = \displaystyle  \left(a\left(a-1\right)y^{a-2}-\left(a+1\right)y^{a}-ay^{a}+y^{a+2}\right)e^{-y^{2}/2}\ \ \ \ \ (18)
\displaystyle  \displaystyle  = \displaystyle  y^{a+2}\left(\frac{a\left(a-1\right)}{y^{4}}-\frac{2a+1}{y^{2}}+1\right)e^{-y^{2}/2} \ \ \ \ \ (19)

As {y\rightarrow\infty}, the last line tends to

\displaystyle  R^{\prime\prime}\rightarrow y^{a+2}e^{-y^{2}/2}=y^{2}R \ \ \ \ \ (20)

so in this limit 16 is a solution. We can therefore propose that {R} has the general form

\displaystyle  R\left(y\right)=y^{\left|m\right|}e^{-y^{2}/2}U\left(y\right) \ \ \ \ \ (21)

where {U} is a function to be determined by solving the exact ODE 10. We can get an ODE for {U} by substituting 21 into 10, although the calculation gets somewhat messy. As Shankar suggests, we can do this in two stages. First, we substitute

\displaystyle  R=y^{\left|m\right|}f\left(y\right) \ \ \ \ \ (22)

where

\displaystyle  f\left(y\right)=e^{-y^{2}/2}U\left(y\right) \ \ \ \ \ (23)

The required derivatives are (To make the notation simpler, I’ll drop the absolute value signs around {m}; you should assume that wherever {m} occurs, it should really be {\left|m\right|}. We can replace the absolute value sign at the end.)

\displaystyle   R^{\prime} \displaystyle  = \displaystyle  my^{m-1}f+y^{m}f^{\prime}\ \ \ \ \ (24)
\displaystyle  R^{\prime\prime} \displaystyle  = \displaystyle  m\left(m-1\right)y^{m-2}f+2my^{m-1}f^{\prime}+y^{m}f^{\prime\prime} \ \ \ \ \ (25)

Plugging these into 10 we have

\displaystyle   -\left(m\left(m-1\right)y^{m-2}f+2my^{m-1}f^{\prime}+y^{m}f^{\prime\prime}\right)-\left(my^{m-2}f+y^{m-1}f^{\prime}\right)+ \displaystyle  \displaystyle  \ldots\ \ \ \ \ (26)
\displaystyle  m^{2}y^{m-2}f-2\varepsilon y^{m}f+y^{m+2}f \displaystyle  = \displaystyle  0 \ \ \ \ \ (27)

Collecting terms and dividing through by {-y^{m}}, we get

\displaystyle  f^{\prime\prime}+f^{\prime}\left(\frac{2m+1}{y}\right)+f\left(2\varepsilon-y^{2}\right)=0 \ \ \ \ \ (28)

We now get the derivatives of {f}:

\displaystyle   f^{\prime} \displaystyle  = \displaystyle  -ye^{-y^{2}/2}U+e^{-y^{2}/2}U^{\prime}\ \ \ \ \ (29)
\displaystyle  \displaystyle  = \displaystyle  e^{-y^{2}/2}\left(U^{\prime}-yU\right)\ \ \ \ \ (30)
\displaystyle  f^{\prime\prime} \displaystyle  = \displaystyle  \left[-y\left(U^{\prime}-yU\right)+U^{\prime\prime}-U-yU^{\prime}\right]e^{-y^{2}/2}\ \ \ \ \ (31)
\displaystyle  \displaystyle  = \displaystyle  \left(U^{\prime\prime}-2yU^{\prime}+\left(y^{2}-1\right)U\right)e^{-y^{2}/2} \ \ \ \ \ (32)

When we plug these into 28, the exponential factor cancels out, so we get

\displaystyle  U^{\prime\prime}-2yU^{\prime}+\left(y^{2}-1\right)U+\frac{2m+1}{y}\left(U^{\prime}-yU\right)+U\left(2\varepsilon-y^{2}\right)=0 \ \ \ \ \ (33)

Collecting terms, we get, upon restoring the absolute values:

\displaystyle  U^{\prime\prime}+\left(\frac{2\left|m\right|+1}{y}-2y\right)U^{\prime}+\left(2\varepsilon-2\left|m\right|-2\right)U=0 \ \ \ \ \ (34)

We can solve this ODE using a power series in {y}, but we’ll leave that till the next post.

Harmonic oscillator in 2-d and 3-d, and in polar and spherical coordinates

Shankar, R. (1994), Principles of Quantum Mechanics, Plenum Press. Chapter 10, Exercises 10.2.2 – 10.2.3.

We’ve seen that the 3-d isotropic harmonic oscillator can be solved in rectangular coordinates using separation of variables. The Hamiltonian is

\displaystyle H=\frac{p_{x}^{2}+p_{y}^{2}+p_{z}^{2}}{2m}+\frac{m\omega^{2}}{2}\left(x^{2}+y^{2}+z^{2}\right) \ \ \ \ \ (1)

The solution to the Schrödinger equation is just the product of three one-dimensional oscillator eigenfunctions, one for each coordinate. That is

\displaystyle \psi_{n}\left(x,y,z\right)=\psi_{n_{x}}\left(x\right)\psi_{n_{y}}\left(y\right)\psi_{n_{z}}\left(z\right) \ \ \ \ \ (2)

Each one-dimensional eigenfunction can be expressed in terms of Hermite polynomials as

\displaystyle \psi_{n_{x}}\left(x\right)=\left(\frac{m\omega}{\pi\hbar}\right)^{1/4}\frac{1}{\sqrt{2^{n_{x}}n_{x}!}}H_{n_{x}}\left(\sqrt{\frac{m\omega}{\hbar}}x\right)e^{-m\omega x^{2}/2\hbar} \ \ \ \ \ (3)

with the functions for {y} and {z} obtained by replacing {x} by {y} or {z} and {n_{x}} by {n_{y}} or {n_{z}}. We also saw earlier that in the 3-d oscillator, the total energy for state {\psi_{n}\left(x,y,z\right)} is given in terms of the quantum numbers of the three 1-d oscillators as

\displaystyle E_{n}=\hbar\omega\left(n+\frac{3}{2}\right)=\hbar\omega\left(n_{x}+n_{y}+n_{z}+\frac{3}{2}\right) \ \ \ \ \ (4)

and that the degeneracy of level {n} is {\frac{1}{2}\left(n+1\right)\left(n+2\right)}.

Since the Hermite polynomial {H_{n_{x}}} has parity {\left(-1\right)^{n_{x}}} (that is, odd (even) polynomials are odd (even) functions), the 3-d wave function {\psi_{n}} has parity {\left(-1\right)^{n_{x}}\left(-1\right)^{n_{y}}\left(-1\right)^{n_{z}}=\left(-1\right)^{n}}.

We can write the one {n=0} state and three {n=1} states in spherical coordinates using the standard transformation

\displaystyle x \displaystyle = \displaystyle r\sin\theta\cos\phi\ \ \ \ \ (5)
\displaystyle y \displaystyle = \displaystyle r\sin\theta\sin\phi\ \ \ \ \ (6)
\displaystyle z \displaystyle = \displaystyle r\cos\phi \ \ \ \ \ (7)

Using the notation {\psi_{n}=\psi_{n_{x}n_{y}n_{z}}=\psi_{n_{x}}\psi_{n_{y}}\psi_{n_{z}}}, we have, using {H_{0}\left(y\right)=1} and {H_{1}\left(y\right)=2y}:

\displaystyle \psi_{000} \displaystyle = \displaystyle \left(\frac{m\omega}{\pi\hbar}\right)^{3/4}e^{-m\omega r^{2}/2\hbar}\ \ \ \ \ (8)
\displaystyle \psi_{100} \displaystyle = \displaystyle \sqrt{\frac{2m\omega}{\hbar}}\left(\frac{m\omega}{\pi\hbar}\right)^{3/4}e^{-m\omega r^{2}/2\hbar}r\sin\theta\cos\phi\ \ \ \ \ (9)
\displaystyle \psi_{010} \displaystyle = \displaystyle \sqrt{\frac{2m\omega}{\hbar}}\left(\frac{m\omega}{\pi\hbar}\right)^{3/4}e^{-m\omega r^{2}/2\hbar}r\sin\theta\sin\phi\ \ \ \ \ (10)
\displaystyle \psi_{001} \displaystyle = \displaystyle \sqrt{\frac{2m\omega}{\hbar}}\left(\frac{m\omega}{\pi\hbar}\right)^{3/4}e^{-m\omega r^{2}/2\hbar}r\cos\theta \ \ \ \ \ (11)

We can check that these are the correct spherical versions of the eigenfunctions by using the Schrödinger equation in spherical coordinates, which is

\displaystyle H\psi=\left[-\frac{\hbar^{2}\nabla^{2}}{2m}+\frac{m\omega^{2}}{2}r^{2}\right]\psi=E\psi \ \ \ \ \ (12)

The spherical laplacian operator is

\displaystyle \nabla^{2}\psi=\frac{1}{r^{2}}\frac{\partial}{\partial r}\left(r^{2}\frac{\partial\psi}{\partial r}\right)+\frac{1}{r^{2}\sin\theta}\frac{\partial}{\partial\theta}\left(\sin\theta\frac{\partial\psi}{\partial\theta}\right)+\frac{1}{r^{2}\sin^{2}\theta}\frac{\partial^{2}\psi}{\partial\phi^{2}} \ \ \ \ \ (13)

You can grind through the derivatives by hand if you like, but I just used Maple to do it, giving the results

\displaystyle H\psi_{000} \displaystyle = \displaystyle \frac{3}{2}\hbar\omega\psi_{000}\ \ \ \ \ (14)
\displaystyle H\psi_{100} \displaystyle = \displaystyle \frac{5}{2}\hbar\omega\psi_{100}\ \ \ \ \ (15)
\displaystyle H\psi_{010} \displaystyle = \displaystyle \frac{5}{2}\hbar\omega\psi_{010}\ \ \ \ \ (16)
\displaystyle H\psi_{001} \displaystyle = \displaystyle \frac{5}{2}\hbar\omega\psi_{001} \ \ \ \ \ (17)

In two dimensions, the analysis is pretty much the same. In the more general case where the masses are equal, but {\omega_{x}\ne\omega_{y}}, the Hamiltonian is

\displaystyle H=\frac{p_{x}^{2}+p_{y}^{2}}{2m}+\frac{m}{2}\left(\omega_{x}^{2}x^{2}+\omega_{y}^{2}y^{2}\right) \ \ \ \ \ (18)

A solution by separation of variables still works, with the result

\displaystyle \psi_{n}\left(x,y\right)=\psi_{n_{x}}\left(x\right)\psi_{n_{y}}\left(y\right) \ \ \ \ \ (19)

The total energy is

\displaystyle E_{n}=E_{n_{x}}+E_{n_{y}}=\hbar\omega\left(n_{x}+\frac{1}{2}+n_{y}+\frac{1}{2}\right)=\hbar\omega\left(n+1\right) \ \ \ \ \ (20)

For a given energy level {n=n_{x}+n_{y}}, there are {n+1} ways of forming {n} out of a sum of 2 non-negative integers, so the degeneracy of level {n} is {n+1}.

The one {n=0} state and two {n=1} states are

\displaystyle \psi_{00} \displaystyle = \displaystyle \left(\frac{m\omega}{\pi\hbar}\right)^{1/2}e^{-m\omega\left(x^{2}+y^{2}\right)/2\hbar}\ \ \ \ \ (21)
\displaystyle \psi_{10} \displaystyle = \displaystyle \sqrt{\frac{2m\omega}{\hbar}}\left(\frac{m\omega}{\pi\hbar}\right)^{1/2}e^{-m\omega\left(x^{2}+y^{2}\right)/2\hbar}x\ \ \ \ \ (22)
\displaystyle \psi_{01} \displaystyle = \displaystyle \sqrt{\frac{2m\omega}{\hbar}}\left(\frac{m\omega}{\pi\hbar}\right)^{1/2}e^{-m\omega\left(x^{2}+y^{2}\right)/2\hbar}y \ \ \ \ \ (23)

To translate to polar coordinates, we use the transformations

\displaystyle x \displaystyle = \displaystyle \rho\cos\phi\ \ \ \ \ (24)
\displaystyle y \displaystyle = \displaystyle \rho\sin\phi \ \ \ \ \ (25)

so we have

\displaystyle \psi_{00} \displaystyle = \displaystyle \left(\frac{m\omega}{\pi\hbar}\right)^{1/2}e^{-m\omega\rho^{2}/2\hbar}\ \ \ \ \ (26)
\displaystyle \psi_{10} \displaystyle = \displaystyle \sqrt{\frac{2m\omega}{\hbar}}\left(\frac{m\omega}{\pi\hbar}\right)^{1/2}e^{-m\omega\rho^{2}/2\hbar}\rho\cos\phi\ \ \ \ \ (27)
\displaystyle \psi_{01} \displaystyle = \displaystyle \sqrt{\frac{2m\omega}{\hbar}}\left(\frac{m\omega}{\pi\hbar}\right)^{1/2}e^{-m\omega\rho^{2}/2\hbar}\rho\sin\phi \ \ \ \ \ (28)

Again, we can check this by plugging these polar formulas into the polar Schrödinger equation, where the 2-d Laplacian is

\displaystyle \nabla^{2}=\frac{\partial^{2}}{\partial\rho^{2}}+\frac{1}{\rho}\frac{\partial}{\partial\rho}+\frac{1}{\rho^{2}}\frac{\partial^{2}}{\partial\phi^{2}} \ \ \ \ \ (29)

The results are

\displaystyle H\psi_{00} \displaystyle = \displaystyle \hbar\omega\psi_{00}\ \ \ \ \ (30)
\displaystyle H\psi_{10} \displaystyle = \displaystyle 2\hbar\omega\psi_{10}\ \ \ \ \ (31)
\displaystyle H\psi_{01} \displaystyle = \displaystyle 2\hbar\omega\psi_{01} \ \ \ \ \ (32)

Uncertainties in the harmonic oscillator and hydrogen atom

Shankar, R. (1994), Principles of Quantum Mechanics, Plenum Press. Chapter 9, Exercises 9.4.1 – 9.4.2.

Here we’ll look at a couple of calculations relevant to the application of the uncertainty principle to the hydrogen atom. When calculating uncertainties, we need to find the average values of various quantities. First, we’ll look at an average in the case of the harmonic oscillator.

The harmonic oscillator eigenstates are

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

where {H_{n}} is the {n}th Hermite polynomial. For {n=1,} we have

\displaystyle H_{1}\left(\sqrt{\frac{m\omega}{\hbar}}x\right)=2\sqrt{\frac{m\omega}{\hbar}}x \ \ \ \ \ (2)

so

\displaystyle \psi_{1}(x)=\frac{\sqrt{2}}{\pi^{1/4}}\left(\frac{m\omega}{\hbar}\right)^{3/4}x\;e^{-m\omega x^{2}/2\hbar} \ \ \ \ \ (3)

For this state, we can calculate the average

\displaystyle \left\langle \frac{1}{X^{2}}\right\rangle \displaystyle = \displaystyle \int_{-\infty}^{\infty}\psi_{1}^{2}(x)\frac{1}{x^{2}}dx\ \ \ \ \ (4)
\displaystyle \displaystyle = \displaystyle \frac{2}{\sqrt{\pi}}\left(\frac{m\omega}{\hbar}\right)^{3/2}\int_{-\infty}^{\infty}e^{-m\omega x^{2}/\hbar}dx\ \ \ \ \ (5)
\displaystyle \displaystyle = \displaystyle \frac{2}{\sqrt{\pi}}\left(\frac{m\omega}{\hbar}\right)^{3/2}\sqrt{\frac{\pi\hbar}{m\omega}}\ \ \ \ \ (6)
\displaystyle \displaystyle = \displaystyle \frac{2m\omega}{\hbar} \ \ \ \ \ (7)

where we evaluated the Gaussian integral in the second line.

We can compare this to {1/\left\langle X^{2}\right\rangle } as follows:

\displaystyle \left\langle X^{2}\right\rangle \displaystyle = \displaystyle \int_{-\infty}^{\infty}\psi_{1}^{2}(x)x^{2}dx\ \ \ \ \ (8)
\displaystyle \displaystyle = \displaystyle \frac{2}{\sqrt{\pi}}\left(\frac{m\omega}{\hbar}\right)^{3/2}\int_{-\infty}^{\infty}e^{-m\omega x^{2}/\hbar}x^{4}dx\ \ \ \ \ (9)
\displaystyle \displaystyle = \displaystyle \frac{2}{\sqrt{\pi}}\left(\frac{m\omega}{\hbar}\right)^{3/2}\frac{3\sqrt{\pi}}{4}\left(\frac{\hbar}{m\omega}\right)^{5/2}\ \ \ \ \ (10)
\displaystyle \displaystyle = \displaystyle \frac{3}{2}\frac{\hbar}{m\omega}\ \ \ \ \ (11)
\displaystyle \frac{1}{\left\langle X^{2}\right\rangle } \displaystyle = \displaystyle \frac{2}{3}\frac{m\omega}{\hbar} \ \ \ \ \ (12)

Thus {\left\langle \frac{1}{X^{2}}\right\rangle } and {\frac{1}{\left\langle X^{2}\right\rangle }} have the same order of magnitude, although they are not equal.

In three dimensions, we consider the ground state of hydrogen

\displaystyle \psi_{100}\left(r\right)=\frac{1}{\sqrt{\pi}a_{0}^{3/2}}e^{-r/a_{0}} \ \ \ \ \ (13)

where {a_{0}} is the Bohr radius

\displaystyle a_{0}\equiv\frac{\hbar^{2}}{me^{2}} \ \ \ \ \ (14)

with {m} and {e} being the mass and charge of the electron. The wave function is normalized as we can see by doing the integral (in 3 dimensions):

\displaystyle \int\psi_{100}^{2}(r)d^{3}\mathbf{r} \displaystyle = \displaystyle \frac{4\pi}{\pi a_{0}^{3}}\int_{0}^{\infty}e^{-2r/a_{0}}r^{2}dr \ \ \ \ \ (15)

We can use the formula (given in Shankar’s Appendix 2)

\displaystyle \int_{0}^{\infty}e^{-r/\alpha}r^{n}dr=\frac{n!}{\alpha^{n+1}} \ \ \ \ \ (16)

We get

\displaystyle \int\psi_{100}^{2}(r)d^{3}\mathbf{r}=\frac{4\pi}{\pi a_{0}^{3}}\frac{2!}{2^{3}}a_{0}^{3}=1 \ \ \ \ \ (17)

as required.

For a spherically symmetric wave function centred at {r=0},

\displaystyle \left(\Delta X\right)^{2}=\left\langle X^{2}\right\rangle -\left\langle X\right\rangle ^{2}=\left\langle X^{2}\right\rangle \ \ \ \ \ (18)

with identical relations for {Y} and {Z}. Since

\displaystyle r^{2} \displaystyle = \displaystyle x^{2}+y^{2}+z^{2}\ \ \ \ \ (19)
\displaystyle \left\langle r^{2}\right\rangle \displaystyle = \displaystyle \left\langle x^{2}\right\rangle +\left\langle y^{2}\right\rangle +\left\langle z^{2}\right\rangle =3\left\langle X^{2}\right\rangle \ \ \ \ \ (20)
\displaystyle \left\langle X^{2}\right\rangle \displaystyle = \displaystyle \frac{1}{3}\left\langle r^{2}\right\rangle \ \ \ \ \ (21)

Thus

\displaystyle \left\langle X^{2}\right\rangle \displaystyle = \displaystyle \frac{1}{3}\int\psi_{100}^{2}(r)r^{2}d^{3}\mathbf{r}\ \ \ \ \ (22)
\displaystyle \displaystyle = \displaystyle \frac{4\pi}{3\pi a_{0}^{3}}\int_{0}^{\infty}e^{-2r/a_{0}}r^{4}dr\ \ \ \ \ (23)
\displaystyle \displaystyle = \displaystyle \frac{4}{3a_{0}^{3}}\frac{4!}{2^{5}}a_{0}^{5}\ \ \ \ \ (24)
\displaystyle \displaystyle = \displaystyle a_{0}^{2}\ \ \ \ \ (25)
\displaystyle \Delta X \displaystyle = \displaystyle a_{0}=\frac{\hbar^{2}}{me^{2}} \ \ \ \ \ (26)

We can also find

\displaystyle \left\langle \frac{1}{r}\right\rangle \displaystyle = \displaystyle \int\psi_{100}^{2}(r)\frac{1}{r}d^{3}\mathbf{r}\ \ \ \ \ (27)
\displaystyle \displaystyle = \displaystyle \frac{4\pi}{\pi a_{0}^{3}}\int_{0}^{\infty}e^{-2r/a_{0}}r\;dr\ \ \ \ \ (28)
\displaystyle \displaystyle = \displaystyle \frac{4}{a_{0}^{3}}\frac{a_{0}^{2}}{4}\ \ \ \ \ (29)
\displaystyle \displaystyle = \displaystyle \frac{1}{a_{0}}\ \ \ \ \ (30)
\displaystyle \left\langle r\right\rangle \displaystyle = \displaystyle \int\psi_{100}^{2}(r)r\;d^{3}\mathbf{r}\ \ \ \ \ (31)
\displaystyle \displaystyle = \displaystyle \frac{4\pi}{\pi a_{0}^{3}}\int_{0}^{\infty}e^{-2r/a_{0}}r^{3}dr\ \ \ \ \ (32)
\displaystyle \displaystyle = \displaystyle \frac{4}{a_{0}^{3}}\frac{6a_{0}^{4}}{16}\ \ \ \ \ (33)
\displaystyle \displaystyle = \displaystyle \frac{3}{2}a_{0} \ \ \ \ \ (34)

Thus both {\left\langle \frac{1}{r}\right\rangle } and {\frac{1}{\left\langle r\right\rangle }} are of the same order of magnitude as {1/a_{0}=me^{2}/\hbar^{2}}.

Harmonic oscillator energies and eigenfunctions derived from the propagator

Shankar, R. (1994), Principles of Quantum Mechanics, Plenum Press. Chapter 8. Section 8.6, Exercise 8.6.3.

Given the propagator for the harmonic oscillator, it is possible to work backwards and deduce the eigenvalues and eigenfunctions of the Hamiltonian, although this isn’t the easiest way to find them. We’ve seen that the propagator for the oscillator is

\displaystyle U\left(x,t;x^{\prime}\right)=A\left(t\right)\exp\left[\frac{im\omega}{2\hbar\sin\omega t}\left(\left(x^{\prime2}+x^{2}\right)\cos\omega t-2x^{\prime}x\right)\right] \ \ \ \ \ (1)

 

where {A\left(t\right)} is some function of time which is found by doing a path integral. Shankar cheats a bit by just telling us what {A} is:

\displaystyle A\left(t\right)=\sqrt{\frac{m\omega}{2\pi i\hbar\sin\omega t}} \ \ \ \ \ (2)

To deduce (some of) the energy levels, we can compare the propagator with its more traditional form

\displaystyle U\left(t\right)=\sum_{n}e^{-iE_{n}t/\hbar}\left|E_{n}\right\rangle \left\langle E_{n}\right| \ \ \ \ \ (3)

where {E_{n}} is the {n}th energy level. In position space this is

\displaystyle U\left(t\right)=\sum_{n}\psi_{n}^*\left(x\right)\psi_{n}\left(x\right)e^{-iE_{n}t/\hbar} \ \ \ \ \ (4)

 

We can try finding the energy levels as follows. We take {x=x^{\prime}=t^{\prime}=0}, which is equivalent to taking the end time {t} to be a multiple of a complete period of the oscillator, so that the particle has returned to its starting point. In that case, 1 becomes

\displaystyle U\left(x,t;x^{\prime}\right)=A\left(t\right)=\sqrt{\frac{m\omega}{2\pi i\hbar\sin\omega t}} \ \ \ \ \ (5)

If we can expand this quantity in powers of {e^{-i\omega t}}, we can compare it with the series 4 and read off the energies from the exponents in the series. To do this, we write

\displaystyle A\left(t\right) \displaystyle = \displaystyle \sqrt{\frac{m\omega}{\pi\hbar\left(e^{i\omega t}-e^{-i\omega t}\right)}}\ \ \ \ \ (6)
\displaystyle \displaystyle = \displaystyle \sqrt{\frac{m\omega}{\pi\hbar}}e^{-i\omega t/2}\frac{1}{\sqrt{1-e^{-2i\omega t}}} \ \ \ \ \ (7)

To save writing, we’ll define the symbol

\displaystyle \eta\equiv e^{-i\omega t} \ \ \ \ \ (8)

so that

\displaystyle A\left(t\right)=\sqrt{\frac{m\omega}{\pi\hbar}}\eta^{1/2}\frac{1}{\sqrt{1-\eta^{2}}} \ \ \ \ \ (9)

We can now expand the last factor using the binomial expansion to get

\displaystyle A\left(t\right)=\sqrt{\frac{m\omega}{\pi\hbar}}\eta^{1/2}\left[1+\frac{1}{2}\eta^{2}+\frac{3}{8}\eta^{4}+\ldots\right] \ \ \ \ \ (10)

In terms of the original variables, we get

\displaystyle A\left(t\right)=\sqrt{\frac{m\omega}{\pi\hbar}}\left[e^{-i\omega t/2}+\frac{1}{2}e^{-5i\omega t/2}+\frac{3}{8}e^{-9i\omega t/2}+\ldots\right] \ \ \ \ \ (11)

 

Comparing with 4, we find energy levels of

\displaystyle E=\frac{\hbar\omega}{2},\frac{5\hbar\omega}{2},\frac{9\hbar\omega}{2},\ldots \ \ \ \ \ (12)

These correspond to {E_{0},E_{2},E_{4},\ldots}. The odd energy levels {\left(\frac{3\hbar\omega}{2},\frac{7\hbar\omega}{2},\ldots\right)} are missing because the corresponding wave functions {\psi_{n}\left(x\right)} are odd functions of {x} and are therefore zero at {x=0}, so the corresponding terms in 4 vanish. The numerical coefficients in 11 give us {\left|\psi_{n}\left(0\right)\right|^{2}} for {n=0,2,4,\ldots}.

To get the other energies, as well as the eigenfunctions, from a comparison of 1 and 4 is possible, but quite messy, even for the lower energies. To do it, we take {t^{\prime}=0} as before, but now we take {x=x^{\prime}\ne0}. That is, we start the oscillator off at some location {x^{\prime}\ne0} and then look at it exactly one period later, when it has returned to the same position. The propagator 1 now becomes

\displaystyle U\left(x,t;x^{\prime}\right) \displaystyle = \displaystyle \sqrt{\frac{m\omega}{2\pi i\hbar\sin\omega t}}\exp\left[\frac{im\omega}{2\hbar\sin\omega t}\left(2x^{2}\left(\cos\omega t-1\right)\right)\right]\ \ \ \ \ (13)
\displaystyle \displaystyle = \displaystyle \sqrt{\frac{m\omega}{\pi\hbar\left(e^{i\omega t}-e^{-i\omega t}\right)}}\exp\left[-\frac{m\omega}{\hbar\left(e^{i\omega t}-e^{-i\omega t}\right)}\left(x^{2}\left(\left(e^{i\omega t}+e^{-i\omega t}\right)-2\right)\right)\right]\ \ \ \ \ (14)
\displaystyle \displaystyle = \displaystyle \sqrt{\frac{m\omega}{\pi\hbar}}\eta^{1/2}\frac{1}{\sqrt{1-\eta^{2}}}\exp\left[-\frac{m\omega x^{2}}{\hbar}\left(\frac{\frac{1}{\eta}+\eta-2}{\frac{1}{\eta}-\eta}\right)\right]\ \ \ \ \ (15)
\displaystyle \displaystyle = \displaystyle \sqrt{\frac{m\omega}{\pi\hbar}}\eta^{1/2}\frac{1}{\sqrt{1-\eta^{2}}}\exp\left[-\frac{m\omega x^{2}}{\hbar}\left(\frac{1+\eta^{2}-2\eta}{1-\eta^{2}}\right)\right] \ \ \ \ \ (16)

We now need to expand this in a power series in {\eta}, which gets very messy so is best handled with software like Maple. Shankar asks only for the first two terms in the series (the terms corresponding to {\eta^{1/2}} and {\eta^{3/2}}) but even doing this by hand can get very tedious. The result from Maple is, for the first two terms:

\displaystyle \eta^{1/2} \displaystyle \rightarrow \displaystyle \sqrt{\frac{m\omega}{\pi\hbar}}e^{-m\omega x^{2}/\hbar}\eta^{1/2}=\sqrt{\frac{m\omega}{\pi\hbar}}e^{-m\omega x^{2}/\hbar}e^{-i\omega t/2\hbar}\ \ \ \ \ (17)
\displaystyle \eta^{3/2} \displaystyle \rightarrow \displaystyle \sqrt{\frac{m\omega}{\pi\hbar}}\frac{2m\omega}{\hbar}e^{-m\omega x^{2}/\hbar}x^{2}\eta^{3/2}=\sqrt{\frac{m\omega}{\pi\hbar}}\frac{2m\omega}{\hbar}e^{-m\omega x^{2}/\hbar}x^{2}e^{-3i\omega t/2\hbar} \ \ \ \ \ (18)

Comparing this with 4, we can read off:

\displaystyle E_{0} \displaystyle = \displaystyle \frac{\hbar\omega}{2}\ \ \ \ \ (19)
\displaystyle \left|\psi_{0}\left(x\right)\right|^{2} \displaystyle = \displaystyle \sqrt{\frac{m\omega}{\pi\hbar}}e^{-m\omega x^{2}/\hbar}\ \ \ \ \ (20)
\displaystyle E_{1} \displaystyle = \displaystyle \frac{3\hbar\omega}{2}\ \ \ \ \ (21)
\displaystyle \left|\psi_{1}\left(x\right)\right|^{2} \displaystyle = \displaystyle \sqrt{\frac{m\omega}{\pi\hbar}}\frac{2m\omega}{\hbar}e^{-m\omega x^{2}/\hbar}x^{2} \ \ \ \ \ (22)

To check this, we recall the eigenfunctions we worked out earlier, using Hermite polynomials

\displaystyle \psi_{n}(x)=\left(\frac{m\omega}{\pi\hbar}\right)^{1/4}\frac{1}{\sqrt{2^{n}n!}}H_{n}\left(\sqrt{\frac{m\omega}{\hbar}}x\right)e^{-m\omega x^{2}/2\hbar} \ \ \ \ \ (23)

 

The first two Hermite polynomials are

\displaystyle H_{0}\left(\sqrt{\frac{m\omega}{\hbar}}x\right) \displaystyle = \displaystyle 1\ \ \ \ \ (24)
\displaystyle H_{1}\left(\sqrt{\frac{m\omega}{\hbar}}x\right) \displaystyle = \displaystyle 2\sqrt{\frac{m\omega}{\hbar}}x \ \ \ \ \ (25)

Plugging these into 23 and comparing with 20 and 22 shows we got the right answer.

Harmonic oscillator: momentum space functions and Hermite polynomial recursion relations from raising and lowering operators

Shankar, R. (1994), Principles of Quantum Mechanics, Plenum Press. Section 7.5, Exercises 7.5.1 – 7.5.3.

Earlier, we found the position space energy eigenfunctions of the harmonic oscillator to be

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

where {y} in the first equation is shorthand for

\displaystyle y=\sqrt{\frac{m\omega}{\hbar}}x \ \ \ \ \ (3)

It turns out that an alternative method for deriving these functions uses the lowering operator {a}. Shankar gives the derivation of {\psi_{n}\left(x\right)} in his section 7.5, but we can use the same technique to derive the momentum space functions. We start with the ground state and use

\displaystyle a\left|0\right\rangle =0 \ \ \ \ \ (4)

In terms of {X} and {P}, we have

\displaystyle a=\sqrt{\frac{m\omega}{2\hbar}}X+i\frac{1}{\sqrt{2m\omega\hbar}}P \ \ \ \ \ (5)

 

To find the momentum space functions, we need to express {X} and {P} in terms of {p}:

\displaystyle X \displaystyle = \displaystyle i\hbar\frac{d}{dp}\ \ \ \ \ (6)
\displaystyle P \displaystyle = \displaystyle p \ \ \ \ \ (7)

We thus have

\displaystyle \left[i\hbar\sqrt{\frac{m\omega}{2\hbar}}\frac{d}{dp}+i\frac{1}{\sqrt{2m\omega\hbar}}p\right]\psi_{0}\left(p\right)=0 \ \ \ \ \ (8)

If we define the auxiliary variable

\displaystyle z\equiv\frac{p}{\sqrt{\hbar m\omega}} \ \ \ \ \ (9)

we get

\displaystyle \left(\frac{d}{dz}+z\right)\psi_{0}\left(z\right)=0 \ \ \ \ \ (10)

This has the solution

\displaystyle \psi_{0}\left(z\right)=Ae^{-z^{2}/2} \ \ \ \ \ (11)

for some normalization constant {A}. Thus in terms of {p} we have

\displaystyle \psi_{0}\left(p\right)=Ae^{-p^{2}/2\hbar m\omega} \ \ \ \ \ (12)

Normalizing in the usual way, making use of the Gaussian integral, we have

\displaystyle \int_{-\infty}^{\infty}\psi_{0}^{2}\left(p\right)dp \displaystyle = \displaystyle A^{2}\int_{-\infty}^{\infty}e^{-p^{2}/\hbar m\omega}dp=1\ \ \ \ \ (13)
\displaystyle A \displaystyle = \displaystyle \frac{1}{\left(\pi\hbar m\omega\right)^{1/4}} \ \ \ \ \ (14)

This agrees with the earlier result which was obtained by solving a second-order differential equation.

We can also use {a} and {a^{\dagger}} to verify a couple of recursion relations for Hermite polynomials. Reverting back to position space we have

\displaystyle X \displaystyle = \displaystyle x\ \ \ \ \ (15)
\displaystyle P \displaystyle = \displaystyle -i\hbar\frac{d}{dx} \ \ \ \ \ (16)

so 5 becomes

\displaystyle a=\sqrt{\frac{m\omega}{2\hbar}}x+\frac{\hbar}{\sqrt{2m\omega\hbar}}\frac{d}{dx} \ \ \ \ \ (17)

Also from 5 we have, since {X} and {P} are both hermitian operators

\displaystyle a^{\dagger} \displaystyle = \displaystyle \sqrt{\frac{m\omega}{2\hbar}}X-i\frac{1}{\sqrt{2m\omega\hbar}}P\ \ \ \ \ (18)
\displaystyle \displaystyle = \displaystyle \sqrt{\frac{m\omega}{2\hbar}}x-\frac{\hbar}{\sqrt{2m\omega\hbar}}\frac{d}{dx} \ \ \ \ \ (19)

Defining

\displaystyle y\equiv\sqrt{\frac{m\omega}{\hbar}}x \ \ \ \ \ (20)

we have

\displaystyle a \displaystyle = \displaystyle \frac{1}{\sqrt{2}}\left(y+\frac{d}{dy}\right)\ \ \ \ \ (21)
\displaystyle a^{\dagger} \displaystyle = \displaystyle \frac{1}{\sqrt{2}}\left(y-\frac{d}{dy}\right) \ \ \ \ \ (22)

We also recall the normalization conditions on the raising and lowering operators:

\displaystyle a\left|n\right\rangle \displaystyle = \displaystyle \sqrt{n}\left|n-1\right\rangle \ \ \ \ \ (23)
\displaystyle a^{\dagger}\left|n\right\rangle \displaystyle = \displaystyle \sqrt{n+1}\left|n+1\right\rangle \ \ \ \ \ (24)

Applying 23 to 1 we have, after cancelling common factors from each side:

\displaystyle \frac{1}{\sqrt{2}}\frac{1}{\sqrt{2^{n}n!}}\left(y+\frac{d}{dy}\right)\left[H_{n}(y)e^{-y^{2}/2}\right] \displaystyle = \displaystyle \frac{\sqrt{n}}{\sqrt{2^{n-1}\left(n-1\right)!}}H_{n-1}\left(y\right)e^{-y^{2}/2}\ \ \ \ \ (25)
\displaystyle \frac{1}{2\sqrt{n}}\frac{1}{\sqrt{2^{n-1}\left(n-1\right)!}}e^{-y^{2}/2}\left[yH_{n}\left(y\right)-yH_{n}\left(y\right)+\frac{dH_{n}}{dy}\right] \displaystyle = \displaystyle \frac{\sqrt{n}}{\sqrt{2^{n-1}\left(n-1\right)!}}H_{n-1}\left(y\right)e^{-y^{2}/2}\ \ \ \ \ (26)
\displaystyle yH_{n}\left(y\right)-yH_{n}\left(y\right)+\frac{dH_{n}}{dy} \displaystyle = \displaystyle 2nH_{n-1}\left(y\right)\ \ \ \ \ (27)
\displaystyle H_{n}^{\prime}\left(y\right) \displaystyle = \displaystyle 2nH_{n-1}\left(y\right) \ \ \ \ \ (28)

Another recursion relation for Hermite polynomials can be found as follows. We start with 22 to get

\displaystyle a+a^{\dagger}=\sqrt{2}y \ \ \ \ \ (29)

We now apply 23 and 24 to 1. We can cancel common factors, including {e^{-y^{2}/2}}, from both sides to get

\displaystyle \left(a+a^{\dagger}\right)\psi_{n} \displaystyle = \displaystyle \sqrt{2}y\psi_{n}\ \ \ \ \ (30)
\displaystyle \frac{\sqrt{2}y}{\sqrt{2^{n}n!}}H_{n}(y) \displaystyle = \displaystyle \frac{\sqrt{n}}{\sqrt{2^{n-1}\left(n-1\right)!}}H_{n-1}(y)+\frac{\sqrt{n+1}}{\sqrt{2^{n+1}\left(n+1\right)!}}H_{n+1}(y)\ \ \ \ \ (31)
\displaystyle \frac{y}{\sqrt{2^{n-1}n\left(n-1\right)!}}H_{n}(y) \displaystyle = \displaystyle \frac{\sqrt{n}}{\sqrt{2^{n-1}\left(n-1\right)!}}H_{n-1}(y)+\frac{1}{2\sqrt{2^{n-1}n\left(n-1\right)!}}H_{n+1}(y)\ \ \ \ \ (32)
\displaystyle yH_{n}\left(y\right) \displaystyle = \displaystyle nH_{n-1}\left(y\right)+\frac{1}{2}H_{n+1}\left(y\right)\ \ \ \ \ (33)
\displaystyle H_{n+1}\left(y\right) \displaystyle = \displaystyle 2yH_{n}\left(y\right)-2nH_{n-1}\left(y\right) \ \ \ \ \ (34)

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.