## Commutators for some symmetry operators

December 16, 2015 phy1520 , , ,

### Q: [1] pr 4.2

If $$\mathcal{T}_\Bd$$, $$\mathcal{D}(\ncap, \phi)$$, and $$\pi$$ denote the translation, rotation, and parity operators respectively. Which of the following commute and why

• (a) $$\mathcal{T}_\Bd$$ and $$\mathcal{T}_{\Bd’}$$, translations in different directions.
• (b) $$\mathcal{D}(\ncap, \phi)$$ and $$\mathcal{D}(\ncap’, \phi’)$$, rotations in different directions.
• (c) $$\mathcal{T}_\Bd$$ and $$\pi$$.
• (d) $$\mathcal{D}(\ncap,\phi)$$ and $$\pi$$.

### A: (a)

Consider
\label{eqn:symmetryOperatorCommutators:20}
\begin{aligned}
\mathcal{T}_\Bd \mathcal{T}_{\Bd’} \ket{\Bx}
&=
\mathcal{T}_\Bd \ket{\Bx + \Bd’} \\
&=
\ket{\Bx + \Bd’ + \Bd},
\end{aligned}

and the reverse application of the translation operators
\label{eqn:symmetryOperatorCommutators:40}
\begin{aligned}
\mathcal{T}_{\Bd’} \mathcal{T}_{\Bd} \ket{\Bx}
&=
\mathcal{T}_{\Bd’} \ket{\Bx + \Bd} \\
&=
\ket{\Bx + \Bd + \Bd’} \\
&=
\ket{\Bx + \Bd’ + \Bd}.
\end{aligned}

so we see that

\label{eqn:symmetryOperatorCommutators:60}
\antisymmetric{\mathcal{T}_\Bd}{\mathcal{T}_{\Bd’}} \ket{\Bx} = 0,

for any position state $$\ket{\Bx}$$, and therefore in general they commute.

### A: (b)

That rotations do not commute when they are in different directions (like any two orthogonal directions) need not be belaboured.

### A: (c)

We have
\label{eqn:symmetryOperatorCommutators:80}
\begin{aligned}
\mathcal{T}_\Bd \pi \ket{\Bx}
&=
\mathcal{T}_\Bd \ket{-\Bx} \\
&=
\ket{-\Bx + \Bd},
\end{aligned}

yet
\label{eqn:symmetryOperatorCommutators:100}
\begin{aligned}
\pi \mathcal{T}_\Bd \ket{\Bx}
&=
\pi \ket{\Bx + \Bd} \\
&=
\ket{-\Bx – \Bd} \\
&\ne
\ket{-\Bx + \Bd}.
\end{aligned}

so, in general $$\antisymmetric{\mathcal{T}_\Bd}{\pi} \ne 0$$.

### A: (d)

We have

\label{eqn:symmetryOperatorCommutators:120}
\begin{aligned}
\pi \mathcal{D}(\ncap, \phi) \ket{\Bx}
&=
\pi \mathcal{D}(\ncap, \phi) \pi^\dagger \pi \ket{\Bx} \\
&=
\pi \mathcal{D}(\ncap, \phi) \pi^\dagger \pi \ket{\Bx} \\
&=
\pi \lr{ \sum_{k=0}^\infty \frac{(-i \BJ \cdot \ncap)^k}{k!} } \pi^\dagger \pi \ket{\Bx} \\
&=
\sum_{k=0}^\infty \frac{(-i (\pi \BJ \pi^\dagger) \cdot (\pi \ncap \pi^\dagger) )^k}{k!} \pi \ket{\Bx} \\
&=
\sum_{k=0}^\infty \frac{(-i \BJ \cdot \ncap)^k}{k!} \pi \ket{\Bx} \\
&=
\mathcal{D}(\ncap, \phi) \pi \ket{\Bx},
\end{aligned}

so $$\antisymmetric{\mathcal{D}(\ncap, \phi)}{\pi} \ket{\Bx} = 0$$, for any position state $$\ket{\Bx}$$, and therefore these operators commute in general.

# References

[1] Jun John Sakurai and Jim J Napolitano. Modern quantum mechanics. Pearson Higher Ed, 2014.

## SHO translation operator expectation

September 2, 2015 phy1520 , , , ,

## Question: SHO translation operator expectation ([1] pr. 2.12)

Using the Heisenberg picture evaluate the expectation of the position operator $$\expectation{x}$$ with respect to the initial time state

\label{eqn:translationExpectation:20}
\ket{\alpha, 0} = e^{-i p_0 a/\Hbar} \ket{0},

where $$p_0$$ is the initial time position operator, and $$a$$ is a constant with dimensions of position.

Recall that the Heisenberg picture position operator expands to

\label{eqn:translationExpectation:40}
x^{\textrm{H}}(t)
= U^\dagger x U
= x_0 \cos(\omega t) + \frac{p_0}{m \omega} \sin(\omega t),

so the expectation of the position operator is
\label{eqn:translationExpectation:60}
\begin{aligned}
\expectation{x}
&=
\bra{0} e^{i p_0 a/\Hbar} \lr{ x_0 \cos(\omega t) + \frac{p_0}{m \omega}
\sin(\omega t) } e^{-i p_0 a/\Hbar} \ket{0} \\
&=
\bra{0} \lr{ e^{i p_0 a/\Hbar} x_0 \cos(\omega t) e^{-i p_0 a/\Hbar} \cos(\omega t) + \frac{p_0}{m \omega} \sin(\omega t) } \ket{0}.
\end{aligned}

The exponential sandwich above can be expanded using the Baker-Campbell-Hausdorff [2] formula

\label{eqn:translationExpectation:80}
\begin{aligned}
e^{i p_0 a/\Hbar} x_0 e^{-i p_0 a/\Hbar}
&=
x_0
+ \frac{i a}{\Hbar} \antisymmetric{p_0}{x_0}
+ \inv{2!} \lr{\frac{i a}{\Hbar}}^2 \antisymmetric{p_0}{\antisymmetric{p_0}{x_0}}
+ \cdots \\
&=
x_0
+ \frac{i a}{\Hbar} \lr{ -i \Hbar }
+ \inv{2!} \lr{\frac{i a}{\Hbar}}^2 \antisymmetric{p_0}{-i \Hbar}
+ \cdots \\
&=
x_0 + a.
\end{aligned}

The position expectation with respect to this translated state is

\label{eqn:translationExpectation:100}
\begin{aligned}
\expectation{x}
&= \bra{0} \lr{ (x_0 + a)\cos(\omega t) + \frac{p_0}{m \omega} \sin(\omega t)
}\ket{0} \\
&= a \cos(\omega t).
\end{aligned}

The final simplification above follows from $$\bra{n} x \ket{n} = \bra{n} p \ket{n} = 0$$.

# References

[1] Jun John Sakurai and Jim J Napolitano. Modern quantum mechanics. Pearson Higher Ed, 2014.

[2] Wikipedia. Baker-campbell-hausdorff formula — wikipedia, the free encyclopedia, 2015. URL https://en.wikipedia.org/w/index.php?title=Baker\%E2\%80\%93Campbell\%E2\%80\%93Hausdorff_formula&oldid=665123858. [Online; accessed 16-August-2015].

## (a)

Evaluate the classical Poisson bracket

\label{eqn:translation:420}
\antisymmetric{x}{F(p)}_{\textrm{classical}}

## (b)

Evaluate the commutator

\label{eqn:translation:440}
\antisymmetric{x}{e^{i p a/\Hbar}}

## (c)

Using the result in \ref{problem:translation:28:b}, prove that
\label{eqn:translation:460}
e^{i p a/\Hbar} \ket{x’},

is an eigenstate of the coordinate operator $$x$$.

## (a)

\label{eqn:translation:480}
\begin{aligned}
\antisymmetric{x}{F(p)}_{\textrm{classical}}
&=
\PD{x}{x} \PD{p}{F(p)} – \PD{p}{x} \PD{x}{F(p)} \\
&=
\PD{p}{F(p)}.
\end{aligned}

## (b)

Having worked backwards through these problems, the answer for this one dimensional problem can be obtained from \ref{eqn:translation:140} and is

\label{eqn:translation:500}
\antisymmetric{x}{e^{i p a/\Hbar}} = a e^{i p a/\Hbar}.

## (c)

\label{eqn:translation:520}
\begin{aligned}
x e^{i p a/\Hbar} \ket{x’}
&=
\lr{
\antisymmetric{x}{e^{i p a/\Hbar}}
e^{i p a/\Hbar} x
+
}
\ket{x’} \\
&=
\lr{ a e^{i p a/\Hbar} + e^{i p a/\Hbar} x ‘ } \ket{x’} \\
&= \lr{ a + x’ } \ket{x’}.
\end{aligned}

This demonstrates that $$e^{i p a/\Hbar} \ket{x’}$$ is an eigenstate of $$x$$ with eigenvalue $$a + x’$$.

## (a)

For power series $$F, G$$, verify

\label{eqn:translation:180}
\antisymmetric{x_k}{G(\Bp)} = i \Hbar \PD{p_k}{G}, \qquad
\antisymmetric{p_k}{F(\Bx)} = -i \Hbar \PD{x_k}{F}.

## (b)

Evaluate $$\antisymmetric{x^2}{p^2}$$, and compare to the classical Poisson bracket $$\antisymmetric{x^2}{p^2}_{\textrm{classical}}$$.

## (a)

Let

\label{eqn:translation:200}
\begin{aligned}
G(\Bp) &= \sum_{k l m} a_{k l m} p_1^k p_2^l p_3^m \\
F(\Bx) &= \sum_{k l m} b_{k l m} x_1^k x_2^l x_3^m.
\end{aligned}

It is simpler to work with a specific $$x_k$$, say $$x_k = y$$. The validity of the general result will still be clear doing so. Expanding the commutator gives

\label{eqn:translation:220}
\begin{aligned}
\antisymmetric{y}{G(\Bp)}
&=
\sum_{k l m} a_{k l m} \antisymmetric{y}{p_1^k p_2^l p_3^m } \\
&=
\sum_{k l m} a_{k l m} \lr{
y p_1^k p_2^l p_3^m – p_1^k p_2^l p_3^m y
} \\
&=
\sum_{k l m} a_{k l m} \lr{
p_1^k y p_2^l p_3^m – p_1^k y p_2^l p_3^m
} \\
&=
\sum_{k l m} a_{k l m}
p_1^k
\antisymmetric{y}{p_2^l}
p_3^m.
\end{aligned}

From \ref{eqn:translation:100}, we have $$\antisymmetric{y}{p_2^l} = l i \Hbar p_2^{l-1}$$, so

\label{eqn:translation:240}
\begin{aligned}
\antisymmetric{y}{G(\Bp)}
&=
\sum_{k l m} a_{k l m}
p_1^k
\antisymmetric{y}{p_2^l}
\lr{ l
i \Hbar p_2^{l-1}
}
p_3^m \\
&=
i \Hbar \PD{y}{G(\Bp)}.
\end{aligned}

It is straightforward to show that
$$\antisymmetric{p}{x^l} = -l i \Hbar x^{l-1}$$, allowing for a similar computation of the momentum commutator

\label{eqn:translation:260}
\begin{aligned}
\antisymmetric{p_y}{F(\Bx)}
&=
\sum_{k l m} b_{k l m} \antisymmetric{p_y}{x_1^k x_2^l x_3^m } \\
&=
\sum_{k l m} b_{k l m} \lr{
p_y x_1^k x_2^l x_3^m – x_1^k x_2^l x_3^m p_y
} \\
&=
\sum_{k l m} b_{k l m} \lr{
x_1^k p_y x_2^l x_3^m – x_1^k p_y x_2^l x_3^m
} \\
&=
\sum_{k l m} b_{k l m}
x_1^k
\antisymmetric{p_y}{x_2^l}
x_3^m \\
&=
\sum_{k l m} b_{k l m}
x_1^k
\lr{ -l i \Hbar x_2^{l-1}}
x_3^m \\
&=
-i \Hbar \PD{p_y}{F(\Bx)}.
\end{aligned}

## (b)

It isn’t clear to me how the results above can be used directly to compute $$\antisymmetric{x^2}{p^2}$$. However, when the first term of such a commutator is a monomomial, it can be expanded in terms of an $$x$$ commutator

\label{eqn:translation:280}
\begin{aligned}
\antisymmetric{x^2}{G(\Bp)}
&= x^2 G – G x^2 \\
&= x \lr{ x G } – G x^2 \\
&= x \lr{ \antisymmetric{ x }{ G } + G x } – G x^2 \\
&= x \antisymmetric{ x }{ G } + \lr{ x G } x – G x^2 \\
&= x \antisymmetric{ x }{ G } + \lr{ \antisymmetric{ x }{ G } + G x } x – G x^2 \\
&= x \antisymmetric{ x }{ G } + \antisymmetric{ x }{ G } x.
\end{aligned}

Similarily,

\label{eqn:translation:300}
\antisymmetric{x^3}{G(\Bp)} = x^2 \antisymmetric{ x }{ G } + x \antisymmetric{ x }{ G } x + \antisymmetric{ x }{ G } x^2.

An induction hypothesis can be formed

\label{eqn:translation:320}
\antisymmetric{x^k}{G(\Bp)} = \sum_{j = 0}^{k-1} x^{k-1-j} \antisymmetric{ x }{ G } x^j,

and demonstrated

\label{eqn:translation:340}
\begin{aligned}
\antisymmetric{x^{k+1}}{G(\Bp)}
&=
x^{k+1} G – G x^{k+1} \\
&=
x \lr{ x^{k} G } – G x^{k+1} \\
&=
x \lr{ \antisymmetric{x^{k}}{G} + G x^k } – G x^{k+1} \\
&=
x \antisymmetric{x^{k}}{G} + \lr{ x G } x^k – G x^{k+1} \\
&=
x \antisymmetric{x^{k}}{G} + \lr{ \antisymmetric{x}{G} + G x } x^k – G x^{k+1} \\
&=
x \antisymmetric{x^{k}}{G} + \antisymmetric{x}{G} x^k \\
&=
x \sum_{j = 0}^{k-1} x^{k-1-j} \antisymmetric{ x }{ G } x^j + \antisymmetric{x}{G} x^k \\
&=
\sum_{j = 0}^{k-1} x^{(k+1)-1-j} \antisymmetric{ x }{ G } x^j + \antisymmetric{x}{G} x^k \\
&=
\sum_{j = 0}^{k} x^{(k+1)-1-j} \antisymmetric{ x }{ G } x^j.
\end{aligned}

That was a bit overkill for this problem, but may be useful later. Application of this to the problem gives

\label{eqn:translation:360}
\begin{aligned}
\antisymmetric{x^2}{p^2}
&=
x \antisymmetric{x}{p^2}
+ \antisymmetric{x}{p^2} x \\
&=
x i \Hbar \PD{x}{p^2}
+ i \Hbar \PD{x}{p^2} x \\
&=
x 2 i \Hbar p
+ 2 i \Hbar p x \\
&= i \Hbar \lr{ 2 x p + 2 p x }.
\end{aligned}

The classical commutator is
\label{eqn:translation:380}
\begin{aligned}
\antisymmetric{x^2}{p^2}_{\textrm{classical}}
&=
\PD{x}{x^2} \PD{p}{p^2} – \PD{p}{x^2} \PD{x}{p^2} \\
&=
2 x 2 p \\
&= 2 x p + 2 p x.
\end{aligned}

This demonstrates the expected relation between the classical and quantum commutators

\label{eqn:translation:400}
\antisymmetric{x^2}{p^2} = i \Hbar \antisymmetric{x^2}{p^2}_{\textrm{classical}}.

## Question: Translation operator and position expectation. ([1] pr. 1.30)

The translation operator for a finite spatial displacement is given by

\label{eqn:translation:20}
J(\Bl) = \exp\lr{ -i \Bp \cdot \Bl/\Hbar },

where $$\Bp$$ is the momentum operator.

## (a)

Evaluate

\label{eqn:translation:40}
\antisymmetric{x_i}{J(\Bl)}.

## (b)

Demonstrate how the expectation value $$\expectation{\Bx}$$ changes under translation.

## (a)

For clarity, let’s set $$x_i = y$$. The general result will be clear despite doing so.

\label{eqn:translation:60}
\antisymmetric{y}{J(\Bl)}
=
\sum_{k= 0} \inv{k!} \lr{\frac{-i}{\Hbar}}
\antisymmetric{y}{
\lr{ \Bp \cdot \Bl }^k
}.

The commutator expands as

\label{eqn:translation:80}
\begin{aligned}
\antisymmetric{y}{
\lr{ \Bp \cdot \Bl }^k
}
+ \lr{ \Bp \cdot \Bl }^k y
&=
y \lr{ \Bp \cdot \Bl }^k \\
&=
y \lr{ p_x l_x + p_y l_y + p_z l_z } \lr{ \Bp \cdot \Bl }^{k-1} \\
&=
\lr{ p_x l_x y + y p_y l_y + p_z l_z y } \lr{ \Bp \cdot \Bl }^{k-1} \\
&=
\lr{ p_x l_x y + l_y \lr{ p_y y + i \Hbar } + p_z l_z y } \lr{ \Bp \cdot \Bl }^{k-1} \\
&=
\lr{ \Bp \cdot \Bl } y \lr{ \Bp \cdot \Bl }^{k-1}
+ i \Hbar l_y \lr{ \Bp \cdot \Bl }^{k-1} \\
&= \cdots \\
&=
\lr{ \Bp \cdot \Bl }^{k-1} y \lr{ \Bp \cdot \Bl }^{k-(k-1)}
+ (k-1) i \Hbar l_y \lr{ \Bp \cdot \Bl }^{k-1} \\
&=
\lr{ \Bp \cdot \Bl }^{k} y
+ k i \Hbar l_y \lr{ \Bp \cdot \Bl }^{k-1}.
\end{aligned}

In the above expansion, the commutation of $$y$$ with $$p_x, p_z$$ has been used. This gives, for $$k \ne 0$$,

\label{eqn:translation:100}
\antisymmetric{y}{
\lr{ \Bp \cdot \Bl }^k
}
=
k i \Hbar l_y \lr{ \Bp \cdot \Bl }^{k-1}.

Note that this also holds for the $$k = 0$$ case, since $$y$$ commutes with the identity operator. Plugging back into the $$J$$ commutator, we have

\label{eqn:translation:120}
\begin{aligned}
\antisymmetric{y}{J(\Bl)}
&=
\sum_{k = 1} \inv{k!} \lr{\frac{-i}{\Hbar}}
k i \Hbar l_y \lr{ \Bp \cdot \Bl }^{k-1} \\
&=
l_y \sum_{k = 1} \inv{(k-1)!} \lr{\frac{-i}{\Hbar}}
\lr{ \Bp \cdot \Bl }^{k-1} \\
&=
l_y J(\Bl).
\end{aligned}

The same pattern clearly applies with the other $$x_i$$ values, providing the desired relation.

\label{eqn:translation:140}
\antisymmetric{\Bx}{J(\Bl)} = \sum_{m = 1}^3 \Be_m l_m J(\Bl) = \Bl J(\Bl).

## (b)

Suppose that the translated state is defined as $$\ket{\alpha_{\Bl}} = J(\Bl) \ket{\alpha}$$. The expectation value with respect to this state is

\label{eqn:translation:160}
\begin{aligned}
\expectation{\Bx’}
&=
\bra{\alpha_{\Bl}} \Bx \ket{\alpha_{\Bl}} \\
&=
\bra{\alpha} J^\dagger(\Bl) \Bx J(\Bl) \ket{\alpha} \\
&=
\bra{\alpha} J^\dagger(\Bl) \lr{ \Bx J(\Bl) } \ket{\alpha} \\
&=
\bra{\alpha} J^\dagger(\Bl) \lr{ J(\Bl) \Bx + \Bl J(\Bl) } \ket{\alpha} \\
&=
\bra{\alpha} J^\dagger J \Bx + \Bl J^\dagger J \ket{\alpha} \\
&=
\bra{\alpha} \Bx \ket{\alpha} + \Bl \braket{\alpha}{\alpha} \\
&=
\expectation{\Bx} + \Bl.
\end{aligned}

# References

[1] Jun John Sakurai and Jim J Napolitano. Modern quantum mechanics. Pearson Higher Ed, 2014.