# Angular momentum and parity

Shankar, R. (1994), Principles of Quantum Mechanics, Plenum Press. Chapter 12, Exercise 12.5.12.

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The parity operator in 3-d reflects every point directly through the origin, so that a position vector ${\mathbf{r}\rightarrow-\mathbf{r}}$. In rectangular coordinates this means replacing each coordinate by its negative. In spherical coordinates, the angular coordinates change according to

 $\displaystyle \theta$ $\displaystyle \rightarrow$ $\displaystyle \pi-\theta\ \ \ \ \ (1)$ $\displaystyle \phi$ $\displaystyle \rightarrow$ $\displaystyle \pi+\phi \ \ \ \ \ (2)$

If this isn’t obvious, picture reflecting a vector ${\mathbf{r}}$ through the origin. If the original vector makes an angle ${\theta}$ with the ${z}$ (vertical) axis, then the reflected vector makes an angle ${\theta}$ with the ${-z}$ axis, which is equivalent to an angle of ${\pi-\theta}$ with the ${+z}$ axis. The azimuthal angle ${\phi}$ just gets rotated by ${\pi}$ to lie on the other side of the ${z}$ axis.

Using this, we can see that the parity operator ${\Pi}$ commutes with both ${L^{2}}$ and ${L_{z}}$, as follows. Since neither of these operators involves the radial coordinate, we can consider their effect on a function ${f\left(\theta,\phi\right)}$. Under parity, we have

$\displaystyle \Pi f\left(\theta,\phi\right)\rightarrow f\left(\pi-\theta,\pi+\phi\right) \ \ \ \ \ (3)$

Thus the derivatives transform under parity according to

 $\displaystyle \frac{\partial f\left(\theta,\phi\right)}{\partial\theta}$ $\displaystyle \rightarrow$ $\displaystyle -\frac{\partial f\left(\pi-\theta,\pi+\phi\right)}{\partial\theta}\ \ \ \ \ (4)$ $\displaystyle \frac{\partial f\left(\theta,\phi\right)}{\partial\phi}$ $\displaystyle \rightarrow$ $\displaystyle \frac{\partial f\left(\pi-\theta,\pi+\phi\right)}{\partial\phi} \ \ \ \ \ (5)$
 $\displaystyle L^{2}$ $\displaystyle =$ $\displaystyle -\hbar^{2}\left[\frac{1}{\sin\theta}\frac{\partial}{\partial\theta}\left(\sin\theta\frac{\partial}{\partial\theta}\right)+\frac{1}{\sin^{2}\theta}\frac{\partial^{2}}{\partial\phi^{2}}\right]\ \ \ \ \ (6)$ $\displaystyle L_{z}$ $\displaystyle =$ $\displaystyle -i\hbar\frac{\partial}{\partial\phi} \ \ \ \ \ (7)$

Thus the combined operation gives

 $\displaystyle L^{2}\Pi f\left(\theta,\phi\right)$ $\displaystyle \rightarrow$ $\displaystyle -\hbar^{2}\left[\frac{1}{\sin\theta}\frac{\partial}{\partial\theta}\left(\sin\theta\frac{\partial}{\partial\theta}\right)+\frac{1}{\sin^{2}\theta}\frac{\partial^{2}}{\partial\phi^{2}}\right]f\left(\pi-\theta,\pi+\phi\right)\ \ \ \ \ (8)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -\hbar^{2}\left[\frac{1}{\sin\theta}\left(-\frac{\partial}{\partial\theta}\right)\left(\sin\theta\left(-\frac{\partial}{\partial\theta}\right)\right)+\frac{1}{\sin^{2}\theta}\frac{\partial^{2}}{\partial\phi^{2}}\right]f\left(\pi-\theta,\pi+\phi\right)\ \ \ \ \ (9)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -\hbar^{2}\left[\frac{1}{\sin\theta}\frac{\partial}{\partial\theta}\left(\sin\theta\frac{\partial}{\partial\theta}\right)+\frac{1}{\sin^{2}\theta}\frac{\partial^{2}}{\partial\phi^{2}}\right]f\left(\pi-\theta,\pi+\phi\right)\ \ \ \ \ (10)$ $\displaystyle$ $\displaystyle =$ $\displaystyle L^{2}f\left(\pi-\theta,\pi+\phi\right) \ \ \ \ \ (11)$

If we apply ${\Pi}$ to ${L^{2}}$, we have

 $\displaystyle \Pi\left[L^{2}f\left(\theta,\phi\right)\right]$ $\displaystyle =$ $\displaystyle -\hbar^{2}\left[\frac{1}{\sin\left(\pi-\theta\right)}\left(-\frac{\partial}{\partial\theta}\right)\left(\sin\left(\pi-\theta\right)\left(-\frac{\partial}{\partial\theta}\right)\right)+\frac{1}{\sin^{2}\left(\pi-\theta\right)}\frac{\partial^{2}}{\partial\phi^{2}}\right]f\left(\pi-\theta,\pi+\phi\right)\ \ \ \ \ (12)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -\hbar^{2}\left[\frac{1}{\sin\theta}\frac{\partial}{\partial\theta}\left(\sin\theta\frac{\partial}{\partial\theta}\right)+\frac{1}{\sin^{2}\theta}\frac{\partial^{2}}{\partial\phi^{2}}\right]f\left(\pi-\theta,\pi+\phi\right)\ \ \ \ \ (13)$ $\displaystyle$ $\displaystyle =$ $\displaystyle L^{2}f\left(\pi-\theta,\pi+\phi\right) \ \ \ \ \ (14)$

Thus

$\displaystyle \left[\Pi,L^{2}\right]=0 \ \ \ \ \ (15)$

where in the first line we used ${\sin\left(\pi-\theta\right)=\sin\theta}$.

Since ${L_{z}}$ involves only a derivative with respect to ${\phi}$ which doesn’t change under parity, we have

$\displaystyle \left[\Pi,L_{z}\right]=0 \ \ \ \ \ (16)$

Since ${\Pi}$ commutes with both ${L^{2}}$ and ${L_{z}}$ it is possible to find a set of functions that are simultaneous eigenfunctions of all three operators. These functions turn out to be the same spherical harmonics that we’ve been using all along. We can show this by starting with the top spherical harmonic

$\displaystyle Y_{l}^{l}=\left(-1\right)^{l}\sqrt{\frac{(2l+1)!}{4\pi}}\frac{1}{2^{l}l!}e^{il\phi}\sin^{l}\theta \ \ \ \ \ (17)$

where we’ve included the ${\left(-1\right)^{l}}$ to be consistent with Shankar’s equation 12.5.32. Under parity, this transforms as

 $\displaystyle \Pi Y_{l}^{l}$ $\displaystyle =$ $\displaystyle \left(-1\right)^{l}\sqrt{\frac{(2l+1)!}{4\pi}}\frac{1}{2^{l}l!}e^{il\left(\pi+\phi\right)}\sin^{l}\left(\pi-\theta\right)\ \ \ \ \ (18)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \left(-1\right)^{l}e^{il\pi}\sqrt{\frac{(2l+1)!}{4\pi}}\frac{1}{2^{l}l!}e^{il\phi}\sin^{l}\theta\ \ \ \ \ (19)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \left(-1\right)^{l}Y_{l}^{l} \ \ \ \ \ (20)$

where we used ${e^{il\pi}=\left(-1\right)^{l}}$ in the second line. Thus ${Y_{l}^{l}}$ is an eigenfunction of ${\Pi}$ with eigenvalue ${\left(-1\right)^{l}}$.

To show that the other spherical harmonics are also eigenfunctions, we can use the lowering operator ${L_{-}}$. In spherical coordinates, we have

$\displaystyle L_{-}Y_{l}^{m}=\hbar\sqrt{(\ell+m)(\ell-m+1)}Y_{l}^{m-1} \ \ \ \ \ (21)$

The operator can be expressed as

$\displaystyle L_{-}=-\hbar e^{-i\phi}\left[\frac{\partial}{\partial\theta}-i\cot\theta\frac{\partial}{\partial\phi}\right] \ \ \ \ \ (22)$

Under parity, we can transform 22 using ${\sin\left(\pi-\theta\right)=\sin\theta}$ and ${\cos\left(\pi-\theta\right)=-\cos\theta}$, so that ${\cot\left(\pi-\theta\right)=-\cot\theta}$. We therefore have

 $\displaystyle \Pi L_{-}$ $\displaystyle =$ $\displaystyle -\hbar e^{-i\left(\pi+\phi\right)}\left[-\frac{\partial}{\partial\theta}+i\cot\theta\frac{\partial}{\partial\phi}\right]\ \ \ \ \ (23)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -\hbar e^{-i\phi}\left[\frac{\partial}{\partial\theta}-i\cot\theta\frac{\partial}{\partial\phi}\right]\ \ \ \ \ (24)$ $\displaystyle$ $\displaystyle =$ $\displaystyle L_{-} \ \ \ \ \ (25)$

Thus ${L_{-}}$ is unchanged by parity, which means that from 21, ${Y_{l}^{m-1}}$ has the same parity as ${Y_{l}^{m}}$. Starting with ${Y_{l}^{l}}$ and using the lowering operator successively to reduce the superscript index, we have therefore

$\displaystyle \Pi Y_{l}^{m}=\left(-1\right)^{l}Y_{l}^{m} \ \ \ \ \ (26)$

Thus all spherical harmonics are also eigenfunctions of parity.

# Spherical harmonics using the lowering operator

Shankar, R. (1994), Principles of Quantum Mechanics, Plenum Press. Chapter 12, Exercise 12.5.11.

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The raising and lowering operators for angular momentum are

$\displaystyle L_{\pm}\equiv L_{x}\pm iL_{y} \ \ \ \ \ (1)$

On a state ${\left|\ell m\right\rangle }$ in the basis of eigenstates of ${L^{2}}$ and ${L_{z}}$, they have the effect:

$\displaystyle L_{\pm}\left|\ell m\right\rangle =\hbar\sqrt{(\ell\mp m)(\ell\pm m+1)}\left|\ell,m\pm1\right\rangle \ \ \ \ \ (2)$

This means that, if we can find the top state ${\left|\ell\ell\right\rangle }$, we can find the state for all lower values of ${m}$ by applying ${L_{-}}$ successively. To illustrate the process we’ll derive the 3 states for ${\ell=1}$. The top state ${\left|11\right\rangle }$ can be obtained by following the derivation given in Shankar from his equation 12.5.28 onwards. In spherical coordinates, the raising and lowering operators have the form

$\displaystyle L_{\pm}=\pm\hbar e^{\pm i\phi}\left[\frac{\partial}{\partial\theta}\pm i\cot\theta\frac{\partial}{\partial\phi}\right] \ \ \ \ \ (3)$

Applying ${L_{+}}$ to the top state ${\left|11\right\rangle }$ must give zero, so if ${\psi_{1}^{1}}$ is the representation of this state in spherical coordinates, we must solve the differential equation

$\displaystyle \left[\frac{\partial}{\partial\theta}+i\cot\theta\frac{\partial}{\partial\phi}\right]\psi_{1}^{1}=0 \ \ \ \ \ (4)$

Since ${\psi_{1}^{1}}$ is also an eigenfunction of ${L_{z}}$ with eigenvalue ${\hbar}$, we know that

$\displaystyle \psi_{1}^{1}=U_{1}^{1}\left(r,\theta\right)e^{i\phi} \ \ \ \ \ (5)$

Thus 4 becomes

$\displaystyle \left(\frac{\partial}{\partial\theta}-\cot\theta\right)U_{1}^{1}=0 \ \ \ \ \ (6)$

This can be solved by writing it in the form

 $\displaystyle \frac{dU_{1}^{1}}{U_{1}^{1}}$ $\displaystyle =$ $\displaystyle \frac{d\left(\sin\theta\right)}{\sin\theta}\ \ \ \ \ (7)$ $\displaystyle \ln U_{1}^{1}$ $\displaystyle =$ $\displaystyle \ln\left(\sin\theta\right)+\ln R\left(r\right)+\ln A \ \ \ \ \ (8)$

where ${R}$ is some unspecified function of ${r}$, and ${A}$ is a constant. We therefore have

 $\displaystyle U_{1}^{1}\left(r,\theta\right)$ $\displaystyle =$ $\displaystyle R\left(r\right)\left(A\sin\theta\right) \ \ \ \ \ (9)$

If we ignore ${R}$ for now, we can normalize over the angular coordinates by requiring

$\displaystyle \int\left|A\sin\theta\right|^{2}d\Omega=1 \ \ \ \ \ (10)$

The element ${d\Omega}$ of solid angle is

$\displaystyle d\Omega=\sin\theta\;d\phi\;d\theta \ \ \ \ \ (11)$

so we have

 $\displaystyle \left|A\right|^{2}\int_{0}^{\pi}\int_{0}^{2\pi}\sin^{3}\theta\;d\phi\;d\theta$ $\displaystyle =$ $\displaystyle 2\pi\left|A\right|^{2}\int_{0}^{\pi}\sin\theta\left(1-\cos^{2}\theta\right)d\theta\ \ \ \ \ (12)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{8\pi}{3}\left|A\right|^{2}\ \ \ \ \ (13)$ $\displaystyle A$ $\displaystyle =$ $\displaystyle \sqrt{\frac{3}{8\pi}} \ \ \ \ \ (14)$

Thus the spherical harmonic ${Y_{1}^{1}}$ is (using Shankar’s normalization convention of multiplying by ${\left(-1\right)^{\ell}}$):

$\displaystyle Y_{1}^{1}=-\sqrt{\frac{3}{8\pi}}\sin\theta e^{i\phi} \ \ \ \ \ (15)$

We can now get ${Y_{1}^{0}}$ by applying ${L_{-}}$ to ${Y_{1}^{1}}$. From 2 we have

 $\displaystyle L_{-}Y_{1}^{1}$ $\displaystyle =$ $\displaystyle \hbar\sqrt{\left(1+1\right)\left(1-1+1\right)}Y_{1}^{0}\ \ \ \ \ (16)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \sqrt{2}\hbar Y_{1}^{0} \ \ \ \ \ (17)$

From 3 we have

 $\displaystyle L_{-}Y_{1}^{1}$ $\displaystyle =$ $\displaystyle -\hbar e^{-i\phi}\left[\frac{\partial}{\partial\theta}-i\cot\theta\frac{\partial}{\partial\phi}\right]Y_{1}^{1}\ \ \ \ \ (18)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -\hbar e^{-i\phi}\left(-\sqrt{\frac{3}{8\pi}}\right)\left[\cos\theta-i\cot\theta\left(i\sin\theta\right)\right]e^{i\phi}\ \ \ \ \ (19)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 2\hbar\sqrt{\frac{3}{8\pi}}\cos\theta \ \ \ \ \ (20)$

Comparing the last two results gives

 $\displaystyle \sqrt{2}\hbar Y_{1}^{0}$ $\displaystyle =$ $\displaystyle 2\sqrt{\frac{3}{8\pi}}\cos\theta\ \ \ \ \ (21)$ $\displaystyle Y_{1}^{0}$ $\displaystyle =$ $\displaystyle \sqrt{\frac{3}{4\pi}}\cos\theta \ \ \ \ \ (22)$

Repeating the process, we get

 $\displaystyle L_{-}Y_{1}^{0}$ $\displaystyle =$ $\displaystyle \hbar\sqrt{\left(1+0\right)\left(1-0+1\right)}Y_{1}^{-1}\ \ \ \ \ (23)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \sqrt{2}\hbar Y_{1}^{-1} \ \ \ \ \ (24)$

Also

 $\displaystyle L_{-}Y_{1}^{0}$ $\displaystyle =$ $\displaystyle -\hbar e^{-i\phi}\left[\frac{\partial}{\partial\theta}-i\cot\theta\frac{\partial}{\partial\phi}\right]Y_{1}^{0}\ \ \ \ \ (25)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -\hbar e^{-i\phi}\sqrt{\frac{3}{4\pi}}\left(-\sin\theta-0\right)\ \ \ \ \ (26)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \hbar\sqrt{\frac{3}{4\pi}}\sin\theta e^{-i\phi} \ \ \ \ \ (27)$

Thus

 $\displaystyle \sqrt{2}\hbar Y_{1}^{-1}$ $\displaystyle =$ $\displaystyle \hbar\sqrt{\frac{3}{4\pi}}\sin\theta e^{-i\phi}\ \ \ \ \ (28)$ $\displaystyle Y_{1}^{-1}$ $\displaystyle =$ $\displaystyle \sqrt{\frac{3}{8\pi}}\sin\theta e^{-i\phi} \ \ \ \ \ (29)$

Comparing these results with Shankar’s equation 12.5.39 we see that they match. [This exercise is similar to one we did earlier, where we used the raising operator to generate spherical harmonics with higher values of ${m}$.]

# Angular momentum – raising and lowering operators from rectangular coordinates

Shankar, R. (1994), Principles of Quantum Mechanics, Plenum Press. Chapter 12, Exercise 12.5.8.

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To calculate the eigenfunctions of angular momentum, we will need expressions for the raising and lowering operators ${L_{\pm}}$ in spherical coordinates. We’ve seen one way of getting these by working with the gradient in spherical coordinates from the start, but it is also possible to convert the rectangular forms of ${L_{\pm}}$ to spherical coordinates by using the chain rule from calculus. This method is similar to one we used earlier in 2-d. To set the scene, we need the conversion formulas between rectangular and spherical coordinates:

 $\displaystyle x$ $\displaystyle =$ $\displaystyle r\sin\theta\cos\phi\ \ \ \ \ (1)$ $\displaystyle y$ $\displaystyle =$ $\displaystyle r\sin\theta\sin\phi\ \ \ \ \ (2)$ $\displaystyle z$ $\displaystyle =$ $\displaystyle r\cos\theta\ \ \ \ \ (3)$ $\displaystyle r$ $\displaystyle =$ $\displaystyle \sqrt{x^{2}+y^{2}+z^{2}}\ \ \ \ \ (4)$ $\displaystyle \theta$ $\displaystyle =$ $\displaystyle \arctan\frac{\sqrt{x^{2}+y^{2}}}{z}\ \ \ \ \ (5)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \arctan\frac{q}{z}\ \ \ \ \ (6)$ $\displaystyle \phi$ $\displaystyle =$ $\displaystyle \arctan\frac{y}{x} \ \ \ \ \ (7)$

To simplify the notation, we’ve defined

$\displaystyle q\equiv\sqrt{x^{2}+y^{2}}=r\sin\theta \ \ \ \ \ (8)$

We’ll also use shorthand notation for sines and cosines so that

 $\displaystyle s_{\theta}$ $\displaystyle \equiv$ $\displaystyle \sin\theta\ \ \ \ \ (9)$ $\displaystyle c_{\theta}$ $\displaystyle \equiv$ $\displaystyle \cos\theta \ \ \ \ \ (10)$

and similarly for ${\phi}$. We’ll also use the notation ${\partial_{r}}$ to mean the partial derivative with respect to ${r}$, with a similar notation for other derivatives.

The required derivatives are

 $\displaystyle \partial_{x}$ $\displaystyle =$ $\displaystyle \partial_{x}r\cdot\partial_{r}+\partial_{x}\theta\cdot\partial_{\theta}+\partial_{x}\phi\cdot\partial_{\phi}\ \ \ \ \ (11)$ $\displaystyle \partial_{y}$ $\displaystyle =$ $\displaystyle \partial_{y}r\cdot\partial_{r}+\partial_{y}\theta\cdot\partial_{\theta}+\partial_{y}\phi\cdot\partial_{\phi}\ \ \ \ \ (12)$ $\displaystyle \partial_{z}$ $\displaystyle =$ $\displaystyle \partial_{z}r\cdot\partial_{r}+\partial_{z}\theta\cdot\partial_{\theta} \ \ \ \ \ (13)$

The required derivatives are

 $\displaystyle \partial_{x}r$ $\displaystyle =$ $\displaystyle \frac{x}{r}\ \ \ \ \ (14)$ $\displaystyle \partial_{y}r$ $\displaystyle =$ $\displaystyle \frac{y}{r}\ \ \ \ \ (15)$ $\displaystyle \partial_{z}r$ $\displaystyle =$ $\displaystyle \frac{z}{r}\ \ \ \ \ (16)$ $\displaystyle \partial_{x}\theta$ $\displaystyle =$ $\displaystyle \frac{x/q}{z\left(1+\frac{q^{2}}{z^{2}}\right)}\ \ \ \ \ (17)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{xz}{qr^{2}}\ \ \ \ \ (18)$ $\displaystyle \partial_{y}\theta$ $\displaystyle =$ $\displaystyle \frac{yz}{qr^{2}}\ \ \ \ \ (19)$ $\displaystyle \partial_{z}\theta$ $\displaystyle =$ $\displaystyle -\frac{q}{r^{2}}\ \ \ \ \ (20)$ $\displaystyle \partial_{x}\phi$ $\displaystyle =$ $\displaystyle \frac{-y/x^{2}}{1+y^{2}/x^{2}}\ \ \ \ \ (21)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -\frac{y}{q^{2}}\ \ \ \ \ (22)$ $\displaystyle \partial_{y}\phi$ $\displaystyle =$ $\displaystyle \frac{x}{q^{2}}\ \ \ \ \ (23)$ $\displaystyle \partial_{z}\phi$ $\displaystyle =$ $\displaystyle 0 \ \ \ \ \ (24)$

Plugging all these into 11 to 13 we have

 $\displaystyle \partial_{x}$ $\displaystyle =$ $\displaystyle \frac{x}{r}\partial_{r}+\frac{xz}{qr^{2}}\partial_{\theta}-\frac{y}{q^{2}}\partial_{\phi}\ \ \ \ \ (25)$ $\displaystyle \partial_{y}$ $\displaystyle =$ $\displaystyle \frac{y}{r}\partial_{r}+\frac{yz}{qr^{2}}\partial_{\theta}+\frac{x}{q^{2}}\partial_{\phi}\ \ \ \ \ (26)$ $\displaystyle \partial_{z}$ $\displaystyle =$ $\displaystyle \frac{z}{r}\partial_{r}-\frac{q}{r^{2}}\partial_{\theta} \ \ \ \ \ (27)$

We can now calculate the components ${L_{x}}$ and ${L_{y}}$:

 $\displaystyle L_{x}$ $\displaystyle =$ $\displaystyle -i\hbar\left(y\partial_{z}-z\partial_{y}\right)\ \ \ \ \ (28)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -i\hbar\left[\frac{yz}{r}\partial_{r}-\frac{yq}{r^{2}}\partial_{\theta}-\frac{yz}{r}\partial_{r}-\frac{yz^{2}}{qr^{2}}\partial_{\theta}-\frac{xz}{q^{2}}\partial_{\phi}\right]\ \ \ \ \ (29)$ $\displaystyle$ $\displaystyle =$ $\displaystyle i\hbar\left[\left(\frac{yq}{r^{2}}+\frac{yz^{2}}{qr^{2}}\right)\partial_{\theta}+\frac{xz}{q^{2}}\partial_{\phi}\right]\ \ \ \ \ (30)$ $\displaystyle$ $\displaystyle =$ $\displaystyle i\hbar\left[\left(s_{\theta}^{2}s_{\phi}+\frac{s_{\theta}s_{\phi}c_{\theta}^{2}}{s_{\theta}}\right)\partial_{\theta}+\frac{s_{\theta}c_{\theta}c_{\phi}}{s_{\theta}^{2}}\partial_{\phi}\right]\ \ \ \ \ (31)$ $\displaystyle$ $\displaystyle =$ $\displaystyle i\hbar\left(\sin\phi\frac{\partial}{\partial\theta}+\cos\phi\cot\theta\frac{\partial}{\partial\phi}\right)\ \ \ \ \ (32)$ $\displaystyle L_{y}$ $\displaystyle =$ $\displaystyle -i\hbar\left(z\partial_{x}-x\partial_{z}\right)\ \ \ \ \ (33)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -i\hbar\left[\frac{xz}{r}\partial_{r}+\frac{xz^{2}}{qr^{2}}\partial_{\theta}-\frac{xz}{r}\partial_{r}+\frac{xq}{r^{2}}\partial_{\theta}-\frac{yz}{q^{2}}\partial_{\phi}\right]\ \ \ \ \ (34)$ $\displaystyle$ $\displaystyle =$ $\displaystyle i\hbar\left[\left(-\frac{xz^{2}}{qr^{2}}-\frac{xq}{r^{2}}\right)\partial_{\theta}+\frac{yz}{q^{2}}\partial_{\phi}\right]\ \ \ \ \ (35)$ $\displaystyle$ $\displaystyle =$ $\displaystyle i\hbar\left[\left(-\frac{s_{\theta}c_{\phi}c_{\theta}^{2}}{s_{\theta}}-s_{\theta}^{2}c_{\phi}\right)\partial_{\theta}+\frac{s_{\theta}c_{\theta}s_{\phi}}{s_{\theta}^{2}}\partial_{\phi}\right]\ \ \ \ \ (36)$ $\displaystyle$ $\displaystyle =$ $\displaystyle i\hbar\left(-\cos\phi\frac{\partial}{\partial\theta}+\sin\phi\cot\theta\frac{\partial}{\partial\phi}\right) \ \ \ \ \ (37)$

From this we get the raising and lowering operators

 $\displaystyle L_{\pm}$ $\displaystyle =$ $\displaystyle L_{x}\pm iL_{y}\ \ \ \ \ (38)$ $\displaystyle$ $\displaystyle =$ $\displaystyle i\hbar\left(\sin\phi\frac{\partial}{\partial\theta}+\cos\phi\cot\theta\frac{\partial}{\partial\phi}\right)\mp\ \ \ \ \ (39)$ $\displaystyle$ $\displaystyle$ $\displaystyle \hbar\left(-\cos\phi\frac{\partial}{\partial\theta}+\sin\phi\cot\theta\frac{\partial}{\partial\phi}\right)\ \ \ \ \ (40)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \hbar e^{\pm i\phi}\frac{\partial}{\partial\theta}\pm i\hbar e^{\pm i\phi}\cot\theta\frac{\partial}{\partial\phi}\ \ \ \ \ (41)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \hbar e^{\pm i\phi}\left(\frac{\partial}{\partial\theta}\pm i\cot\theta\frac{\partial}{\partial\phi}\right) \ \ \ \ \ (42)$

[Admittedly, it’s probably easier and more elegant to use spherical coordinates from the start, but it’s instructive to see how it’s done starting with rectangular coordinates.]

# 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.

# Spin matrices: general case

Required math: algebra

Required physics: 3-d Schrödinger equation

Reference: Griffiths, David J. (2005), Introduction to Quantum Mechanics, 2nd Edition; Pearson Education – Problem 4.53.

We can generalize the calculation for spin 3/2 to get the spin matrices for any spin ${s}$.

We first find ${S_{\pm}}$ from the equation

$\displaystyle S_{\pm}|s\, m\rangle=\hbar\sqrt{s(s+1)-m(m\pm1)}|s\, m\pm1\rangle \ \ \ \ \ (1)$

First, consider ${S_{+}}$. We know that ${S_{+}|s\, s\rangle=0}$ and that ${m=s-k}$ for ${k=1\ldots2s}$ for the remaining eigenstates of ${S_{z}}$. We can also see from the spin 3/2 case that the ${S_{+}}$matrix contains non-zero entries only on the diagonal above the main diagonal (since the raising operator always raises an eigenstate ${m}$ of ${S_{z}}$ to the state ${m+1}$), and that these entries consist of the square root factor given in 1 for values of ${m}$ starting with ${m=s-1}$ as matrix element ${S_{+(1,2)}}$ (i.e. the element in row 1, column 2 of matrix ${S_{+}}$), and then working down the diagonal, with element ${S_{+(k,k+1)}}$ being the square root factor for ${m=s-k}$.

Letting ${m=s-k}$ for ${k=1\ldots2s}$ we get

 $\displaystyle \sqrt{s(s+1)-m(m+1)}$ $\displaystyle =$ $\displaystyle \sqrt{s(s+1)-(s-k)(s-k+1)}\ \ \ \ \ (2)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \sqrt{2sk+k(1-k)}\ \ \ \ \ (3)$ $\displaystyle$ $\displaystyle \equiv$ $\displaystyle c_{k} \ \ \ \ \ (4)$

We can therefore write

$\displaystyle S_{+}=\hbar\left[\begin{array}{ccccccc} 0 & c_{1} & 0 & 0 & \ldots & 0 & 0\\ 0 & 0 & c_{2} & 0 & \ldots & 0 & 0\\ 0 & 0 & 0 & c_{3} & \ldots & 0 & 0\\ 0 & 0 & 0 & 0 & \ldots & 0 & 0\\ \vdots & \vdots & \vdots & \vdots & \ddots & \vdots & \vdots\\ 0 & 0 & 0 & 0 & \ldots & 0 & c_{2s}\\ 0 & 0 & 0 & 0 & \ldots & 0 & 0 \end{array}\right] \ \ \ \ \ (5)$

For those readers wanting to get the answer as given in Griffiths’s book, we can make the substitution

$\displaystyle j\equiv-k+1+s \ \ \ \ \ (6)$

so that ${j=s\ldots-s+1}$ as ${k=1\ldots2s}$. Then

 $\displaystyle \sqrt{2sk+k(1-k)}$ $\displaystyle =$ $\displaystyle \sqrt{2(1+s-j)s+(1+s-j)(j-s)}\ \ \ \ \ (7)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \sqrt{(1+s-j)(j+s)}\ \ \ \ \ (8)$ $\displaystyle$ $\displaystyle \equiv$ $\displaystyle b_{j} \ \ \ \ \ (9)$

as given in the book.

The calculation for ${S_{-}}$ is similar, giving ${S_{-(k+1,k)}=c_{k}}$ for ${k=1\ldots2s}$, so we get

$\displaystyle S_{-}=\hbar\left[\begin{array}{ccccccc} 0 & 0 & 0 & 0 & \ldots & 0 & 0\\ c_{1} & 0 & 0 & 0 & \ldots & 0 & 0\\ 0 & c_{2} & 0 & 0 & \ldots & 0 & 0\\ 0 & 0 & c_{3} & 0 & \ldots & 0 & 0\\ \vdots & \vdots & \vdots & \vdots & \ddots & \vdots & \vdots\\ 0 & 0 & 0 & 0 & \ldots & 0 & 0\\ 0 & 0 & 0 & 0 & \ldots & c_{2s} & 0 \end{array}\right] \ \ \ \ \ (10)$

In Griffiths’s notation, ${S_{-(k+1,k)}=b_{j}}$, where ${j=-k+1+s}$ as before, and we can then use the formulas ${S_{x}=(S_{+}+S_{-})/2}$ and ${S_{y}=(S_{+}-S_{-})/2i}$ to get the actual spin matrices, which will have non-zero entries on the two diagonals on either side of the main diagonal.

# Spin 3/2

Required math: algebra

Required physics: 3-d Schrödinger equation

Reference: Griffiths, David J. (2005), Introduction to Quantum Mechanics, 2nd Edition; Pearson Education – Problem 4.52.

Continuing our exploration of higher spin states, we can work out the spin matrix ${S_{x}}$ for spin 3/2. We use the raising and lowering operators first:

 $\displaystyle S_{\pm}|s\, m\rangle$ $\displaystyle =$ $\displaystyle \hbar\sqrt{s(s+1)-m(m\pm1)}|s\, m\pm1\rangle \ \ \ \ \ (1)$

Here, ${|s\, m\rangle}$ is an eigenstate of ${S^{2}}$ with eigenvalue ${s\left(s+1\right)}$ and of ${S_{z}}$ with eigenvalue ${m}$. We can work out the effects of ${S_{\pm}}$ on the various eigenstates of ${S_{z}}$ for ${s=3/2}$ and get

 $\displaystyle S_{+}\Big|\frac{3}{2}\,\frac{3}{2}\Big\rangle$ $\displaystyle =$ $\displaystyle 0\ \ \ \ \ (2)$ $\displaystyle S_{+}\Big|\frac{3}{2}\,\frac{1}{2}\Big\rangle$ $\displaystyle =$ $\displaystyle \sqrt{3}\hbar\Big|\frac{3}{2}\,\frac{3}{2}\Big\rangle\ \ \ \ \ (3)$ $\displaystyle S_{+}\Big|\frac{3}{2}\,-\frac{1}{2}\Big\rangle$ $\displaystyle =$ $\displaystyle 2\hbar\Big|\frac{3}{2}\,\frac{1}{2}\Big\rangle\ \ \ \ \ (4)$ $\displaystyle S_{+}\Big|\frac{3}{2}\,-\frac{3}{2}\Big\rangle$ $\displaystyle =$ $\displaystyle \sqrt{3}\hbar\Big|\frac{3}{2}\,-\frac{1}{2}\Big\rangle\ \ \ \ \ (5)$ $\displaystyle S_{-}\Big|\frac{3}{2}\,\frac{3}{2}\Big\rangle$ $\displaystyle =$ $\displaystyle \sqrt{3}\hbar\Big|\frac{3}{2}\,\frac{1}{2}\Big\rangle\ \ \ \ \ (6)$ $\displaystyle S_{-}\Big|\frac{3}{2}\,\frac{1}{2}\Big\rangle$ $\displaystyle =$ $\displaystyle 2\hbar\Big|\frac{3}{2}\,-\frac{1}{2}\Big\rangle\ \ \ \ \ (7)$ $\displaystyle S_{-}\Big|\frac{3}{2}\,-\frac{1}{2}\Big\rangle$ $\displaystyle =$ $\displaystyle \sqrt{3}\hbar\Big|\frac{3}{2}\,-\frac{3}{2}\Big\rangle\ \ \ \ \ (8)$ $\displaystyle S_{-}\Big|\frac{3}{2}\,-\frac{3}{2}\Big\rangle$ $\displaystyle =$ $\displaystyle 0 \ \ \ \ \ (9)$

Combining these conditions, we get the matrix forms for ${S_{\pm}}$:

$\displaystyle S_{+}=\hbar\left(\begin{array}{cccc} 0 & \sqrt{3} & 0 & 0\\ 0 & 0 & 2 & 0\\ 0 & 0 & 0 & \sqrt{3}\\ 0 & 0 & 0 & 0 \end{array}\right) \ \ \ \ \ (10)$

$\displaystyle S_{-}=\hbar\left(\begin{array}{cccc} 0 & 0 & 0 & 0\\ \sqrt{3} & 0 & 0 & 0\\ 0 & 2 & 0 & 0\\ 0 & 0 & \sqrt{3} & 0 \end{array}\right) \ \ \ \ \ (11)$

so, since ${S_{x}=(S_{+}+S_{-})/2}$

$\displaystyle S_{x}=\frac{\hbar}{2}\left(\begin{array}{cccc} 0 & \sqrt{3} & 0 & 0\\ \sqrt{3} & 0 & 2 & 0\\ 0 & 2 & 0 & \sqrt{3}\\ 0 & 0 & \sqrt{3} & 0 \end{array}\right) \ \ \ \ \ (12)$

The characteristic equation for the matrix part of ${S_{x}}$ is

 $\displaystyle -\lambda(-\lambda(\lambda^{2}-3)-2(-2\lambda))-\sqrt{3}(\sqrt{3}(\lambda^{2}-3))$ $\displaystyle =$ $\displaystyle 0\ \ \ \ \ (13)$ $\displaystyle \lambda^{4}-10\lambda^{2}+9$ $\displaystyle =$ $\displaystyle 0\ \ \ \ \ (14)$ $\displaystyle \lambda$ $\displaystyle =$ $\displaystyle \pm3,\,\,\pm1 \ \ \ \ \ (15)$

Thus the eigenvalues of ${S_{x}}$ are ${\pm3\hbar/2}$ and ${\pm\hbar/2}$ as expected.

# Harmonic oscillator: coherent states

Required math: calculus

Required physics: Schrödinger equation

References: Griffiths, David J. (2005), Introduction to Quantum Mechanics, 2nd Edition; Pearson Education – Problem 3.35.

The uncertainty principle for the nth stationary state of the harmonic oscillator satisfies the condition

$\displaystyle \sigma_{p}\sigma_{x}=\hbar\left(n+\frac{1}{2}\right) \ \ \ \ \ (1)$

Thus only the ground state (where ${n=0}$) satisfies the uncertainty limit of ${\sigma_{p}\sigma_{x}=\hbar/2}$. However, some linear combinations of the stationary states do satisfy this limit. These are called coherent states and are eigenstates of the lowering operator ${a_{-}}$. That is, they can be written as

$\displaystyle a_{-}|\alpha\rangle=\alpha|\alpha\rangle \ \ \ \ \ (2)$

where ${\alpha}$ is a complex eigenvalue and the eigenstate ${|\alpha\rangle}$ is a linear combination of the harmonic oscillator stationary states (well, actually, since these states form a complete set, the last condition is always true, but never mind).

(a) We can calculate the means for position and momentum for these coherent states. We review the equations involving the raising and lowering operators first:

 $\displaystyle x$ $\displaystyle =$ $\displaystyle \sqrt{\frac{\hbar}{2m\omega}}(a_{+}+a_{-})\ \ \ \ \ (3)$ $\displaystyle p$ $\displaystyle =$ $\displaystyle i\sqrt{\frac{\hbar m\omega}{2}}(a_{+}-a_{-})\ \ \ \ \ (4)$ $\displaystyle \left(a_{+}\right)^{\dagger}$ $\displaystyle =$ $\displaystyle a_{-} \ \ \ \ \ (5)$

We have

 $\displaystyle \langle x\rangle$ $\displaystyle =$ $\displaystyle \langle\alpha|x|\alpha\rangle\ \ \ \ \ (6)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \sqrt{\frac{\hbar}{2m\omega}}(\langle\alpha|a_{+}|\alpha\rangle+\langle\alpha|a_{-}|\alpha\rangle)\ \ \ \ \ (7)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \sqrt{\frac{\hbar}{2m\omega}}(\left\langle a_{-}\alpha\left|\alpha\right.\right\rangle +\langle\alpha|a_{-}|\alpha\rangle)\ \ \ \ \ (8)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \sqrt{\frac{\hbar}{2m\omega}}(\alpha^*+\alpha)\ \ \ \ \ (9)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \sqrt{\frac{2\hbar}{m\omega}}\Re(\alpha) \ \ \ \ \ (10)$

where ${\Re(\alpha)}$ is the real part of ${\alpha}$. In getting the third line, we used the hermitian conjugate property of ${a_{+}}$ above.

Doing a similar calculation we can find ${\langle x^{2}\rangle}$:

 $\displaystyle \langle x^{2}\rangle$ $\displaystyle =$ $\displaystyle \frac{\hbar}{2m\omega}\langle\alpha|a_{+}^{2}+a_{-}^{2}+a_{+}a_{-}+a_{-}a_{+}|\alpha\rangle\ \ \ \ \ (11)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{\hbar}{2m\omega}\langle\alpha|a_{+}^{2}+a_{-}^{2}+a_{+}a_{-}+(1+a_{+}a_{-})|\alpha\rangle\ \ \ \ \ (12)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{\hbar}{2m\omega}(\alpha{}^{*2}+\alpha^{2}+2\alpha\alpha^*+1)\ \ \ \ \ (13)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{\hbar}{m\omega}\left(2\Re(\alpha)^{2}+\frac{1}{2}\right) \ \ \ \ \ (14)$

where in the second line we have used the commutator ${[a_{-},a_{+}]=1}$.

The calculations for ${\langle p\rangle}$ and ${\langle p^{2}\rangle}$ are similar and the results are

 $\displaystyle \langle p\rangle$ $\displaystyle =$ $\displaystyle -\sqrt{2\hbar m\omega}\Im(\alpha)\ \ \ \ \ (15)$ $\displaystyle \langle p^{2}\rangle$ $\displaystyle =$ $\displaystyle \hbar m\omega\left(2\Im(\alpha)^{2}+\frac{1}{2}\right) \ \ \ \ \ (16)$

where ${\Im(\alpha)}$ is the imaginary part of ${\alpha}$.

(b) The standard deviations are

 $\displaystyle \sigma_{x}$ $\displaystyle =$ $\displaystyle \sqrt{\langle x^{2}\rangle-\langle x\rangle^{2}}\ \ \ \ \ (17)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \sqrt{\frac{\hbar}{2m\omega}}\ \ \ \ \ (18)$ $\displaystyle \sigma_{p}$ $\displaystyle =$ $\displaystyle \sqrt{\langle p^{2}\rangle-\langle p\rangle^{2}}\ \ \ \ \ (19)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \sqrt{\frac{\hbar m\omega}{2}} \ \ \ \ \ (20)$

Therefore, the uncertainty principle here is ${\sigma_{x}\sigma_{p}=\hbar/2}$ as required.

(c) The expansion of ${|\alpha\rangle}$ in terms of the stationary states is ${|\alpha\rangle=\sum_{n=0}^{\infty}c_{n}|n\rangle}$, so we have, using the property of the lowering operator ${a_{-}|n\rangle=\sqrt{n}|n-1\rangle}$:

 $\displaystyle a_{-}|\alpha\rangle$ $\displaystyle =$ $\displaystyle \sum_{n=1}^{\infty}c_{n}\sqrt{n}|n-1\rangle \ \ \ \ \ (21)$

Note that the sum now starts at ${n=1}$, since ${a_{-}|0\rangle=0}$. Also, since ${a_{-}|\alpha\rangle=\alpha|\alpha\rangle}$:

 $\displaystyle a_{-}|\alpha\rangle$ $\displaystyle =$ $\displaystyle \sum_{n=0}^{\infty}c_{n}\alpha|n\rangle\ \ \ \ \ (22)$ $\displaystyle \sum_{n=1}^{\infty}c_{n}\sqrt{n}|n-1\rangle$ $\displaystyle =$ $\displaystyle \sum_{n=0}^{\infty}c_{n}\alpha|n\rangle \ \ \ \ \ (23)$

Since the energy eigenstates form an orthonormal set, we can equate coefficients of each eigenstate. We start with the first term in each series. We have

$\displaystyle c_{1}=\alpha c_{0} \ \ \ \ \ (24)$

The next term gives

 $\displaystyle c_{2}$ $\displaystyle =$ $\displaystyle \frac{\alpha}{\sqrt{2}}c_{1}\ \ \ \ \ (25)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{\alpha^{2}}{\sqrt{2}}c_{0} \ \ \ \ \ (26)$

The next term is

$\displaystyle c_{3}=\frac{\alpha^{3}}{\sqrt{3\times2}}c_{0}$

At this point, we can guess the general pattern, which we can propose to be

$\displaystyle c_{n}=\frac{\alpha^{n}}{\sqrt{n!}}c_{0} \ \ \ \ \ (27)$

We can prove the formula in the question using mathematical induction. First, we establish the anchor step by equating coefficients of ${|0\rangle}$:

$\displaystyle c_{1}=\alpha c_{0}=\frac{\alpha^{1}}{\sqrt{1!}}c_{0} \ \ \ \ \ (28)$

Now we prove that if ${c_{n}=(\alpha^{n}/\sqrt{n!})c_{0}}$ then ${c_{n+1}=(\alpha^{n+1}/\sqrt{(n+1)!})c_{0}}$. By equating the cofficients of ${|n\rangle}$ we get

 $\displaystyle \sqrt{n+1}c_{n+1}$ $\displaystyle =$ $\displaystyle \alpha c_{n}\ \ \ \ \ (29)$ $\displaystyle c_{n+1}$ $\displaystyle =$ $\displaystyle \frac{\alpha}{\sqrt{n+1}}\frac{\alpha^{n}}{\sqrt{n!}}c_{0}\ \ \ \ \ (30)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{\alpha^{n+1}}{\sqrt{(n+1)!}}c_{0} \ \ \ \ \ (31)$

(d) We can find ${c_{0}}$ by normalizing ${|\alpha\rangle}$. Writing it out in terms of the series:

 $\displaystyle \langle\alpha|\alpha\rangle$ $\displaystyle =$ $\displaystyle \sum_{n=0}^{\infty}c_{n}^*c_{n}\langle n|n\rangle\ \ \ \ \ (32)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \sum_{n=0}^{\infty}|c_{n}|^{2}\ \ \ \ \ (33)$ $\displaystyle$ $\displaystyle =$ $\displaystyle |c_{0}|^{2}\sum_{n=0}^{\infty}\frac{|\alpha|^{2n}}{n!}\ \ \ \ \ (34)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 1 \ \ \ \ \ (35)$

In the first line, we have used the orthogonality of the eigenstates to eliminate all terms containing ${\langle n|m\rangle}$ where ${n\neq m}$. In the second line we use the fact that all the energy eigenstates are normalized.

The series is the expansion of ${e^{|\alpha|^{2}}}$ so assuming ${c_{0}}$ is real, we get

$\displaystyle c_{0}=e^{-|\alpha|^{2}/2} \ \ \ \ \ (36)$

(e) Adding in the time dependence, we get

 $\displaystyle a_{-}\sum_{n=0}^{\infty}c_{n}e^{-iE_{n}t/\hbar}|n\rangle$ $\displaystyle =$ $\displaystyle \sum_{n=1}^{\infty}c_{n}\sqrt{n}e^{-iE_{n}t/\hbar}|n-1\rangle\ \ \ \ \ (37)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \sum_{n=1}^{\infty}c_{n}\sqrt{n}e^{-iE_{n-1}t/\hbar}e^{-i\omega t}|n-1\rangle \ \ \ \ \ (38)$

In the second line we have used the fact that for the harmonic oscillator, ${E_{n}=E_{n-1}+\hbar\omega}$. This expression can be made equal to ${\alpha(t)\sum_{n=0}^{\infty}c_{n}|n\rangle}$ if we define ${\alpha(t)\equiv e^{-i\omega t}\alpha}$. Thus the time-dependent state is still an eigenstate of ${a_{-}}$, but now the eigenvalue is time-dependent.

(f) The harmonic oscillator ground state is a coherent state with eigenvalue ${\alpha=0}$, since the lowering operator produces zero when applied to the ground state. We already know that the uncertainty principle is satisfied exactly for the ground state.

# Harmonic oscillator: matrix elements

Required math: calculus, matrix algebra

Required physics: Schrödinger equation

References: Griffiths, David J. (2005), Introduction to Quantum Mechanics, 2nd Edition; Pearson Education – Problem 3.33.

Shankar, R. (1994), Principles of Quantum Mechanics, Plenum Press. Section 7.4, Exercise 7.4.1.

In analyzing the harmonic oscillator, we used the raising and lowering operators to calculate ${\left\langle x\right\rangle }$ and ${\left\langle p\right\rangle }$, finding that they are both zero for all stationary states. These quantities are really the diagonal elements of the matrices ${\mathsf{X}}$ and ${\mathsf{P}}$. That is

 $\displaystyle \left\langle x\right\rangle _{nn}$ $\displaystyle =$ $\displaystyle \langle n|x|n\rangle\ \ \ \ \ (1)$ $\displaystyle$ $\displaystyle =$ $\displaystyle X_{nn} \ \ \ \ \ (2)$

We can use the same technique to calculate the off-diagonal elements.

We review the equations involving the raising and lowering operators first:

 $\displaystyle x$ $\displaystyle =$ $\displaystyle \sqrt{\frac{\hbar}{2m\omega}}(a_{+}+a_{-})\ \ \ \ \ (3)$ $\displaystyle p$ $\displaystyle =$ $\displaystyle i\sqrt{\frac{\hbar m\omega}{2}}(a_{+}-a_{-})\ \ \ \ \ (4)$ $\displaystyle a_{+}\psi_{n}$ $\displaystyle =$ $\displaystyle \sqrt{n+1}\psi_{n+1}\ \ \ \ \ (5)$ $\displaystyle a_{-}\psi_{n}$ $\displaystyle =$ $\displaystyle \sqrt{n}\psi_{n-1} \ \ \ \ \ (6)$

The general matrix elements for the operator ${x}$ can then be calculated:

 $\displaystyle \langle n|x|n'\rangle$ $\displaystyle =$ $\displaystyle \sqrt{\frac{\hbar}{2m\omega}}(\sqrt{n'+1}\langle n|n'+1\rangle+\sqrt{n'}\langle n|n'-1\rangle)\ \ \ \ \ (7)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \sqrt{\frac{\hbar}{2m\omega}}(\sqrt{n'+1}\delta_{n,n'+1}+\sqrt{n'}\delta_{n,n'-1}) \ \ \ \ \ (8)$

By similar reasoning we get the matrix elements for ${p}$:

$\displaystyle \langle n|p|n'\rangle=i\sqrt{\frac{\hbar m\omega}{2}}(\sqrt{n'+1}\delta_{n,n'+1}-\sqrt{n'}\delta_{n,n'-1}) \ \ \ \ \ (9)$

These results agree with those found by doing the integrals involving Hermite polynomials.

We now have all the matrix elements of ${\mathsf{X}}$ and ${\mathsf{P}}$ so it would be interesting to calculate the full hamiltonian matrix, which is

$\displaystyle \textsf{H}=\frac{1}{2m}\textsf{P}^{2}+\frac{m\omega^{2}}{2}\textsf{X}^{2} \ \ \ \ \ (10)$

In order to calculate the squares of the two matrices, we observe that both ${\textsf{X }}$ and ${\textsf{P}}$ are tridiagonal matrices with the added condition that their main diagonals are all zero. That is, the two diagonals above and below the main diagonal are the only places with non-zero elements. The square of such a matrix will have non-zero elements only on the main diagonal, and on the diagonals two above and below the main diagonal (you can verify this by drawing out such a matrix and seeing where the non-zero elements lie, or by doing tedious calculations with indices).

We can demonstrate how these elements can be calculated by considering the diagonal elements of ${\textsf{X}^{2}}$.

 $\displaystyle \textsf{X}_{nn}^{2}$ $\displaystyle =$ $\displaystyle \sum_{n'}\langle n|x|n'\rangle\langle n'|x|n\rangle\ \ \ \ \ (11)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{\hbar}{2m\omega}\sum_{n'}[\sqrt{n'+1}\delta_{n,n'+1}+\sqrt{n'}\delta_{n,n'-1}][\sqrt{n+1}\delta_{n',n+1}+\sqrt{n}\delta_{n',n-1}]\ \ \ \ \ (12)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{\hbar}{2m\omega}(2n+1) \ \ \ \ \ (13)$

The last line is obtained by noting that all the terms in the sum contain the product of two Kronecker deltas, so only in those cases where both deltas are non-zero is there a non-zero contribution to the sum. This happens only in the terms involving the product of the first and fourth terms (where ${n'=n-1}$) and the second and third terms (where ${n'=n+1}$).

By a similar argument, we get

$\displaystyle \textsf{P}_{nn}^{2}=\frac{\hbar m\omega}{2}(2n+1) \ \ \ \ \ (14)$

Therefore the diagonal elements of ${(1/2m)\textsf{P}^{2}+(m\omega^{2}/2)\textsf{X}^{2}}$ are

$\displaystyle \textsf{H}_{nn}=\hbar\omega\left(n+\frac{1}{2}\right) \ \ \ \ \ (15)$

which is what you would expect, as these are the energy levels of the harmonic oscillator.

It remains only to show that the off-diagonal elements of ${\textsf{H}}$ are zero.

 $\displaystyle \textsf{X}_{nm}^{2}$ $\displaystyle =$ $\displaystyle \sum_{n'}\langle n|x|n'\rangle\langle n'|x|m\rangle\ \ \ \ \ (16)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{\hbar}{2m\omega}\sum_{n'}[\sqrt{n'+1}\delta_{n,n'+1}+\sqrt{n'}\delta_{n,n'-1}][\sqrt{m+1}\delta_{n',m+1}+\sqrt{m}\delta_{n',m-1}] \ \ \ \ \ (17)$

To see which non-zero elements exist on row n, we note that for a given value of n, we must have either ${n'=n-1}$ or ${n'=n+1}$ in order for one of the deltas in the first term to be non-zero. If ${n'=n-1}$, then in the second term, we must have either ${n-1=m+1}$ or ${n-1=m-1}$. The second case results in a diagonal element which we have already considered, so we need consider only the case ${m=n-2}$. In this case, the matrix element is

$\displaystyle \textsf{X}_{n,n-2}^{2}=\frac{\hbar}{2m\omega}\sqrt{n(n-1)} \ \ \ \ \ (18)$

Similarly, if ${n'=n+1}$, the non-diagonal term is ${n+1=m-1}$ or ${m=n+2}$, and we get

$\displaystyle \textsf{X}_{n,n+2}^{2}=\frac{\hbar}{2m\omega}\sqrt{(n+1)(n+2)} \ \ \ \ \ (19)$

Similar reasoning gives us the elements from ${\textsf{P}^{2}}$:

 $\displaystyle \textsf{P}_{n,n-2}^{2}$ $\displaystyle =$ $\displaystyle -\frac{\hbar m\omega}{2}\sqrt{n(n-1)}\ \ \ \ \ (20)$ $\displaystyle \textsf{P}_{n,n+2}^{2}$ $\displaystyle =$ $\displaystyle -\frac{\hbar m\omega}{2}\sqrt{(n+1)(n+2)} \ \ \ \ \ (21)$

Combining these two results, we see that the non-diagonal elements of ${(1/2m)\textsf{P}^{2}+(m\omega^{2}/2)\textsf{X}^{2}}$ are all zero.