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Show that the operator $U(\alpha, \beta) = e^{i(\alpha \hat{x}^2 + \beta \hat{p}_{x}^2)}$ can represent the space reflection of the 1D Galilei group: $x \to -x; t \to t$.

I don't really know anything about group theory and the concept of "representing" something isn't clear. It sort of makes sense in the case where you have rotations in 3D and you have matrices that "represent" the rotations but I don't know what to do in this case. I would think showing that $\hat{x} [U(\alpha, \beta) | x' \rangle] = -x' [U(\alpha, \beta)|x' \rangle]$ would do it, but that doesn't even seem to be true.

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  • $\begingroup$ I think that $t \to t$ should be replaced by $p\to -p$. So you are asked to look for a choice of $\alpha$ $\beta$ producing a unitary operator whose square is the identity up to a phase and such that it implements the said transformations of $x$ and $p$. The answer is clearly positive for physical reasons, fixing identical the two parameters of the form $i t$. For some $t$, $x$and $p$ are reversed because $U(it,it)$ is the evolutor of a quantum oscillator. As $U(it,it)U(it,it)$, for that $t$, leaves fixed $x$ and $p$, it must be a $ c$ number due to the Schur lemma... $\endgroup$ – Valter Moretti Oct 31 '14 at 7:14
  • $\begingroup$ Sorry, replace $it$ for $t$ in my comment, I missed the presence of a $i$ in the exponent of your formula of $U$. $\endgroup$ – Valter Moretti Oct 31 '14 at 7:24
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I expand my comment into an answer. The idea is to fix $\alpha, \beta \in \mathbb R$ in order that, if $P:= U(\alpha, \beta)$ (which is automatically unitary), we have

(i) $PP=e^{ik}I$ for some $k\in \mathbb R$,

(ii) $P\hat{x}P^\dagger = -\hat{x}$,

(iii) $P\hat{p}P^\dagger = -\hat{p}$

Since $\hat{x}$ and $\hat{p}$ has to be treated symmetrically, we assume $\alpha=\beta = t$. From (i) and (ii), it arises $PP\hat{x}= \hat{x}PP$ and $PP\hat{p}= \hat{p}PP$. Since the representation of the CCR is irreducible (actually one should deal with the associate Weyl algebra, but it is irrelevant here), Schur lemma implies that $PP= cI$ for some complex $c$. Since $PP$ is unitary (i) must hold automatically, if (ii) and (iii) are satisfied. However it is technically more convenient using (iii) in the following to fix $t$ and next check if that choice satisfies (ii) and (iii).

The Hamiltonian $H = \hat{x}^2 +\hat{p}^2$ has spectrum ($\hbar=1$) $$E_n = 2n+1\:, \quad n=0,1,\ldots$$ So that $$e^{it H} = e^{it}\sum_{n=0}^{+\infty} e^{i2nt}|n\rangle \langle n|$$ Let us fix $t$ such that (i) is valid. Condition (i) reads here $$e^{it H}e^{it H} = e^{2it}\sum_{n=0}^{+\infty} e^{i4nt}|n\rangle \langle n|= e^{ik}\sum_{n=0}^{+\infty} |n\rangle \langle n|$$ This is possible only if $t= t_m= \frac{\pi m}{2}$ for some $m=0,1,\ldots$. However, for $m=0,2,4,\ldots$ one has $$e^{it_mH}= I\:,$$ which trivially does not satisfies (ii) and (iii). It remains the case $m=1,3,\ldots$ which produces $$e^{it_m H} = -\sum_{n=0}^{+\infty} e^{in \pi}|n\rangle \langle n|\:.$$ Therefore, our candidate is the one for $t=\pi/2$ (the other choices produce the same result): $$P:= U(\pi/2,\pi/2) = e^{\frac{i\pi}{2}(\hat{x}^2+ \hat{p}^2)}= -\sum_{n=0}^{+\infty} e^{in \pi}|n\rangle \langle n| = \sum_{n=0}^{+\infty} (-1)^{n+1}|n\rangle \langle n|\:.$$ With this choice $PP=I$. It is worth noticing that, with this definition, $P$ would be also self-adjoint that is an observable.

To check if (i) and (ii) are true define $$\hat{x}(t) = e^{itH}\hat{x}e^{-itH} \quad \mbox{and}\quad \hat{p}(t) = e^{itH}\hat{p}e^{-itH}$$ Using twice commutation relations you easily see that $$\frac{d^2\hat{x}}{dt^2} + 4 \hat{x}(t)=0 \quad \mbox{and} \quad \frac{d^2\hat{p}}{dt^2} + 4 \hat{p}(t)=0\:.$$ The solutions are (using the fact that $d\hat{x}/dt = 2\hat{p}(t)$ and $d\hat{p}/dt = -2\hat{x}(t)$ again from CCR) $$\hat{x}(t) = \hat{x} \cos(2t) + \hat{p} \sin(2t) \quad \mbox{and} \quad \hat{p}(t) = \hat{p} \cos(2t) - \hat{x} \sin(2t)\:.$$ You see that, indeed, $$\hat{x}\left(\frac{\pi}{2}\right)= -\hat{x} \quad \mbox{and}\quad \hat{p}\left(\frac{\pi}{2}\right)= -\hat{p}$$ which are equivalent to (ii) and (iii) for $P:= U(\pi/2,\pi/2)$.

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  • $\begingroup$ Thanks. This seems to work, but how come if I just start by computing $U \hat{x} U^{\dagger}$ with the original $U(\alpha, \beta)$ I get $2 \hbar \beta \hat{p}_{x} - \hat{x}$ ? $\endgroup$ – Mr. G Oct 31 '14 at 13:12
  • $\begingroup$ Sorry I did not understand. Are you asking how to explicitly compute $U(\alpha,\beta)x U(\alpha, \beta)^\dagger$? $\endgroup$ – Valter Moretti Oct 31 '14 at 14:36
  • $\begingroup$ Essentially, yes. It doesn't seem to work for me. I wrote $xU^{\dagger} = [x, U^{\dagger}] - U^{\dagger}x$ and used $[x, F(p)]$ $\endgroup$ – Mr. G Oct 31 '14 at 14:52
  • $\begingroup$ The action of a symmetry, represented by a unitary (or antiunitary) operator $U$, on an observable $A$ is by defintion $A \to U^\dagger A U$ (or $UAU^\dagger$ depending on the preferred convention). I do not understand well instead how you implement the action of a symmetry... $\endgroup$ – Valter Moretti Oct 31 '14 at 14:57
  • $\begingroup$ I see I think. Start for the action on vectors which is $U|\psi\rangle$. In our case requiring that $\hat{x}U |x\rangle = -x U|x\rangle$ is equivalent to require $U^\dagger\hat{x}U |x\rangle = -x|x\rangle$, which in turn means $U^\dagger\hat{x}U = -\hat{x}$ because the vectors $|x\rangle$ form a basis and $-x|x\rangle = -\hat{x}|x\rangle$ $\endgroup$ – Valter Moretti Oct 31 '14 at 15:02

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