# Translational invariance and conservation of momentum

Shankar, R. (1994), Principles of Quantum Mechanics, Plenum Press. Chapter 11.

One consequence of the invariance of the Hamiltonian under translation is that the momentum and Hamiltonian commute:

$\displaystyle \left[P,H\right]=0 \ \ \ \ \ (1)$

In quantum mechanics, commuting quantities are simultaneously observable, and we can find a basis for the Hilbert space consisting of eigenstates of both ${P}$ and ${H}$. We’ve seen that Ehrenfest’s theorem allows us to conclude that for such a system, the average momentum is conserved so that ${\left\langle \dot{P}\right\rangle =0}$. We can go a step further and state that if a system starts out in an eigenstate of ${P}$, then it remains in that eigenstate for all time.

First, we need to make a rather subtle observation, which is that

$\displaystyle \left[P,H\right]=0\rightarrow\left[P,U\left(t\right)\right]=0 \ \ \ \ \ (2)$

That is, if ${P}$ and ${H}$ commute, then ${P}$ also commutes with the propagator ${U\left(t\right)}$. For a time-independent Hamiltonian, the propagator is

$\displaystyle U\left(t\right)=e^{-iHt/\hbar} \ \ \ \ \ (3)$

Since this can be expanded in a power series in the Hamiltonian, condition 2 follows easily enough. What if the Hamiltonian is time-dependent? In this case, the propagator comes out to a time-ordered integral

$\displaystyle U\left(t\right)=T\left\{ \exp\left[-\frac{i}{\hbar}\int_{0}^{t}H\left(t^{\prime}\right)dt^{\prime}\right]\right\} \equiv\lim_{N\rightarrow\infty}\prod_{n=0}^{N-1}e^{-i\Delta H\left(n\Delta\right)/\hbar} \ \ \ \ \ (4)$

Here the time interval ${\left[0,t\right]}$ is divided into ${N}$ time slices, each of length ${\Delta=t/N}$. As explained in the earlier post, the reason we can’t just integrate the RHS directly by summing the exponents is that such a procedure works only if the operators in the exponents all commute with each other. If ${H}$ is time-dependent, its forms at different times may not commute, so we can’t get a simple closed form for ${U\left(t\right)}$.

However, if ${\left[P,H\left(t\right)\right]=0}$ for all times, then ${P}$ commutes with all the exponents on the RHS of 4, so we still get ${\left[P,U\left(t\right)\right]=0}$. Another way of looking at this is by imposing the condition ${\left[P,H\left(t\right)\right]=0}$ we’re saying that if ${H\left(t\right)}$ can be expanded in a power series in ${X}$ and ${P}$, it depends only on ${P}$, and not on ${X}$. This follows from the fact that

$\displaystyle \left[X^{n},P\right]=i\hbar nX^{n-1} \ \ \ \ \ (5)$

so that ${P}$ does not commute with any power of ${X}$.

Given that 2 is valid for all Hamiltonians, then if we start in a eigenstate ${\left|p\right\rangle }$ of ${P}$, then

 $\displaystyle P\left|p\right\rangle$ $\displaystyle =$ $\displaystyle p\left|p\right\rangle \ \ \ \ \ (6)$ $\displaystyle PU\left(t\right)\left|p\right\rangle$ $\displaystyle =$ $\displaystyle U\left(t\right)P\left|p\right\rangle \ \ \ \ \ (7)$ $\displaystyle$ $\displaystyle =$ $\displaystyle U\left(t\right)p\left|p\right\rangle \ \ \ \ \ (8)$ $\displaystyle$ $\displaystyle =$ $\displaystyle pU\left(t\right)\left|p\right\rangle \ \ \ \ \ (9)$

Thus ${U\left(t\right)\left|p\right\rangle }$ remains an eigenstate of ${P}$ with the same eigenvalue ${p}$ for all time. For a single particle moving in one dimension, the state ${\left|p\right\rangle }$ describes a free particle with momentum ${p}$ (and thus a completely undetermined position).

# Correspondence between classical and quantum transformations

Shankar, R. (1994), Principles of Quantum Mechanics, Plenum Press. Chapter 11.

When we consider infinitesimal transformations of some dynamical variable, there is a correspondence between classical and quantum mechanics which we can see as follows. First, we’ll summarize the results from classical mechanics. We can define a canonical transformation generated by a variable ${g}$ as

 $\displaystyle \bar{q}_{i}$ $\displaystyle =$ $\displaystyle q_{i}+\varepsilon\frac{\partial g}{\partial p_{i}}\equiv q_{i}+\delta q_{i}\ \ \ \ \ (1)$ $\displaystyle \bar{p}_{i}$ $\displaystyle =$ $\displaystyle p_{i}-\varepsilon\frac{\partial g}{\partial q_{i}}\equiv p_{i}+\delta p_{i} \ \ \ \ \ (2)$

Here, ${\varepsilon}$ is an infintesimal amount and ${\delta q_{i}}$ and ${\delta p_{i}}$ are the infinitesimal amounts by which the coordinates and momenta vary. It follows from these definitions that, for any dynamical variable ${\omega}$, its variation ${\delta\omega}$ is given by a Poisson bracket

$\displaystyle \delta\omega=\omega\left(\bar{q}_{i},\bar{p}_{i}\right)-\omega\left(q_{i},p_{i}\right)=\varepsilon\left\{ \omega,g\right\} \ \ \ \ \ (3)$

For the special cases of coordinates and momenta, this is

 $\displaystyle \delta q_{i}$ $\displaystyle =$ $\displaystyle \varepsilon\left\{ q_{i},g\right\} \ \ \ \ \ (4)$ $\displaystyle \delta p_{i}$ $\displaystyle =$ $\displaystyle \varepsilon\left\{ p_{i},g\right\} \ \ \ \ \ (5)$

If the generator is the momentum ${p_{j}}$, then

 $\displaystyle \delta q_{i}$ $\displaystyle =$ $\displaystyle \varepsilon\left\{ q_{i},p_{j}\right\} =\varepsilon\delta_{ij}\ \ \ \ \ (6)$ $\displaystyle \delta p_{i}$ $\displaystyle =$ $\displaystyle \varepsilon\left\{ p_{i},p_{j}\right\} =0 \ \ \ \ \ (7)$

Thus, in classical mechanics, ${p_{j}}$ is the generator of translations in direction ${j}$.

If ${\omega=H}$ (the Hamiltonian) and if ${\left\{ H,g\right\} =0}$, then ${g}$ is conserved (it doesn’t vary with time). Because the transformation 1 and 2 is canonical, it preserves the Poisson brackets so that

 $\displaystyle \left\{ \overline{q}_{i},\overline{q}_{j}\right\}$ $\displaystyle =$ $\displaystyle \left\{ \overline{p}_{i},\overline{p}_{j}\right\} =0\ \ \ \ \ (8)$ $\displaystyle \left\{ \overline{q}_{i},\overline{p}_{j}\right\}$ $\displaystyle =$ $\displaystyle \delta_{ij} \ \ \ \ \ (9)$

What do these things correspond to in quantum mechanics? [I find Shankar’s treatment in section 11.2 to be almost tautological, since it merely repeats the derivation given earlier. I’ll try to be a bit more general.]

Suppose we have some infinitesimal transformation given by a unitary operator ${U\left(\varepsilon\right)}$. We can then define the changes in ${X}$ and ${P}$ by

 $\displaystyle \delta X$ $\displaystyle =$ $\displaystyle U^{\dagger}\left(\varepsilon\right)XU\left(\varepsilon\right)-X\ \ \ \ \ (10)$ $\displaystyle \delta P$ $\displaystyle =$ $\displaystyle U^{\dagger}\left(\varepsilon\right)PU\left(\varepsilon\right)-P \ \ \ \ \ (11)$

Since ${U\left(\varepsilon\right)}$ describes an infinitesimal transformation, we can expand it to first order in ${\varepsilon}$:

$\displaystyle U\left(\varepsilon\right)=I-\frac{i\varepsilon}{\hbar}G \ \ \ \ \ (12)$

where ${G=G^{\dagger}}$ is some Hermitian operator known as the generator of the transformation. (We’ve seen a proof that the translation operator ${T\left(\varepsilon\right)}$ (a special case of ${U\left(\varepsilon\right)}$) is unitary and that its generator is Hermitian earlier, and the current case follows the same reasoning.) Using this form we have from 10 and 11, to order ${\varepsilon}$:

 $\displaystyle \delta X$ $\displaystyle =$ $\displaystyle \left(I+\frac{i\varepsilon}{\hbar}G\right)X\left(I-\frac{i\varepsilon}{\hbar}G\right)-X\ \ \ \ \ (13)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -\frac{i\varepsilon}{\hbar}\left[X,G\right]\ \ \ \ \ (14)$ $\displaystyle \delta P$ $\displaystyle =$ $\displaystyle \left(I+\frac{i\varepsilon}{\hbar}G\right)P\left(I-\frac{i\varepsilon}{\hbar}G\right)-P\ \ \ \ \ (15)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -\frac{i\varepsilon}{\hbar}\left[P,G\right] \ \ \ \ \ (16)$

If ${G=P}$, then

 $\displaystyle \delta X$ $\displaystyle =$ $\displaystyle -\frac{i\varepsilon}{\hbar}\left[X,P\right]=\varepsilon I\ \ \ \ \ (17)$ $\displaystyle \delta P$ $\displaystyle =$ $\displaystyle -\frac{i\varepsilon}{\hbar}\left[P,P\right]=0 \ \ \ \ \ (18)$

Comparing this with 6 and 7 we see that (in one dimension, where the classical coordinate is given by ${x}$ and momentum by ${p}$) there is a correspondence between the classical Poisson bracket and quantum commutator:

$\displaystyle \left\{ x,p\right\} \leftrightarrow-\frac{i}{\hbar}\left[X,P\right] \ \ \ \ \ (19)$

The momentum operator ${P}$ in quantum mechanics is thus the generator of translations, just as ${p}$ generates translations in classical mechanics.

More generally, we can define the variation in some arbitrary dynamical operator ${\Omega}$ in a similar way, using 12 to expand the RHS:

 $\displaystyle \delta\Omega$ $\displaystyle =$ $\displaystyle U^{\dagger}\left(\varepsilon\right)\Omega U\left(\varepsilon\right)-\Omega\ \ \ \ \ (20)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -\frac{i\varepsilon}{\hbar}\left[\Omega,G\right] \ \ \ \ \ (21)$

The correspondence with classical mechanics is then

$\displaystyle \left\{ \omega,g\right\} \leftrightarrow-\frac{i}{\hbar}\left[\Omega,G\right] \ \ \ \ \ (22)$

The general rule is that a quantum commutator is ${i\hbar}$ times the corresponding classical Poisson bracket.

If ${\Omega=H}$ and ${\left[H,G\right]=0}$, then by Ehrenfest’s theorem, ${\left\langle \dot{G}\right\rangle =0}$ and the average value of ${G}$ is conserved.

The correspondence is a bit odd in that the generator ${g}$ in classical mechanics enters as a derivative in 1 and 2 while the generator ${G}$ in quantum mechanics enters as an operator (no derivatives) in 12.

One other feature is worth noting. A canonical transformation preserves the Poisson brackets 8 in the new coordinate system. In quantum mechanics, it is the commutators that get preserved. For example, using the fact that ${U}$ is unitary so that ${UU^{\dagger}=I}$:

 $\displaystyle U^{\dagger}\left[X,P\right]U$ $\displaystyle =$ $\displaystyle U^{\dagger}XPU-U^{\dagger}PXU\ \ \ \ \ (23)$ $\displaystyle$ $\displaystyle =$ $\displaystyle U^{\dagger}XUU^{\dagger}PU-U^{\dagger}PUU^{\dagger}XU\ \ \ \ \ (24)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \left[U^{\dagger}XU,U^{\dagger}PU\right] \ \ \ \ \ (25)$

# Translation operator from passive transformations

Shankar, R. (1994), Principles of Quantum Mechanics, Plenum Press. Chapter 11.

We’ve seen that the translation operator ${T\left(\varepsilon\right)}$ in quantum mechanics can be derived by considering the translation to be an active transformation, that is, a transformation where the state vectors, rather than the operators, get transformed according to

$\displaystyle T\left(\varepsilon\right)\left|\psi\right\rangle =\left|\psi_{\varepsilon}\right\rangle \ \ \ \ \ (1)$

Using this approach, we found that

$\displaystyle T\left(\varepsilon\right)=I-\frac{i\varepsilon}{\hbar}P \ \ \ \ \ (2)$

so that the momentum ${P}$ is the generator of the transformation.

We can also derive ${T}$ using a passive transformation, where the state vectors remain the same but the operators are transformed according to

 $\displaystyle T^{\dagger}\left(\varepsilon\right)XT\left(\varepsilon\right)$ $\displaystyle =$ $\displaystyle X+\varepsilon I\ \ \ \ \ (3)$ $\displaystyle T^{\dagger}\left(\varepsilon\right)PT\left(\varepsilon\right)$ $\displaystyle =$ $\displaystyle P \ \ \ \ \ (4)$

This is equivalent to an active transformation since

 $\displaystyle \left\langle \psi\left|T^{\dagger}\left(\varepsilon\right)XT\left(\varepsilon\right)\right|\psi\right\rangle$ $\displaystyle =$ $\displaystyle \left\langle T\left(\varepsilon\right)\psi\left|X\right|T\left(\varepsilon\right)\psi\right\rangle \ \ \ \ \ (5)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \left\langle \psi_{\varepsilon}\left|X\right|\psi_{\varepsilon}\right\rangle \ \ \ \ \ (6)$ $\displaystyle$ $\displaystyle =$ $\displaystyle x+\varepsilon \ \ \ \ \ (7)$

As before we start by taking

$\displaystyle T\left(\varepsilon\right)=I-\frac{i\varepsilon}{\hbar}G \ \ \ \ \ (8)$

where ${G}$ is some Hermitian operator, so that ${G^{\dagger}=G}$. Plugging this into 3 we get, keeping only terms up to order ${\varepsilon}$:

 $\displaystyle T^{\dagger}\left(\varepsilon\right)XT\left(\varepsilon\right)$ $\displaystyle =$ $\displaystyle \left(I+\frac{i\varepsilon}{\hbar}G\right)X\left(I-\frac{i\varepsilon}{\hbar}G\right)\ \ \ \ \ (9)$ $\displaystyle$ $\displaystyle =$ $\displaystyle X+\frac{i\varepsilon}{\hbar}I\left(GX-XG\right)\ \ \ \ \ (10)$ $\displaystyle$ $\displaystyle =$ $\displaystyle X-\frac{i\varepsilon}{\hbar}\left[X,G\right]\ \ \ \ \ (11)$ $\displaystyle$ $\displaystyle =$ $\displaystyle X+\varepsilon I \ \ \ \ \ (12)$

Therefore

 $\displaystyle -\frac{i\varepsilon}{\hbar}\left[X,G\right]$ $\displaystyle =$ $\displaystyle \varepsilon I\ \ \ \ \ (13)$ $\displaystyle \left[X,G\right]$ $\displaystyle =$ $\displaystyle i\hbar I \ \ \ \ \ (14)$

Since ${\left[X,P\right]=i\hbar}$ we see that

$\displaystyle G=P+f\left(X\right) \ \ \ \ \ (15)$

The extra ${f\left(X\right)}$ is there because any function of ${X}$ alone commutes with ${X}$, so

$\displaystyle \left[X,G\right]=\left[X,P\right]+\left[X,f\left(X\right)\right]=i\hbar I+0 \ \ \ \ \ (16)$

We can eliminate ${f\left(X\right)}$ by considering 4.

 $\displaystyle T^{\dagger}\left(\varepsilon\right)PT\left(\varepsilon\right)$ $\displaystyle =$ $\displaystyle \left(I+\frac{i\varepsilon}{\hbar}G\right)P\left(I-\frac{i\varepsilon}{\hbar}G\right)\ \ \ \ \ (17)$ $\displaystyle$ $\displaystyle =$ $\displaystyle P+\frac{i\varepsilon}{\hbar}I\left(GP-PG\right)\ \ \ \ \ (18)$ $\displaystyle$ $\displaystyle =$ $\displaystyle P-\frac{i\varepsilon}{\hbar}\left[P,G\right]\ \ \ \ \ (19)$ $\displaystyle$ $\displaystyle =$ $\displaystyle P \ \ \ \ \ (20)$

Thus we must have ${\left[P,G\right]=0}$, which means that ${G}$ must be a function of ${P}$ alone. This means that the most general form for ${f\left(X\right)}$ is ${f\left(X\right)=\mbox{constant}}$, but there’s nothing to be gained by adding some non-zero constant to ${G}$, so we can take ${f\left(X\right)=0}$. Thus we end up with the same form 2 that we got from the active transformation.

Translational invariance is the condition that the Hamiltonian is unaltered by a translation. In the passive representation this is stated by the condition

$\displaystyle T^{\dagger}\left(\varepsilon\right)HT\left(\varepsilon\right)=H \ \ \ \ \ (21)$

Since translation is unitary, we can apply a theorem that is valid for any operator ${\Omega}$ which can be expanded in powers of ${X}$ and ${P}$. For any unitary operator ${U}$, we have

$\displaystyle U^{\dagger}\Omega\left(X,P\right)U=\Omega\left(U^{\dagger}XU,U^{\dagger}PU\right) \ \ \ \ \ (22)$

This follows because for a unitary operator ${U^{\dagger}U=UU^{\dagger}=I}$ so we can insert the product ${UU^{\dagger}}$ anywhere we like. In particular, we can insert it between each pair of factors in every term of the power series expansion of ${\Omega}$, for example

 $\displaystyle U^{\dagger}X^{2}P^{2}U$ $\displaystyle =$ $\displaystyle U^{\dagger}XXPPU\ \ \ \ \ (23)$ $\displaystyle$ $\displaystyle =$ $\displaystyle U^{\dagger}XUU^{\dagger}XUU^{\dagger}PUU^{\dagger}PU\ \ \ \ \ (24)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \left(U^{\dagger}XU\right)^{2}\left(U^{\dagger}PU\right)^{2} \ \ \ \ \ (25)$

For 21 this means that

$\displaystyle T^{\dagger}\left(\varepsilon\right)H\left(X,P\right)T\left(\varepsilon\right)=H\left(X+\varepsilon I,P\right)=H\left(X,P\right) \ \ \ \ \ (26)$

As before, this leads to the condition

$\displaystyle \left[P,H\right]=0 \ \ \ \ \ (27)$

which means that ${P}$ is conserved, according to Ehrenfest’s theorem.

# M65 & M66 in Leo (25/3/2017)

M65 (upper right) and M66 (lower left) are two galaxies in Leo. They are about 35 million light years away. (Click to enlarge.)

Photo location: Monifieth (near Dundee), Scotland, UK.

Date: 25 March 2017; 2100 UCT.

Telescope: 11-inch Celestron SCT.

Camera: Pentax K3

Exposure: ISO 1600, 93 30-second exposures stacked with DeepSkyStacker.

# Translational invariance in quantum mechanics

Shankar, R. (1994), Principles of Quantum Mechanics, Plenum Press. Chapter 11, Exercises 11.2.1 – 11.2.2.

In classical mechanics, we’ve seen that if a dynamical variable ${g}$ is used to generate a transformation of the variables ${q_{i}}$ and ${p_{i}}$ (the coordinates and canonical momenta), then if the Hamiltonian is invariant under this transformation, the quantity ${g}$ is conserved, meaning that it remains constant over time. We’d like to extend these results to quantum mechanics, but in doing so, there is one large obstacle. In classical mechanics, we can specify the exact position (given by ${q_{i}}$) and the exact momentum (${p_{i}}$) at every instant in time for every particle. In other words, every particle has a precisely defined trajectory through phase space. Due to the uncertainty principle, we cannot do this in quantum mechanics, since we cannot specify the position and momentum of any particle with arbitrary precision, so we can’t define a precise trajectory for any particle.

The way in which this problem is usually handled is to examine the effects of changes in the expectation values of dynamical variables, rather than with their precise values at any given time. In the case of a single particle moving in one dimension, we can apply this idea to investigate how we might invoke translational invariance. Classically, where ${x}$ is the position variable and ${p}$ is the momentum, an infinitesimal translation by a distance ${\varepsilon}$ is given by

 $\displaystyle x$ $\displaystyle \rightarrow$ $\displaystyle x+\varepsilon\ \ \ \ \ (1)$ $\displaystyle p$ $\displaystyle \rightarrow$ $\displaystyle p \ \ \ \ \ (2)$

In quantum mechanics, the equivalent translation is reflected in the expectation values:

 $\displaystyle \left\langle X\right\rangle$ $\displaystyle \rightarrow$ $\displaystyle \left\langle X\right\rangle +\varepsilon\ \ \ \ \ (3)$ $\displaystyle \left\langle P\right\rangle$ $\displaystyle \rightarrow$ $\displaystyle \left\langle P\right\rangle \ \ \ \ \ (4)$

In order to find the expectation values ${\left\langle X\right\rangle }$ and ${\left\langle P\right\rangle }$ we need to use the state vector ${\left|\psi\right\rangle }$. There are two ways of interpreting the transformation. The first, known as the active transformation picture, is to say that translating the position generates a new state vector ${\left|\psi_{\varepsilon}\right\rangle }$ with the properties

 $\displaystyle \left\langle \psi_{\varepsilon}\left|X\right|\psi_{\varepsilon}\right\rangle$ $\displaystyle =$ $\displaystyle \left\langle \psi\left|X\right|\psi\right\rangle +\varepsilon\ \ \ \ \ (5)$ $\displaystyle \left\langle \psi_{\varepsilon}\left|P\right|\psi_{\varepsilon}\right\rangle$ $\displaystyle =$ $\displaystyle \left\langle \psi\left|P\right|\psi\right\rangle \ \ \ \ \ (6)$

Since ${\left|\psi_{\varepsilon}\right\rangle }$ is another state vector in the same vector space as ${\left|\psi\right\rangle }$, there must be an operator ${T\left(\varepsilon\right)}$ which we call the translation operator, and which maps one vector onto the other:

$\displaystyle T\left(\varepsilon\right)\left|\psi\right\rangle =\left|\psi_{\varepsilon}\right\rangle \ \ \ \ \ (7)$

In terms of the translation operator, the translation becomes

 $\displaystyle \left\langle \psi\left|T^{\dagger}\left(\varepsilon\right)XT\left(\varepsilon\right)\right|\psi\right\rangle$ $\displaystyle =$ $\displaystyle \left\langle \psi\left|X\right|\psi\right\rangle +\varepsilon\ \ \ \ \ (8)$ $\displaystyle \left\langle \psi\left|T^{\dagger}\left(\varepsilon\right)PT\left(\varepsilon\right)\right|\psi\right\rangle$ $\displaystyle =$ $\displaystyle \left\langle \psi\left|P\right|\psi\right\rangle \ \ \ \ \ (9)$

These relations allow us to define the second interpretation, called the passive transformation picture, in which the state vectors do not change, but rather the position and momentum operators change. That is, we can transform the operators according to

 $\displaystyle X$ $\displaystyle \rightarrow$ $\displaystyle T^{\dagger}\left(\varepsilon\right)XT\left(\varepsilon\right)=X+\varepsilon I\ \ \ \ \ (10)$ $\displaystyle P$ $\displaystyle \rightarrow$ $\displaystyle T^{\dagger}\left(\varepsilon\right)PT\left(\varepsilon\right)=P \ \ \ \ \ (11)$

We need to find the explicit form for ${T}$. To begin, we consider its effect on a position eigenket ${\left|x\right\rangle }$. One possibility is

$\displaystyle T\left(\varepsilon\right)\left|x\right\rangle =\left|x+\varepsilon\right\rangle \ \ \ \ \ (12)$

However, to be completely general, we should consider the case where ${T}$ not only shifts ${x}$ by ${\varepsilon}$, but also introduces a phase factor. That is, the most general effect of ${T}$ is

$\displaystyle T\left(\varepsilon\right)\left|x\right\rangle =e^{i\varepsilon g\left(x\right)/\hbar}\left|x+\varepsilon\right\rangle \ \ \ \ \ (13)$

where ${g\left(x\right)}$ is some arbitrary real function of ${x}$. Using this form, we have, for some arbitrary state vector ${\left|\psi\right\rangle }$:

 $\displaystyle \left|\psi_{\varepsilon}\right\rangle$ $\displaystyle =$ $\displaystyle T\left(\varepsilon\right)\left|\psi\right\rangle \ \ \ \ \ (14)$ $\displaystyle$ $\displaystyle =$ $\displaystyle T\left(\varepsilon\right)\int_{-\infty}^{\infty}\left|x\right\rangle \left\langle x\left|\psi\right.\right\rangle dx\ \ \ \ \ (15)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \int_{-\infty}^{\infty}e^{i\varepsilon g\left(x\right)/\hbar}\left|x+\varepsilon\right\rangle \left\langle x\left|\psi\right.\right\rangle dx\ \ \ \ \ (16)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \int_{-\infty}^{\infty}e^{i\varepsilon g\left(x^{\prime}-\varepsilon\right)/\hbar}\left|x^{\prime}\right\rangle \left\langle x^{\prime}-\varepsilon\left|\psi\right.\right\rangle dx^{\prime} \ \ \ \ \ (17)$

To get the last line, we changed the integration variable to ${x^{\prime}=x+\varepsilon}$. Multiplying by the bra ${\left\langle x\right|}$ gives, using ${\left\langle x\left|x^{\prime}\right.\right\rangle =\delta\left(x-x^{\prime}\right)}$:

 $\displaystyle \left\langle x\left|T\left(\varepsilon\right)\right|\psi\right\rangle =\left\langle x\left|\psi_{\varepsilon}\right.\right\rangle$ $\displaystyle =$ $\displaystyle e^{i\varepsilon g\left(x-\varepsilon\right)/\hbar}\left\langle x-\varepsilon\left|\psi\right.\right\rangle \ \ \ \ \ (18)$ $\displaystyle$ $\displaystyle =$ $\displaystyle e^{i\varepsilon g\left(x-\varepsilon\right)/\hbar}\psi\left(x-\varepsilon\right) \ \ \ \ \ (19)$

That is, the action of ${T\left(\varepsilon\right)}$ is to move the coordinate axis a distance ${\varepsilon}$ to the right, which means that the new state vector ${\left|\psi_{\varepsilon}\right\rangle }$ becomes the old state vector at position ${x-\varepsilon}$. Alternatively, we can leave the coordinate axis alone and shift the wave function a distance ${\varepsilon}$ to the right, so that the new vector at position ${x}$ is the old vector at position ${x-\varepsilon}$ (multiplied by a phase factor).

We can now use this result to calculate 8 and 9:

 $\displaystyle \left\langle \psi_{\varepsilon}\left|X\right|\psi_{\varepsilon}\right\rangle$ $\displaystyle =$ $\displaystyle \int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\left\langle \psi_{\varepsilon}\left|x\right.\right\rangle \left\langle x\left|X\right|x^{\prime}\right\rangle \left\langle x^{\prime}\left|\psi_{\varepsilon}\right.\right\rangle dx\;dx^{\prime}\ \ \ \ \ (20)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\left\langle \psi_{\varepsilon}\left|x\right.\right\rangle x^{\prime}\delta\left(x-x^{\prime}\right)\left\langle x^{\prime}\left|\psi_{\varepsilon}\right.\right\rangle dx\;dx^{\prime}\ \ \ \ \ (21)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \int_{-\infty}^{\infty}\left\langle \psi_{\varepsilon}\left|x\right.\right\rangle x\left\langle x\left|\psi_{\varepsilon}\right.\right\rangle dx\ \ \ \ \ (22)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \int_{-\infty}^{\infty}e^{-i\varepsilon g\left(x-\varepsilon\right)/\hbar}\psi^*\left(x-\varepsilon\right)xe^{i\varepsilon g\left(x-\varepsilon\right)/\hbar}\psi\left(x-\varepsilon\right)dx\ \ \ \ \ (23)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \int_{-\infty}^{\infty}\psi^*\left(x-\varepsilon\right)x\psi\left(x-\varepsilon\right)dx\ \ \ \ \ (24)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \int_{-\infty}^{\infty}\psi^*\left(x^{\prime}\right)\left(x^{\prime}+\varepsilon\right)\psi\left(x^{\prime}\right)dx^{\prime}\ \ \ \ \ (25)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \left\langle \psi\left|X\right|\psi\right\rangle +\varepsilon \ \ \ \ \ (26)$

In the second line, we used the matrix element of ${X}$

$\displaystyle \left\langle x\left|X\right|x^{\prime}\right\rangle =x^{\prime}\delta\left(x-x^{\prime}\right) \ \ \ \ \ (27)$

and in the penultimate line, we again used the change of integration variable to ${x^{\prime}=x-\varepsilon}$. Thus we regain 8.

The momentum transforms as follows.

 $\displaystyle \left\langle \psi_{\varepsilon}\left|P\right|\psi_{\varepsilon}\right\rangle$ $\displaystyle =$ $\displaystyle \int_{-\infty}^{\infty}\psi^*\left(x-\varepsilon\right)e^{-i\varepsilon g\left(x-\varepsilon\right)/\hbar}\left(-i\hbar\frac{d}{dx}\right)\left(e^{i\varepsilon g\left(x-\varepsilon\right)/\hbar}\psi\left(x-\varepsilon\right)\right)dx\ \ \ \ \ (28)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \int_{-\infty}^{\infty}\psi^*\left(x-\varepsilon\right)\left(\varepsilon\frac{d}{dx}\left(g\left(x-\varepsilon\right)\right)\psi\left(x-\varepsilon\right)-i\hbar\frac{d}{dx}\left(\psi\left(x-\varepsilon\right)\right)\right)dx\ \ \ \ \ (29)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \int_{-\infty}^{\infty}\psi^*\left(x^{\prime}\right)\left(\varepsilon\psi\left(x^{\prime}\right)\frac{d}{dx^{\prime}}g\left(x^{\prime}\right)-i\hbar\frac{d}{dx^{\prime}}\psi\left(x^{\prime}\right)\right)dx^{\prime}\ \ \ \ \ (30)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \varepsilon\left\langle \frac{d}{dx}g\left(x\right)\right\rangle +\left\langle P\right\rangle \ \ \ \ \ (31)$

In the third line, we again transformed the integration variable to ${x^{\prime}=x-\varepsilon}$, and used the fact that ${dx=dx^{\prime}}$, so a derivative with respect to ${x}$ is the same as a derivative with respect to ${x^{\prime}}$. [This derivation is condensed a bit compared to the derivation of ${\left\langle \psi_{\varepsilon}\left|X\right|\psi_{\varepsilon}\right\rangle }$, but you can insert a couple of sets of complete states and do the extra integrals if you like.]

If we now impose the condition 9 so that the momentum is unchanged by the translation, this is equivalent to choosing the phase function ${g\left(x\right)=0}$, and this is what is done in most applications.

Having explored the properties of the translation operator, we can now define what we mean by translational invariance in quantum mechanics. This is the requirement that the expectation value of the Hamiltonian is unchanged under the transformation. That is

$\displaystyle \left\langle \psi\left|H\right|\psi\right\rangle =\left\langle \psi_{\varepsilon}\left|H\right|\psi_{\varepsilon}\right\rangle \ \ \ \ \ (32)$

For this, we need the explicit form of ${T\left(\varepsilon\right)}$. Since ${\varepsilon=0}$ corresponds to no translation, we require ${T\left(0\right)=I}$. To first order in ${\varepsilon}$, we can then write

$\displaystyle T\left(\varepsilon\right)=I-\frac{i\varepsilon}{\hbar}G \ \ \ \ \ (33)$

where ${G}$ is some operator, called the generator of translations, that is to be determined. From 13 (with ${g=0}$ from now on), we have

$\displaystyle \left\langle x^{\prime}+\varepsilon\left|x+\varepsilon\right.\right\rangle =\left\langle x^{\prime}\left|T^{\dagger}\left(\varepsilon\right)T\left(\varepsilon\right)\right|x\right\rangle =\delta\left(x^{\prime}-x\right)=\left\langle x^{\prime}\left|x\right.\right\rangle \ \ \ \ \ (34)$

so we must have

$\displaystyle T^{\dagger}\left(\varepsilon\right)T\left(\varepsilon\right)=I \ \ \ \ \ (35)$

so that ${T}$ is unitary. Applying this condition to 33 up to order ${\varepsilon}$, we have

 $\displaystyle T^{\dagger}\left(\varepsilon\right)T\left(\varepsilon\right)$ $\displaystyle =$ $\displaystyle \left(I+\frac{i\varepsilon}{\hbar}G^{\dagger}\right)\left(I-\frac{i\varepsilon}{\hbar}G\right)\ \ \ \ \ (36)$ $\displaystyle$ $\displaystyle =$ $\displaystyle I+\frac{i\varepsilon}{\hbar}\left(G^{\dagger}-G\right)+\mathcal{O}\left(\varepsilon^{2}\right) \ \ \ \ \ (37)$

Requiring 35 shows that ${G=G^{\dagger}}$ so ${G}$ is Hermitian. Now, from 19 (${g=0}$ again) we have

$\displaystyle \left\langle x\left|T\left(\varepsilon\right)\right|\psi\right\rangle =\psi\left(x-\varepsilon\right) \ \ \ \ \ (38)$

We expand both sides to order ${\varepsilon}$:

$\displaystyle \left\langle x\left|I\right|\psi\right\rangle -\frac{i\varepsilon}{\hbar}\left\langle x\left|G\right|\psi\right\rangle =\psi\left(x\right)-\varepsilon\frac{d\psi}{dx} \ \ \ \ \ (39)$

Since ${\left\langle x\left|I\right|\psi\right\rangle =\left\langle x\left|\psi\right.\right\rangle =\psi\left(x\right)}$, we have

$\displaystyle \left\langle x\left|G\right|\psi\right\rangle =-i\hbar\frac{d\psi}{dx}=\left\langle x\left|P\right|\psi\right\rangle \ \ \ \ \ (40)$

so ${G=P}$ and the momentum operator is the generator of translations, and the translation operator is, to order ${\varepsilon}$

$\displaystyle T\left(\varepsilon\right)=I-\frac{i\varepsilon}{\hbar}P \ \ \ \ \ (41)$

By plugging this into 32 and expanding the RHS, we find that in order for the Hamiltonian to be invariant, the expectation value of the commutator ${\left[P,H\right]}$ must be zero (the derivation is done in Shankar’s eqn 11.2.15). Using Ehrenfest’s theorem we then find that the expectation value ${\left\langle \dot{P}\right\rangle =\left\langle \left[P,H\right]\right\rangle =0}$, so that the expectation value of ${P}$ is conserved over time.

Note that we cannot say that the momentum itself (rather than just its expectation value) is conserved since, due to the uncertainty principle, we never know what the exact momentum is at any given time.

# Fermions and bosons in the infinite square well

Shankar, R. (1994), Principles of Quantum Mechanics, Plenum Press. Chapter 10, Exercise 10.3.4.

Suppose we have two identical particles in an infinite square well. The energy levels in a well of width ${L}$ are

$\displaystyle E=\frac{\left(\pi n\hbar\right)^{2}}{2mL^{2}} \ \ \ \ \ (1)$

where ${n=1,2,3,\ldots}$ The corresponding wave functions are given by

$\displaystyle \psi_{n}\left(x\right)=\sqrt{\frac{2}{L}}\sin\frac{n\pi x}{L} \ \ \ \ \ (2)$

If the total energy of the two particles is ${\pi^{2}\hbar^{2}/mL^{2}}$, the only possible configuration is for both particles to be in the ground state ${n=1}$. This means the particles must be bosons, so the state vector is

$\displaystyle \left|x_{1},x_{2}\right\rangle =\frac{2}{L}\sin\frac{\pi x_{1}}{L}\sin\frac{\pi x_{2}}{L} \ \ \ \ \ (3)$

If the total energy is ${5\pi^{2}\hbar^{2}/2mL^{2}}$, then one particle is in the state ${n=1}$ and the other is in ${n=2}$. Since the states are different, the particles can be either bosons or fermions. For bosons, the state vector is

 $\displaystyle \left|x_{1},x_{2}\right\rangle$ $\displaystyle =$ $\displaystyle \frac{1}{\sqrt{2}}\left[\frac{2}{L}\sin\frac{\pi x_{1}}{L}\sin\frac{2\pi x_{2}}{L}+\frac{2}{L}\sin\frac{2\pi x_{1}}{L}\sin\frac{\pi x_{2}}{L}\right]\ \ \ \ \ (4)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{\sqrt{2}}{L}\left[\sin\frac{\pi x_{1}}{L}\sin\frac{2\pi x_{2}}{L}+\sin\frac{2\pi x_{1}}{L}\sin\frac{\pi x_{2}}{L}\right] \ \ \ \ \ (5)$

For fermions, the state must be antisymmetric, so we have

$\displaystyle \left|x_{1},x_{2}\right\rangle =\frac{\sqrt{2}}{L}\left[\sin\frac{\pi x_{1}}{L}\sin\frac{2\pi x_{2}}{L}-\sin\frac{2\pi x_{1}}{L}\sin\frac{\pi x_{2}}{L}\right] \ \ \ \ \ (6)$

# Invariance of symmetric and antisymmetric states; exchange operators

Shankar, R. (1994), Principles of Quantum Mechanics, Plenum Press. Chapter 10, Exercise 10.3.5.

In a system with two particles, the state in the ${X}$ basis is given by ${\left|x_{1},x_{2}\right\rangle }$ where ${x_{i}}$ is the position of particle ${i}$. We can define the exchange operator ${P_{12}}$ as an operator that swaps the two particles, so that

$\displaystyle P_{12}\left|x_{1},x_{2}\right\rangle =\left|x_{2},x_{1}\right\rangle \ \ \ \ \ (1)$

To find the eigenvalues and eigenvectors of ${P_{12}}$ we have

$\displaystyle P_{12}\left|\psi\left(x_{1},x_{2}\right)\right\rangle =\alpha\left|\psi\left(x_{1},x_{2}\right)\right\rangle =\psi\left(x_{2},x_{1}\right) \ \ \ \ \ (2)$

where ${\alpha}$ is the eigenvalue and ${\left|\psi\left(x_{1},x_{2}\right)\right\rangle }$ is the eigenvector. Using the same argument as before, we can write

 $\displaystyle \left|\psi\left(x_{1},x_{2}\right)\right\rangle$ $\displaystyle =$ $\displaystyle \beta\left|x_{1},x_{2}\right\rangle +\gamma\left|x_{2},x_{1}\right\rangle \ \ \ \ \ (3)$ $\displaystyle \left|\psi\left(x_{2},x_{1}\right)\right\rangle$ $\displaystyle =$ $\displaystyle \beta\left|x_{2},x_{1}\right\rangle +\gamma\left|x_{1},x_{2}\right\rangle \ \ \ \ \ (4)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \alpha\left[\beta\left|x_{1},x_{2}\right\rangle +\gamma\left|x_{2},x_{1}\right\rangle \right] \ \ \ \ \ (5)$

Equating coefficients in the first and third lines, we arrive at

$\displaystyle \alpha=\pm1 \ \ \ \ \ (6)$

which gives the same symmetric and antisymmetric eigenfunctions that we had before:

 $\displaystyle \psi_{S}\left(x_{1},x_{2}\right)$ $\displaystyle =$ $\displaystyle \frac{1}{\sqrt{2}}\left(\left|x_{1},x_{2}\right\rangle +\left|x_{2},x_{1}\right\rangle \right)\ \ \ \ \ (7)$ $\displaystyle \psi_{A}\left(x_{1},x_{2}\right)$ $\displaystyle =$ $\displaystyle \frac{1}{\sqrt{2}}\left(\left|x_{1},x_{2}\right\rangle -\left|x_{2},x_{1}\right\rangle \right) \ \ \ \ \ (8)$

We can derive a couple of other properties of the exchange operator by noting that if it is applied twice in succession, we get the original state back, so that

 $\displaystyle P_{12}^{2}$ $\displaystyle =$ $\displaystyle I\ \ \ \ \ (9)$ $\displaystyle P_{12}$ $\displaystyle =$ $\displaystyle P_{12}^{-1} \ \ \ \ \ (10)$

Thus the operator is its own inverse.

Consider also the two states ${\left|x_{1}^{\prime},x_{2}^{\prime}\right\rangle }$ and ${\left|x_{1},x_{2}\right\rangle }$. Then

 $\displaystyle \left\langle x_{1}^{\prime},x_{2}^{\prime}\left|P_{12}^{\dagger}P_{12}\right|x_{1},x_{2}\right\rangle$ $\displaystyle =$ $\displaystyle \left\langle P_{12}x_{1}^{\prime},x_{2}^{\prime}\left|P_{12}x_{1},x_{2}\right.\right\rangle \ \ \ \ \ (11)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \left\langle x_{2}^{\prime},x_{1}^{\prime}\left|x_{2},x_{1}\right.\right\rangle \ \ \ \ \ (12)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \left(\left\langle x_{2}^{\prime}\right|\otimes\left\langle x_{1}^{\prime}\right|\right)\left(\left|x_{2}\right\rangle \otimes\left|x_{1}\right\rangle \right)\ \ \ \ \ (13)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \delta\left(x_{2}^{\prime}-x_{2}\right)\delta\left(x_{1}^{\prime}-x_{1}\right) \ \ \ \ \ (14)$

However, the last line is just equal to the inner product of the original states, that is

$\displaystyle \left\langle x_{1}^{\prime},x_{2}^{\prime}\left|x_{1},x_{2}\right.\right\rangle =\delta\left(x_{2}-x_{2}^{\prime}\right)\delta\left(x_{1}-x_{1}^{\prime}\right)=\delta\left(x_{2}^{\prime}-x_{2}\right)\delta\left(x_{1}^{\prime}-x_{1}\right) \ \ \ \ \ (15)$

This means that

 $\displaystyle P_{12}^{\dagger}P_{12}$ $\displaystyle =$ $\displaystyle I\ \ \ \ \ (16)$ $\displaystyle P_{12}^{\dagger}$ $\displaystyle =$ $\displaystyle P_{12}^{-1}=P_{12} \ \ \ \ \ (17)$

Thus ${P_{12}}$ is both Hermitian and unitary.

Shankar asks us to show that, for a general basis vector ${\left|\omega_{1},\omega_{2}\right\rangle }$, ${P_{12}\left|\omega_{1},\omega_{2}\right\rangle =\left|\omega_{2},\omega_{1}\right\rangle }$. One argument could be that, since the ${X}$ basis spans the space, we can express any other vector such as ${\left|\omega_{1},\omega_{2}\right\rangle }$ as a linear combination of the ${\left|x_{1},x_{2}\right\rangle }$ vectors, so that applying ${P_{12}}$ to ${\left|\omega_{1},\omega_{2}\right\rangle }$ means applying it to a sum of ${\left|x_{1},x_{2}\right\rangle }$ vectors, which swaps the two particles in every term. I’m not sure if this is a rigorous result. In any case, if we accept this result it shows that if we start in a state that is totally symmetric (that is, a boson state), this state is an eigenvector of ${P_{12}}$ with eigenvalue ${+1}$. Similarly, if we start in an antisymmetric (fermion) state, this state is an eigenvector of ${P_{12}}$ with eigenvalue ${-1}$.

Now we can look at some other properties of ${P_{12}}$. Consider

 $\displaystyle P_{12}X_{1}P_{12}\left|x_{1},x_{2}\right\rangle$ $\displaystyle =$ $\displaystyle P_{12}X_{1}\left|x_{2},x_{1}\right\rangle \ \ \ \ \ (18)$ $\displaystyle$ $\displaystyle =$ $\displaystyle x_{2}P_{12}\left|x_{2},x_{1}\right\rangle \ \ \ \ \ (19)$ $\displaystyle$ $\displaystyle =$ $\displaystyle x_{2}\left|x_{1},x_{2}\right\rangle \ \ \ \ \ (20)$ $\displaystyle$ $\displaystyle =$ $\displaystyle X_{2}\left|x_{1},x_{2}\right\rangle \ \ \ \ \ (21)$

This follows because the operator ${X_{1}}$ operates on the first particle in the state ${\left|x_{2},x_{1}\right\rangle }$ which on the RHS of the first line is at position ${x_{2}}$. Thus ${X_{1}\left|x_{2},x_{1}\right\rangle =x_{2}\left|x_{2},x_{1}\right\rangle }$, that is, ${X_{1}}$ returns the numerical value of the position of the first particle, which is ${x_{2}}$. This means that in terms of the operators alone

 $\displaystyle P_{12}X_{1}P_{12}$ $\displaystyle =$ $\displaystyle X_{2}\ \ \ \ \ (22)$ $\displaystyle P_{12}X_{2}P_{12}$ $\displaystyle =$ $\displaystyle X_{1}\ \ \ \ \ (23)$ $\displaystyle P_{12}P_{1}P_{12}$ $\displaystyle =$ $\displaystyle P_{2}\ \ \ \ \ (24)$ $\displaystyle P_{12}P_{2}P_{12}$ $\displaystyle =$ $\displaystyle P_{1} \ \ \ \ \ (25)$

In the last two lines, the operator ${P_{i}}$ is the momentum of particle ${i}$, and the result follows by applying the operators to the momentum basis state ${\left|p_{1},p_{2}\right\rangle }$.

For some general operator which can be expanded in a power series of terms containing powers of ${X_{i}}$ and/or ${P_{i}}$, we can use 10 to insert ${P_{12}P_{12}}$ between every factor of ${X_{i}}$ or ${P_{i}}$. For example

 $\displaystyle P_{12}P_{1}X_{2}^{2}X_{1}P_{12}$ $\displaystyle =$ $\displaystyle P_{12}P_{1}P_{12}P_{12}X_{2}P_{12}P_{12}X_{2}P_{12}P_{12}X_{1}P_{12}\ \ \ \ \ (26)$ $\displaystyle$ $\displaystyle =$ $\displaystyle P_{2}X_{1}^{2}X_{2} \ \ \ \ \ (27)$

That is, for any operator ${\Omega\left(X_{1},P_{1};X_{2},P_{2}\right)}$ we have

$\displaystyle P_{12}\Omega\left(X_{1},P_{1};X_{2},P_{2}\right)P_{12}=\Omega\left(X_{2},P_{2};X_{1},P_{1}\right) \ \ \ \ \ (28)$

The Hamiltonian for a system of two identical particles must be symmetric under exchange of the particles, since it represents an observable (the energy), and this observable must remain unchanged if we swap the particles. (In the case of two fermions, the wave function is antisymmetric, but the wave function itself is not an observable. The wave function gets multiplied by ${-1}$ if we swap the particles, but the square modulus of the wave function, which contains the physics, remains the same.) Thus we have

$\displaystyle P_{12}H\left(X_{1},P_{1};X_{2},P_{2}\right)P_{12}=H\left(X_{2},P_{2};X_{1},P_{1}\right)=H\left(X_{1},P_{1};X_{2},P_{2}\right) \ \ \ \ \ (29)$

[Note that this condition doesn’t necessarily follow if the two particles are not identical, since exchanging them in this case leads to an observably different system. For example, exchanging the proton and electron in a hydrogen atom leads to a different system.]

The propagator is defined as

$\displaystyle U\left(t\right)=e^{-iHt/\hbar} \ \ \ \ \ (30)$

and the propagator dictates how a state evolves according to

$\displaystyle \left|\psi\left(t\right)\right\rangle =U\left(t\right)\left|\psi\left(0\right)\right\rangle \ \ \ \ \ (31)$

Since the only operator on which ${U}$ depends is ${H}$, then ${U}$ is also invariant, so that

$\displaystyle P_{12}U\left(X_{1},P_{1};X_{2},P_{2}\right)P_{12}=U\left(X_{2},P_{2};X_{1},P_{1}\right)=U\left(X_{1},P_{1};X_{2},P_{2}\right) \ \ \ \ \ (32)$

Multiplying from the left by ${P_{12}}$ and subtracting, we get the commutator

$\displaystyle \left[U,P_{12}\right]=0 \ \ \ \ \ (33)$

For a symmetric state ${\left|\psi_{S}\right\rangle }$ or antisymmetric state ${\left|\psi_{A}\right\rangle }$, we have

 $\displaystyle UP_{12}\left|\psi_{S}\left(0\right)\right\rangle$ $\displaystyle =$ $\displaystyle U\left|\psi_{S}\left(0\right)\right\rangle =\left|\psi_{S}\left(t\right)\right\rangle =P_{12}U\left|\psi_{S}\left(0\right)\right\rangle \ \ \ \ \ (34)$ $\displaystyle UP_{12}\left|\psi_{A}\left(0\right)\right\rangle$ $\displaystyle =$ $\displaystyle -U\left|\psi_{A}\left(0\right)\right\rangle =-\left|\psi_{A}\left(t\right)\right\rangle =P_{12}U\left|\psi_{A}\left(0\right)\right\rangle \ \ \ \ \ (35)$

This means that states that begin as symmetric or antisymmetric remain symmetric or antisymmetric for all time. In other words, a system that starts in an eigenstate of ${P_{12}}$ remains in the same eigenstate as time passes.

# Compound systems of fermions and bosons

Shankar, R. (1994), Principles of Quantum Mechanics, Plenum Press. Chapter 10, Exercise 10.3.6.

In a system of identical particles, we’ve seen that if the particles are bosons, the state vector is symmetric with respect to the exchange of any two particles (that is, ${\psi\left(a,b\right)=\psi\left(b,a\right)}$ where ${a}$ and ${b}$ are any two of the particles in the system), while for fermions, the state vector is antisymmetric, meaning that ${\psi\left(a,b\right)=-\psi\left(a,b\right)}$. What happens if we have a compound object such as a hydrogen atom that is composed of a collection of fermions and/or bosons?

Suppose we look at the hydrogen atom in particular. It is composed of a proton and an electron, both of which are fermions. The proton and electron are not, of course, identical particles, but now suppose we have two hydrogen atoms. The two protons are identical fermions, just as are the two electrons. However, when analyzing a system of two hydrogen atoms, the relevant question is what happens to the state vector if we exchange the two atoms. In doing so, we exchange both the two protons and the two electrons. Each exchange multiplies the state vector by ${-1}$, so the net effect of exchanging both protons and both electrons is to multiply the state vector by ${\left(-1\right)^{2}=1}$. In other words, a hydrogen atom acts as a boson, even though it is composed of two fermions.

In general, if we have a compound object containing ${n}$ fermions, then the state vector for a system of two such objects is multiplied by ${\left(-1\right)^{n}}$ when these two objects are exchanged. That is, a compound object containing an even number of fermions behaves as a boson, while if it contains an odd number of fermions, it behaves as a fermion.

A compound object consisting entirely of bosons will always behave as a boson, no matter how many such bosonic particles it contains, since interchanging all ${n}$ bosons just multiplies the state vector by ${\left(+1\right)^{n}=1}$.

# Identical particles – bosons and fermions revisited

Shankar, R. (1994), Principles of Quantum Mechanics, Plenum Press. Chapter 10, Exercises 10.3.1 – 10.3.3.

Although we’ve looked at the quantum treatment of identical particles as done by Griffiths, it’s worth summarizing Shankar’s treatment of the topic as it provides a few more insights.

In classical physics, suppose we have two identical particles, where ‘identical’ here means that all their physical properties such as mass, size, shape, charge and so on are the same. Suppose we do an experiment in which these two particles collide and rebound in some way. Can we tell which particle ends up in which location? We’re not allowed to label the particles by writing on them, for example, since then they would no longer be identical. In classical physics, we can determine which particle is which by tracing their histories. For example, if we start with particle 1 at position ${\mathbf{r}_{1}}$ and particle 2 at position ${\mathbf{r}_{2}}$, then let them collide, and finally measure their locations at some time after the collision, we might find that one particle ends up at position ${\mathbf{r}_{3}}$ and the other at position ${\mathbf{r}_{4}}$. If we videoed the collision event, we would see the two particles follow well-defined paths before and after the collision, so by observing which particle followed the path that leads from ${\mathbf{r}_{1}}$ to the collision and then out again, we can tell whether it ends up at ${\mathbf{r}_{3}}$ or ${\mathbf{r}_{4}}$. That is, the identification of a particle depends on our ability to watch it as it travels through space.

In quantum mechanics, because of the uncertainty principle, a particle does not have a well-defined trajectory, since in order to define such a trajectory, we would need to specify its position and momentum precisely at each instant of time as it travels. In terms of our collision experiment, if we measured one particle to be at starting position ${\mathbf{r}_{1}}$ at time ${t=0}$ then we know nothing about its momentum, because we specified the position exactly. Thus we can’t tell what trajectory this particle will follow. If we measure the two particles at positions ${\mathbf{r}_{1}}$ and ${\mathbf{r}_{2}}$ at ${t=0}$, and then at ${\mathbf{r}_{3}}$ and ${\mathbf{r}_{4}}$ at some later time, we have no way of knowing which particle ends up at ${\mathbf{r}_{3}}$ and which at ${\mathbf{r}_{4}}$. In terms of the state vector, this means that the physics in the state vector must be the same if we exchange the two particles within the wave function. Since multiplying a state vector ${\psi}$ by some complex constant ${\alpha}$ leaves the physics unchanged, this means that we require

$\displaystyle \psi\left(a,b\right)=\alpha\psi\left(b,a\right) \ \ \ \ \ (1)$

where ${a}$ and ${b}$ represent the two particles.

For a two-particle system, the vector space is spanned by a direct product of the two one-particle vector spaces. Thus the two basis vectors in this vector space that can describe the two particle ${a}$ and ${b}$ are ${\left|ab\right\rangle }$ and ${\left|ba\right\rangle }$. If these two particles are identical, then ${\psi}$ must be some linear combination of these two vectors that satisfies 1. That is

 $\displaystyle \psi\left(b,a\right)$ $\displaystyle =$ $\displaystyle \beta\left|ab\right\rangle +\gamma\left|ba\right\rangle \ \ \ \ \ (2)$ $\displaystyle \psi\left(a,b\right)$ $\displaystyle =$ $\displaystyle \alpha\psi\left(b,a\right)\ \ \ \ \ (3)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \alpha\left(\beta\left|ab\right\rangle +\gamma\left|ba\right\rangle \right) \ \ \ \ \ (4)$

However, ${\psi\left(a,b\right)}$ is also just ${\psi\left(b,a\right)}$ with ${a}$ swapped with ${b}$, that is

$\displaystyle \psi\left(a,b\right)=\beta\left|ba\right\rangle +\gamma\left|ab\right\rangle \ \ \ \ \ (5)$

Since ${\left|ab\right\rangle }$ and ${\left|ba\right\rangle }$ are independent, we can equate their coefficients in the last two equations to get

 $\displaystyle \alpha\beta$ $\displaystyle =$ $\displaystyle \gamma\ \ \ \ \ (6)$ $\displaystyle \alpha\gamma$ $\displaystyle =$ $\displaystyle \beta \ \ \ \ \ (7)$

Inserting the second equation into the first, we get

 $\displaystyle \alpha^{2}\gamma$ $\displaystyle =$ $\displaystyle \gamma\ \ \ \ \ (8)$ $\displaystyle \alpha^{2}$ $\displaystyle =$ $\displaystyle 1\ \ \ \ \ (9)$ $\displaystyle \alpha$ $\displaystyle =$ $\displaystyle \pm1 \ \ \ \ \ (10)$

Thus the two possible state functions 1 are combinations of ${\left|ab\right\rangle }$ and ${\left|ba\right\rangle }$ such that

$\displaystyle \psi\left(a,b\right)=\pm\psi\left(b,a\right) \ \ \ \ \ (11)$

The plus sign gives the symmetric state, which can be written as

$\displaystyle \psi\left(ab,S\right)=\frac{1}{\sqrt{2}}\left(\left|ab\right\rangle +\left|ba\right\rangle \right) \ \ \ \ \ (12)$

and the minus sign gives the antisymmetric state

$\displaystyle \psi\left(ab,A\right)=\frac{1}{\sqrt{2}}\left(\left|ab\right\rangle -\left|ba\right\rangle \right) \ \ \ \ \ (13)$

The ${\frac{1}{\sqrt{2}}}$ factor normalizes the states so that

 $\displaystyle \left\langle \psi\left(ab,S\right)\left|\psi\left(ab,S\right)\right.\right\rangle$ $\displaystyle =$ $\displaystyle 1\ \ \ \ \ (14)$ $\displaystyle \left\langle \psi\left(ab,A\right)\left|\psi\left(ab,A\right)\right.\right\rangle$ $\displaystyle =$ $\displaystyle 1 \ \ \ \ \ (15)$

This follows because the basis vectors ${\left|ab\right\rangle }$ and ${\left|ba\right\rangle }$ are orthonormal vectors.

Particles with symmetric states are called bosons and particles with antisymmetric states are called fermions. The Pauli exclusion principle for fermions follows directly from 13, since if we set the state variables of the two particles to be the same, that is, ${a=b}$, then

$\displaystyle \psi\left(aa,A\right)=\frac{1}{\sqrt{2}}\left(\left|aa\right\rangle -\left|aa\right\rangle \right)=0 \ \ \ \ \ (16)$

The symmetry or antisymmetry rules apply to all the properties of the particle taken as an aggregate. That is, the labels ${a}$ and ${b}$ can refer to the particle’s location plus its other quantum numbers such as spin, charge, and so on. In order for two fermions to be excluded, the states of the two fermions must be identical in all their quantum numbers, so that two fermions with the same orbital location (as two electrons in the same orbital within an atom, for example) are allowed if their spins are different.

Example 1 Suppose we have 2 identical bosons that are measured to be in states ${\left|\phi\right\rangle }$ and ${\left|\chi\right\rangle }$ where ${\left\langle \phi\left|\chi\right.\right\rangle \ne0}$. What is their combined state vector? Since they are bosons, their state vector must be symmetric, so we must have

 $\displaystyle \psi\left(\phi,\chi\right)$ $\displaystyle =$ $\displaystyle A\left|\phi\chi\right\rangle +B\left|\chi\phi\right\rangle \ \ \ \ \ (17)$

Because ${\psi}$ must be symmetric, we must have ${A=B}$, so that ${\psi\left(\phi,\chi\right)=\psi\left(\chi,\phi\right)}$. The 2-particle states can be written as direct products, so we have

$\displaystyle \psi\left(\phi,\chi\right)=A\left(\left|\phi\right\rangle \otimes\left|\chi\right\rangle +\left|\chi\right\rangle \otimes\left|\phi\right\rangle \right) \ \ \ \ \ (18)$

To normalize, we have, assuming that ${\left|\phi\right\rangle }$ and ${\left|\chi\right\rangle }$ are normalized:

 $\displaystyle \left|\psi\right|^{2}$ $\displaystyle =$ $\displaystyle 1\ \ \ \ \ (19)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \left|A\right|^{2}\left(\left\langle \phi\right|\otimes\left\langle \chi\right|+\left\langle \chi\right|\otimes\left\langle \phi\right|\right)\left(\left|\phi\right\rangle \otimes\left|\chi\right\rangle +\left|\chi\right\rangle \otimes\left|\phi\right\rangle \right)\ \ \ \ \ (20)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \left|A\right|^{2}\left(1+1+\left|\left\langle \phi\left|\chi\right.\right\rangle \right|^{2}+\left|\left\langle \chi\left|\phi\right.\right\rangle \right|^{2}\right)\ \ \ \ \ (21)$ $\displaystyle A$ $\displaystyle =$ $\displaystyle \frac{\pm1}{\sqrt{2\left(1+\left|\left\langle \phi\left|\chi\right.\right\rangle \right|^{2}\right)}} \ \ \ \ \ (22)$

Thus the normalized state vector is (choosing the + sign):

 $\displaystyle \psi\left(\phi,\chi\right)$ $\displaystyle =$ $\displaystyle \frac{1}{\sqrt{2\left(1+\left|\left\langle \phi\left|\chi\right.\right\rangle \right|^{2}\right)}}\left(\left|\phi\chi\right\rangle +\left|\chi\phi\right\rangle \right) \ \ \ \ \ (23)$

Notice that this reduces to 12 if ${\left\langle \phi\left|\chi\right.\right\rangle =0}$.

For more than 2 particles, we need to form state vectors that are either totally symmetric or totally antisymmetric.

Example 2 Suppose we have 3 identical bosons, and they are measured to be in states 3, 3 and 4. Since two of them are in the same state, there are 3 possible combinations, which we can write as ${\left|334\right\rangle ,}$ ${\left|343\right\rangle }$ and ${\left|433\right\rangle }$. Assuming these states are orthonormal, the full normalized state vector is

$\displaystyle \psi\left(3,3,4\right)=\frac{1}{\sqrt{3}}\left(\left|334\right\rangle +\left|343\right\rangle +\left|433\right\rangle \right) \ \ \ \ \ (24)$

The ${\frac{1}{\sqrt{3}}}$ ensures that ${\left|\psi\left(3,3,4\right)\right|^{2}=1}$.

Incidentally, for ${N\ge3}$ particles, it turns out to be impossible to construct a linear combination of the basis states such that the overall state vector is symmetric with respect to the interchange of some pairs of particles and antisymmetric with respect to the interchange of other pairs. A general proof for all ${N}$ requires group theory, but for ${N=3}$ we can show this by brute force. There are ${3!=6}$ basis vectors

$\displaystyle \left|123\right\rangle ,\left|231\right\rangle ,\left|312\right\rangle ,\left|132\right\rangle ,\left|321\right\rangle ,\left|213\right\rangle \ \ \ \ \ (25)$

Suppose we require the compound state vector to be symmetric with respect to exchanging 1 and 2. We then must have

$\displaystyle \psi=A\left(\left|123\right\rangle +\left|213\right\rangle \right)+B\left(\left|231\right\rangle +\left|132\right\rangle \right)+C\left(\left|312\right\rangle +\left|321\right\rangle \right) \ \ \ \ \ (26)$

If we now try to make ${\psi}$ antisymmetric with respect to exchanging 2 and 3, we must have

$\displaystyle \psi=D\left(\left|123\right\rangle -\left|132\right\rangle \right)+E\left(\left|231\right\rangle -\left|321\right\rangle \right)+F\left(\left|312\right\rangle -\left|213\right\rangle \right) \ \ \ \ \ (27)$

Comparing the two, we see that

 $\displaystyle A$ $\displaystyle =$ $\displaystyle D=-F\ \ \ \ \ (28)$ $\displaystyle B$ $\displaystyle =$ $\displaystyle E=-D\ \ \ \ \ (29)$ $\displaystyle C$ $\displaystyle =$ $\displaystyle F=-E \ \ \ \ \ (30)$

Eliminating ${A,B}$, and ${C}$ we have, combining the 3 equations:

$\displaystyle D=-E=F \ \ \ \ \ (31)$

But from the first equation, we have ${D=-F}$, so ${F=-F=0}$. From the other equations, this implies that ${D=-F=0}$ and ${E=-F=0}$, and thus that ${A=B=C=0}$. So there is no non-trivial solution that allows both a symmetric and antisymmetric particle exchange within the same state vector.

Example 3 Suppose we have 3 particles and only 3 distinct states that each particle can have. If the particles are distinguishable (not identical) the total number of states is found by considering the possibilities. If all 3 particles are in different states, then there are ${3!=6}$ possible overall states. If two particles are in one state and one particle in another, there are ${\binom{3}{2}=3}$ ways of choosing the two states, for each of which there are 2 ways of partitioning these two states (that is, which state has 2 particles and which has the other one), and for each of those there are 3 possible configurations, so there are ${3\times2\times3=18}$ possible configurations. Finally, if all 3 particles are in the same state, there are 3 possibilities. Thus the total for distinguishable particles is ${6+18+3=27}$.

If the particles are bosons, then if all 3 are in different states, there is only 1 symmetric combination of the 6 basis states. If two particles are in one state and one particle in another, there are ${3\times2=6}$ ways of partitioning the states, each of which contributes only one symmetric overall state. Finally, if all 3 particles are in the same state, there are 3 possibilities. Thus the total for bosons is ${1+6+3=10}$.

For fermions, all three particles must be in different states, so there is only 1 possibility.

# 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

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