# Combining translations and rotations

Shankar, R. (1994), Principles of Quantum Mechanics, Plenum Press. Chapter 12, Exercise 12.2.4.

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When it comes to symmetries in quantum mechanics, we’ve looked at translations and rotations in two dimensions, and found that the generators are the momenta ${P_{x}}$ and ${P_{y}}$ for translations, and the angular momentum ${L_{z}}$ for rotations.

From the fact that ${L_{z}}$ does not commute with either momentum or position operators, you might guess that if we performed some sequence of translations and rotations on a system that the order in which these operations are done matters. In fact, you can see this by considering simple two-dimensional geometry, without reference to quantum mechanics. Consider the ${x}$ and ${y}$ axes on a sheet of graph paper. First, translate these axes by adding the vector ${\mathbf{r}}$ to all points, so that the new origin of coordinates lies at position ${\mathbf{r}}$ as referenced in the original coordinates. Next, do a rotation about the original origin by some angle ${\phi}$. This will move the new origin around the original ${z}$ axis. Now, do the inverse of the original translation by adding ${-\mathbf{r}}$ to all points. Finally, do the inverse of the rotation by rotating the system by ${-\phi}$ around the original ${z}$ axis. You’ll find that the ${xy}$ axes that have undergone this sequence of transformations does not coincide with the original ${xy}$ axes. However, if you did the same set of four transformations in the order: translate by ${\mathbf{r}}$, translate by ${-\mathbf{r}}$, rotate by ${\phi}$, rotate by ${-\phi}$, the transformed axes would coincide with the original axes.

To see how this works in quantum mechanics, we can again consider infinitesimal translations and rotations. If we start with a point at location ${\left[x,y\right]}$ and apply the four transformations described above, but now for an infinitesimal translation ${\boldsymbol{\varepsilon}=\varepsilon_{x}\hat{\mathbf{x}}+\varepsilon_{y}\hat{\mathbf{y}}}$ and rotation ${\varepsilon_{z}\hat{\mathbf{z}}}$, then the successive transformations work as follows. In each case, we’ll retain terms up to order ${\varepsilon_{x}\varepsilon_{z}}$ and ${\varepsilon_{y}\varepsilon_{z}}$ but discard terms of order ${\varepsilon_{x}^{2}}$, ${\varepsilon_{y}^{2}}$, ${\varepsilon_{z}^{2}}$ and higher. [I’m not quite sure of the rationale that allows us to do this, apart from the fact that it gives the right answer.]

 $\displaystyle \left[\begin{array}{c} x\\ y \end{array}\right]$ $\displaystyle {\longrightarrow\atop T\left(\boldsymbol{\varepsilon}\right)}$ $\displaystyle \left[\begin{array}{c} x+\varepsilon_{x}\\ y+\varepsilon_{y} \end{array}\right]\ \ \ \ \ (1)$ $\displaystyle \left[\begin{array}{c} x+\varepsilon_{x}\\ y+\varepsilon_{y} \end{array}\right]$ $\displaystyle {\longrightarrow\atop R\left(\varepsilon_{z}\hat{\mathbf{z}}\right)}$ $\displaystyle \left[\begin{array}{c} x+\varepsilon_{x}-\left(y+\varepsilon_{y}\right)\varepsilon_{z}\\ y+\varepsilon_{y}+\left(x+\varepsilon_{x}\right)\varepsilon_{z} \end{array}\right]\ \ \ \ \ (2)$ $\displaystyle \left[\begin{array}{c} x+\varepsilon_{x}-\left(y+\varepsilon_{y}\right)\varepsilon_{z}\\ y+\varepsilon_{y}+\left(x+\varepsilon_{x}\right)\varepsilon_{z} \end{array}\right]$ $\displaystyle {\longrightarrow\atop T\left(-\boldsymbol{\varepsilon}\right)}$ $\displaystyle \left[\begin{array}{c} x+\varepsilon_{x}-\left(y+\varepsilon_{y}\right)\varepsilon_{z}-\varepsilon_{x}\\ y+\varepsilon_{y}+\left(x+\varepsilon_{x}\right)\varepsilon_{z}-\varepsilon_{y} \end{array}\right]\ \ \ \ \ (3)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \left[\begin{array}{c} x-\left(y+\varepsilon_{y}\right)\varepsilon_{z}\\ y+\left(x+\varepsilon_{x}\right)\varepsilon_{z} \end{array}\right]\ \ \ \ \ (4)$ $\displaystyle \left[\begin{array}{c} x-\left(y+\varepsilon_{y}\right)\varepsilon_{z}\\ y+\left(x+\varepsilon_{x}\right)\varepsilon_{z} \end{array}\right]$ $\displaystyle {\longrightarrow\atop R\left(-\varepsilon_{z}\hat{\mathbf{z}}\right)}$ $\displaystyle \left[\begin{array}{c} x-\left(y+\varepsilon_{y}\right)\varepsilon_{z}+\left[y+\left(x+\varepsilon_{x}\right)\varepsilon_{z}\right]\varepsilon_{z}\\ y+\left(x+\varepsilon_{x}\right)\varepsilon_{z}-\left[x-\left(y+\varepsilon_{y}\right)\varepsilon_{z}\right]\varepsilon_{z} \end{array}\right]\ \ \ \ \ (5)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \left[\begin{array}{c} x-\varepsilon_{y}\varepsilon_{z}\\ y+\varepsilon_{x}\varepsilon_{z} \end{array}\right] \ \ \ \ \ (6)$

Thus, to this order in the infinitesimals, the combination of translation-rotation-translation-rotation is equivalent to a single translation by a distance ${\left[-\varepsilon_{y}\varepsilon_{z},\varepsilon_{x}\varepsilon_{z}\right]}$. We can write this in terms of the unitary quantum operators for translations and rotations as

$\displaystyle U\left[R\left(-\varepsilon_{z}\hat{\mathbf{z}}\right)\right]T\left(-\boldsymbol{\varepsilon}\right)U\left[R\left(\varepsilon_{z}\hat{\mathbf{z}}\right)\right]T\left(\boldsymbol{\varepsilon}\right)=T\left(-\varepsilon_{y}\varepsilon_{z}\hat{\mathbf{x}}+\varepsilon_{x}\varepsilon_{z}\hat{\mathbf{y}}\right) \ \ \ \ \ (7)$

Using the forms of these operators for infinitesimal transformations, we can expand both sides to give

 $\displaystyle \left(I+\frac{i\varepsilon_{z}}{\hbar}L_{z}\right)\left[I+\frac{i}{\hbar}\left(\varepsilon_{x}P_{x}+\varepsilon_{y}P_{y}\right)\right]\times$ $\displaystyle \left(I-\frac{i\varepsilon_{z}}{\hbar}L_{z}\right)\left[I-\frac{i}{\hbar}\left(\varepsilon_{x}P_{x}+\varepsilon_{y}P_{y}\right)\right]$ $\displaystyle = \ \ \ \ \ (8)$ $\displaystyle I-\frac{i}{\hbar}\left(-\varepsilon_{y}\varepsilon_{z}P_{x}+\varepsilon_{x}\varepsilon_{z}P_{y}\right)$

Since the infinitesimal displacements are arbitrary, this equation can be valid only if the coefficients of each combination of ${\varepsilon_{x},\varepsilon_{y}}$ and ${\varepsilon_{z}}$ are equal on both sides. As above, we’ll discard any terms of order ${\varepsilon_{x}^{2}}$, ${\varepsilon_{y}^{2}}$, ${\varepsilon_{z}^{2}}$ and higher. The algebra is straightforward although a bit tedious, so I’ll just give a couple of examples here.

The coefficient of ${\varepsilon_{z}}$ on its own is, on the LHS

$\displaystyle \frac{i\varepsilon_{z}}{\hbar}L_{z}-\frac{i\varepsilon_{z}}{\hbar}L_{z}=0 \ \ \ \ \ (9)$

On the RHS, there is no term in ${\varepsilon_{z}}$, so we get 0 on the RHS. In this case, we see the equation is consistent.

For the ${\varepsilon_{x}\varepsilon_{z}}$ term, we get on the LHS:

$\displaystyle \varepsilon_{x}\varepsilon_{z}\frac{i^{2}}{\hbar^{2}}\left(L_{z}P_{x}-L_{z}P_{x}-P_{x}L_{z}+L_{z}P_{x}\right)=-\varepsilon_{x}\varepsilon_{z}\frac{i^{2}}{\hbar^{2}}\left[P_{x},L_{z}\right] \ \ \ \ \ (10)$

On the RHS, the term is

$\displaystyle -\frac{i}{\hbar}\varepsilon_{x}\varepsilon_{z}P_{y} \ \ \ \ \ (11)$

Thus the condition here becomes

$\displaystyle \left[P_{x},L_{z}\right]=-i\hbar P_{y} \ \ \ \ \ (12)$

which agrees with the commutation relation we found earlier. By considering the coefficient of ${\varepsilon_{y}\varepsilon_{z}}$, we arrive at the other condition, which is

$\displaystyle \left[P_{y},L_{z}\right]=i\hbar P_{x} \ \ \ \ \ (13)$

The result of this calculation doesn’t tell us anything new about the translation or rotation operators, but it does show that the condition 7 is consistent with what we already know about the commutators of position, momentum and angular momentum.

As Shankar points out, we might think that we need to verify the conditions for an infinite number of combinations of rotations and translations, since each such combination gives rise to a different overall transformation. He says that it has actually been shown that the example above is sufficient to guarantee that all such combinations do in fact give valid results, although he doesn’t give the details. We are, however, given the exercise of verifying this claim for one special case, which we’ll consider now.

In this example, we’ll consider the same four transformations, in the same order, as above except that we’ll take the translation to be entirely in the ${x}$ direction so that ${\varepsilon_{y}=0}$. This time, we’ll retain terms up to ${\varepsilon_{x}\varepsilon_{z}^{2}}$ and see what we get. We start by repeating the calculations in 1 through 6. However, because we’re saving higher order terms, we need to represent the infinitesimal rotations by

$\displaystyle R\left(\varepsilon_{z}\hat{\mathbf{z}}\right)=\left[\begin{array}{cc} 1-\frac{\varepsilon_{z}^{2}}{2} & -\varepsilon_{z}\\ \varepsilon_{z} & 1-\frac{\varepsilon_{z}^{2}}{2} \end{array}\right] \ \ \ \ \ (14)$

That is, we’re approximating ${\cos\varepsilon_{z}}$ by the first two terms in its expansion. Using this, we have

 $\displaystyle \left[\begin{array}{c} x\\ y \end{array}\right]$ $\displaystyle {\longrightarrow\atop T\left(\boldsymbol{\varepsilon}\right)}$ $\displaystyle \left[\begin{array}{c} x+\varepsilon_{x}\\ y \end{array}\right]\ \ \ \ \ (15)$ $\displaystyle \left[\begin{array}{c} x+\varepsilon_{x}\\ y \end{array}\right]$ $\displaystyle {\longrightarrow\atop R\left(\varepsilon_{z}\hat{\mathbf{z}}\right)}$ $\displaystyle \left[\begin{array}{c} \left(x+\varepsilon_{x}\right)\left(1-\frac{\varepsilon_{z}^{2}}{2}\right)-y\varepsilon_{z}\\ y\left(1-\frac{\varepsilon_{z}^{2}}{2}\right)+\left(x+\varepsilon_{x}\right)\varepsilon_{z} \end{array}\right]\ \ \ \ \ (16)$ $\displaystyle \left[\begin{array}{c} x+\varepsilon_{x}-y\varepsilon_{z}\\ y+\left(x+\varepsilon_{x}\right)\varepsilon_{z} \end{array}\right]$ $\displaystyle {\longrightarrow\atop T\left(-\boldsymbol{\varepsilon}\right)}$ $\displaystyle \left[\begin{array}{c} \left(x+\varepsilon_{x}\right)\left(1-\frac{\varepsilon_{z}^{2}}{2}\right)-y\varepsilon_{z}-\varepsilon_{x}\\ y\left(1-\frac{\varepsilon_{z}^{2}}{2}\right)+\left(x+\varepsilon_{x}\right)\varepsilon_{z} \end{array}\right]\ \ \ \ \ (17)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \left[\begin{array}{c} x\left(1-\frac{\varepsilon_{z}^{2}}{2}\right)-y\varepsilon_{z}-\frac{1}{2}\varepsilon_{x}\varepsilon_{z}^{2}\\ y\left(1-\frac{\varepsilon_{z}^{2}}{2}\right)+\varepsilon_{z}x+\varepsilon_{x}\varepsilon_{z} \end{array}\right]\ \ \ \ \ (18)$ $\displaystyle \left[\begin{array}{c} x-y\varepsilon_{z}\\ y+\left(x+\varepsilon_{x}\right)\varepsilon_{z} \end{array}\right]$ $\displaystyle {\longrightarrow\atop R\left(-\varepsilon_{z}\hat{\mathbf{z}}\right)}$ $\displaystyle \left[\begin{array}{c} \left[x\left(1-\frac{\varepsilon_{z}^{2}}{2}\right)-y\varepsilon_{z}-\frac{1}{2}\varepsilon_{x}\varepsilon_{z}^{2}\right]\left(1-\frac{\varepsilon_{z}^{2}}{2}\right)+\left[y\left(1-\frac{\varepsilon_{z}^{2}}{2}\right)+\varepsilon_{z}x+\varepsilon_{x}\varepsilon_{z}\right]\varepsilon_{z}\\ \left[y\left(1-\frac{\varepsilon_{z}^{2}}{2}\right)+\varepsilon_{z}x+\varepsilon_{x}\varepsilon_{z}\right]\left(1-\frac{\varepsilon_{z}^{2}}{2}\right)-\left[x\left(1-\frac{\varepsilon_{z}^{2}}{2}\right)-y\varepsilon_{z}-\frac{1}{2}\varepsilon_{x}\varepsilon_{z}^{2}\right]\varepsilon_{z} \end{array}\right]\ \ \ \ \ (19)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \left[\begin{array}{c} x\left(1+\frac{\varepsilon_{z}^{4}}{4}\right)+\frac{1}{2}\varepsilon_{x}\varepsilon_{z}^{2}+\frac{1}{4}\varepsilon_{x}\varepsilon_{z}^{4}\\ y\left(1+\frac{\varepsilon_{z}^{4}}{4}\right)+\varepsilon_{x}\varepsilon_{z} \end{array}\right] \ \ \ \ \ (20)$

To get the last line, I used Maple to do the algebra in multiplying out the terms. At this point, we can neglect the terms in ${\varepsilon_{z}^{4}}$, leaving us with the overall transformation:

$\displaystyle \left[\begin{array}{c} x\\ y \end{array}\right]\longrightarrow\left[\begin{array}{c} x+\frac{1}{2}\varepsilon_{x}\varepsilon_{z}^{2}\\ y+\varepsilon_{x}\varepsilon_{z} \end{array}\right] \ \ \ \ \ (21)$

This is equivalent to a translation by ${\boldsymbol{\varepsilon}=\frac{1}{2}\varepsilon_{x}\varepsilon_{z}^{2}\hat{\mathbf{x}}+\varepsilon_{x}\varepsilon_{z}\hat{\mathbf{y}}}$, so by analogy with 7, we have the condition

$\displaystyle U\left[R\left(-\varepsilon_{z}\hat{\mathbf{z}}\right)\right]T\left(-\boldsymbol{\varepsilon}\right)U\left[R\left(\varepsilon_{z}\hat{\mathbf{z}}\right)\right]T\left(\boldsymbol{\varepsilon}\right)=T\left(\frac{1}{2}\varepsilon_{x}\varepsilon_{z}^{2}\hat{\mathbf{x}}+\varepsilon_{x}\varepsilon_{z}\hat{\mathbf{y}}\right) \ \ \ \ \ (22)$

To expand the operators on the LHS and retain terms up to ${\varepsilon_{x}\varepsilon_{z}^{2}}$, we need to expand the rotation operators up to order ${\varepsilon_{z}^{2}}$. Treating the rotation operator as an exponential, this expansion is

$\displaystyle R\left(\varepsilon_{z}\hat{\mathbf{z}}\right)=I-\frac{i\varepsilon_{z}}{\hbar}L_{z}+\frac{i^{2}\varepsilon_{z}^{2}}{2\hbar^{2}}L_{z}^{2}+\ldots \ \ \ \ \ (23)$

Using this approximation gives us

 $\displaystyle \left(I+\frac{i\varepsilon_{z}}{\hbar}L_{z}+\frac{i^{2}\varepsilon_{z}^{2}}{2\hbar^{2}}L_{z}^{2}\right)\left[I+\frac{i}{\hbar}\varepsilon_{x}P_{x}\right]\left(I-\frac{i\varepsilon_{z}}{\hbar}L_{z}+\frac{i^{2}\varepsilon_{z}^{2}}{2\hbar^{2}}L_{z}^{2}\right)\left[I-\frac{i}{\hbar}\varepsilon_{x}P_{x}\right]$ $\displaystyle =$ $\displaystyle I-\frac{i}{\hbar}\left(\frac{1}{2}\varepsilon_{x}\varepsilon_{z}^{2}P_{x}+\varepsilon_{x}\varepsilon_{z}P_{y}\right) \ \ \ \ \ (24)$

By equating the coefficients of ${\varepsilon_{x}\varepsilon_{z}}$ we regain 12, so that condition checks out.

Extracting the coefficient of ${\varepsilon_{x}\varepsilon_{z}^{2}}$ on the LHS gives

 $\displaystyle \frac{i^{3}}{\hbar^{3}}\varepsilon_{x}\varepsilon_{z}^{2}\left(-L_{z}P_{x}L_{z}+\frac{L_{z}^{2}P_{x}}{2}-\frac{L_{z}^{2}P_{x}}{2}+\frac{P_{x}L_{z}^{2}}{2}-\frac{L_{z}^{2}P_{x}}{2}+L_{z}^{2}P_{x}\right)$ $\displaystyle =$ $\displaystyle \frac{i^{3}}{\hbar^{3}}\varepsilon_{x}\varepsilon_{z}^{2}\left(-L_{z}P_{x}L_{z}+\frac{L_{z}^{2}P_{x}}{2}+\frac{P_{x}L_{z}^{2}}{2}\right) \ \ \ \ \ (25)$

Matching this to the ${\varepsilon_{x}\varepsilon_{z}^{2}}$ term on the RHS of 24, we get the condition specified in Shankar’s problem:

$\displaystyle -2L_{z}P_{x}L_{z}+L_{z}^{2}P_{x}+P_{x}L_{z}^{2}=\hbar^{2}P_{x} \ \ \ \ \ (26)$

We can show that this condition reduces to the already-known commutators by using the identity

 $\displaystyle \left[\Lambda,\left[\Lambda,\Omega\right]\right]$ $\displaystyle =$ $\displaystyle \Lambda\left(\Lambda\Omega-\Omega\Lambda\right)-\left(\Lambda\Omega-\Omega\Lambda\right)\Lambda\ \ \ \ \ (27)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -2\Lambda\Omega\Lambda+\Lambda^{2}\Omega+\Omega\Lambda^{2} \ \ \ \ \ (28)$

Applying this to 26 we have

 $\displaystyle -2L_{z}P_{x}L_{z}+L_{z}^{2}P_{x}+P_{x}L_{z}^{2}$ $\displaystyle =$ $\displaystyle \left[L_{z},\left[L_{z},P_{x}\right]\right]\ \ \ \ \ (29)$ $\displaystyle$ $\displaystyle =$ $\displaystyle i\hbar\left[L_{z},P_{y}\right]\ \ \ \ \ (30)$ $\displaystyle$ $\displaystyle =$ $\displaystyle i\hbar\left(-i\hbar P_{x}\right)\ \ \ \ \ (31)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \hbar^{2}P_{x} \ \ \ \ \ (32)$

Thus the more complicated condition 26 actually reduces to existing commutators.

# Translation invariance in two dimensions

Shankar, R. (1994), Principles of Quantum Mechanics, Plenum Press. Chapter 12, Exercise 12.1.1.

In preparation for an examination of rotation invariance, we’ll have a look at translational invariance in two dimensions. We can apply much of what we did with translation in one dimension, where we showed that the momentum ${P}$ is the generator of translations. In particular, the translation operator ${T\left(\varepsilon\right)}$ for an infinitesimal translation ${\varepsilon}$ is

$\displaystyle T\left(\varepsilon\right)=I-\frac{i\varepsilon}{\hbar}P \ \ \ \ \ (1)$

In two dimensions, we can write an infinitesimal translation as ${\boldsymbol{\delta}a}$ where

$\displaystyle \boldsymbol{\delta}a=\delta a_{x}\hat{\mathbf{x}}+\delta a_{y}\hat{\mathbf{y}} \ \ \ \ \ (2)$

In one dimension, we showed earlier that

$\displaystyle \left\langle x\left|T\left(\varepsilon\right)\right|\psi\right\rangle =\psi\left(x-\varepsilon\right) \ \ \ \ \ (3)$

The analogous relation in two dimensions is

$\displaystyle \left\langle x,y\left|T\left(\boldsymbol{\delta}a\right)\right|\psi\right\rangle =\psi\left(x-\delta a_{x},y-\delta a_{y}\right) \ \ \ \ \ (4)$

We can verify that the correct form for ${T\left(\boldsymbol{\delta}a\right)}$ is

 $\displaystyle T\left(\boldsymbol{\delta}a\right)$ $\displaystyle =$ $\displaystyle I-\frac{i}{\hbar}\boldsymbol{\delta}a\cdot\mathbf{P}\ \ \ \ \ (5)$ $\displaystyle$ $\displaystyle =$ $\displaystyle I-\frac{i}{\hbar}\left(\delta a_{x}P_{x}+\delta a_{y}P_{y}\right) \ \ \ \ \ (6)$

Using the representation of momentum in the position basis, which is

 $\displaystyle P_{x}$ $\displaystyle =$ $\displaystyle -i\hbar\frac{\partial}{\partial x}\ \ \ \ \ (7)$ $\displaystyle P_{y}$ $\displaystyle =$ $\displaystyle -i\hbar\frac{\partial}{\partial y} \ \ \ \ \ (8)$

the LHS of 4 is, using ${\left\langle x,y\left|\psi\right.\right\rangle =\psi\left(x,y\right)}$:

 $\displaystyle \left\langle x,y\left|T\left(\boldsymbol{\delta}a\right)\right|\psi\right\rangle$ $\displaystyle =$ $\displaystyle \left\langle x,y\left|I-\frac{i}{\hbar}\left(\delta a_{x}P_{x}+\delta a_{y}P_{y}\right)\right|\psi\right\rangle \ \ \ \ \ (9)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \psi\left(x,y\right)-\delta a_{x}\frac{\partial\psi}{\partial x}-\delta a_{y}\frac{\partial\psi}{\partial y} \ \ \ \ \ (10)$

The last line is also what we get if we expand the RHS of 4 to first order in ${\boldsymbol{\delta}a}$, which verifies that 5 is correct, so that the two-dimensional momentum ${\mathbf{P}}$ is the generator of two-dimensional translations.

We can apply the exponentiation technique we used in the one-dimensional case to obtain the translation operator for a finite translation in two dimensions. We need to be careful that we don’t run into problems with non-commuting operators, but in view of 7 and 8 and the fact that derivatives with respect to different independent variables commute, we see that

$\displaystyle \left[P_{x},P_{y}\right]=0 \ \ \ \ \ (11)$

We can divide a finite translation ${\mathbf{a}}$ into ${N}$ small steps, each of size ${\frac{\mathbf{a}}{N}}$, so that the translation is

$\displaystyle T\left(\mathbf{a}\right)=\left(I-\frac{i}{\hbar N}\mathbf{a}\cdot\mathbf{P}\right)^{N} \ \ \ \ \ (12)$

Because the two components of momentum commute, we can take the limit of this expression to get the exponential form:

$\displaystyle T\left(\mathbf{a}\right)=\lim_{N\rightarrow\infty}\left(I-\frac{i}{\hbar N}\mathbf{a}\cdot\mathbf{P}\right)^{N}=e^{-i\mathbf{a}\cdot\mathbf{P}/\hbar} \ \ \ \ \ (13)$

Again, because the two components of momentum commute, we can combine two translations, by ${\mathbf{a}}$ and then by ${\mathbf{b}}$, to get

$\displaystyle T\left(\mathbf{b}\right)T\left(\mathbf{a}\right)=e^{-i\mathbf{b}\cdot\mathbf{P}/\hbar}e^{-i\mathbf{a}\cdot\mathbf{P}/\hbar}=e^{-i\left(\mathbf{a}+\mathbf{b}\right)\cdot\mathbf{P}/\hbar}=T\left(\mathbf{b}+\mathbf{a}\right) \ \ \ \ \ (14)$

# 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.

# 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.

# Translations in space and time

Required math: calculus

Required physics: Schrödinger equation

References: Griffiths, David J. (2005), Introduction to Quantum Mechanics, 2nd Edition; Pearson Education – Problem 3.39.

Since ${\hat{p}=(\hbar/i)\partial/\partial x}$, we can write the exponential of the momentum operator by using a Taylor expansion:

$\displaystyle e^{i\hat{p}x_{0}/\hbar}=1+x_{0}\frac{\partial}{\partial x}+\frac{x_{0}^{2}}{2!}\frac{\partial^{2}}{\partial x^{2}}+\ldots \ \ \ \ \ (1)$

where ${x_{0}}$ is a constant. If we apply this operator to a function, we get

$\displaystyle e^{i\hat{p}x_{0}/\hbar}f(x)=f(x)+x_{0}\frac{\partial f}{\partial x}+\frac{x_{0}^{2}}{2!}\frac{\partial^{2}f}{\partial x^{2}}+\ldots \ \ \ \ \ (2)$

which is the Taylor expansion of ${f(x+x_{0})}$ about the point ${x}$, provided that the derivatives are evaluated at that point. Since this exponential operator effectively shifts the function by a distance ${x_{0}}$, the operator ${p/\hbar}$ is called the generator of translations in space.

Similarly, for a function ${\Psi(x,t)}$ that satisfies the time-dependent Schrodinger equation, we have

$\displaystyle i\hbar\frac{\partial\Psi(x,t)}{\partial t}=\hat{H}\Psi(x,t) \ \ \ \ \ (3)$

so

$\displaystyle e^{-i\hat{H}t_{0}/\hbar}=1+t_{0}\frac{\partial}{\partial t}+\frac{t_{0}^{2}}{2!}\frac{\partial^{2}}{\partial t^{2}}+\ldots \ \ \ \ \ (4)$

and

 $\displaystyle e^{-i\hat{H}t_{0}/\hbar}\Psi(x,t)$ $\displaystyle =$ $\displaystyle \Psi(x,t)+t_{0}\frac{\partial\Psi}{\partial t}+\frac{t_{0}^{2}}{2!}\frac{\partial^{2}\Psi}{\partial t^{2}}+\ldots\ \ \ \ \ (5)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \Psi(x,t+t_{0}) \ \ \ \ \ (6)$

which is the Taylor expansion with respect to time. The operator ${-H/\hbar}$ is therefore known as the generator of translations in time.

The mean of an observable at time ${t+t_{0}}$ is

 $\displaystyle \langle Q\rangle_{t+t_{0}}$ $\displaystyle =$ $\displaystyle \langle\Psi(t+t_{0})|Q(t+t_{0})|\Psi(t+t_{0})\rangle\ \ \ \ \ (7)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \langle e^{-i\hat{H}t_{0}/\hbar}\Psi(t)|Q(t+t_{0})|e^{-i\hat{H}t_{0}/\hbar}\Psi(t)\rangle\ \ \ \ \ (8)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \langle\Psi(t)|e^{i\hat{H}t_{0}/\hbar}Q(t+t_{0})e^{-i\hat{H}t_{0}/\hbar}|\Psi(t)\rangle \ \ \ \ \ (9)$

Expanding the two exponentials and the operator to first order in ${t_{0}}$, we get

 $\displaystyle \langle Q\rangle_{t+t_{0}}$ $\displaystyle =$ $\displaystyle \left\langle \Psi(t)|\left(1+\frac{i}{\hbar}t_{0}\hat{H}\right)\left(Q(t)+t_{0}\frac{\partial Q}{\partial t}\right)\left(1-\frac{i}{\hbar}t_{0}\hat{H}\right)|\Psi(t)\right\rangle \ \ \ \ \ (10)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \left\langle \Psi(t)|Q(t)+t_{0}\frac{\partial Q}{\partial t}+\frac{i}{\hbar}t_{0}(\hat{H}Q(t)-Q(t)\hat{H})|\Psi(t)\right\rangle +\mathcal{O}(t_{0}^{2})\ \ \ \ \ (11)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \langle Q\rangle_{t}+\left\langle \frac{\partial Q}{\partial t}\right\rangle t_{0}+\frac{i}{\hbar}t_{0}\langle[\hat{H},Q]\rangle+\mathcal{O}(t_{0}^{2}) \ \ \ \ \ (12)$

where ${\mathcal{O}(t_{0}^{2})}$ represents terms of second and higher order in ${t_{0}}$. Moving ${\langle Q\rangle_{t}}$ to the LHS, dividing through by ${t_{0}}$ and taking the limit as ${t_{0}\rightarrow0}$ gives us the relation we got while studying the energy-time uncertainty principle:

$\displaystyle \frac{d\langle Q\rangle}{dt}=\left\langle \frac{\partial Q}{\partial t}\right\rangle +\frac{i}{\hbar}\langle[\hat{H},Q]\rangle \ \ \ \ \ (13)$