I am reading this reference. In this article, the author define the Heisenberg-Weyl operators (Eqs. [148] in the reference):

$$X(s)=e^{-is\hat{p}/2} \tag{1},$$ $$Z(t)=e^{it\hat{q}/2} \tag{2},$$

where $s,t \in \mathbb{R}$, and $\hat{q}, \hat{p}$ are the canonical pair (position ans momentum quadratures of an electromagnetic field) whose eigenstates $\left| q\right>$ and $\left| p\right>$ are conected through a Fourier transform (see Eqs. [9] in the reference), that is,

$$\left| q\right>=(2\sqrt{\pi})^{-1}\int_{-\infty}^{+\infty}dp~e^{-iqp/2} \left| p\right>\tag{3},$$ $$\left| p\right>=(2\sqrt{\pi})^{-1}\int_{-\infty}^{+\infty}dq~e^{+iqp/2} \left| q\right>\tag{4}$$

Then, the author establishes the action of the operators $X(s)$ and $Z(t)$ on the quadrature eigenstates as (Eqs. [149] and [150] in the article)

$$X(s)\left| q\right>=\left| q + s\right>, \tag{5}$$ $$Z(t)\left| q\right>=e^{itq/2}\left| q \right>, \tag{6}$$ $$X(s)\left| p\right>=e^{-isp/2}\left| p\right>, \tag{7}$$ $$Z(t)\left| p\right>=\left| p + t \right>. \tag{8}$$

Then, I proceed to give a proof for Eq. (5) as follows

$$\begin{aligned} X(s)\left| q\right>=&(2\sqrt{\pi})^{-1} e^{-is\hat{p}/2}\int_{-\infty}^{+\infty}dp~e^{-iqp/2} \left| p\right>\\ =& (2\sqrt{\pi})^{-1} \int_{-\infty}^{+\infty}dp~e^{-iqp/2} e^{-is\hat{p}/2}\left| p\right> \\ =&(2\sqrt{\pi})^{-1} \int_{-\infty}^{+\infty}dp~e^{-iqp/2} \sum_{k=0}^{\infty} \frac{1}{k!}\left( \frac{is}{2}\right)^k \hat{p}^{k}\left| p\right>\\ =&(2\sqrt{\pi})^{-1} \int_{-\infty}^{+\infty}dp~e^{-iqp/2} \sum_{k=0}^{\infty} \frac{1}{k!}\left( \frac{is}{2}\right)^k p^{k}\left| p\right>\\ =& (2\sqrt{\pi})^{-1} \int_{-\infty}^{+\infty}dp~e^{-iqp/2} e^{-isp/2}\left| p\right> \\ =& (2\sqrt{\pi})^{-1} \int_{-\infty}^{+\infty}dp~e^{-i(q+s)p/2} \left| p \right> \\ =& \left| q + s \right>. \end{aligned} $$ where in the first line I expand $\left| q \right>$ in the momentum basis as Eq. (3) establishes and in the third line I expand the exponential in McLaurin series.

For the proof of Eq. (8), I give a similar procedure:

$$ \begin{aligned} Z(t)\left| p\right>=&(2\sqrt{\pi})^{-1} e^{it\hat{q}/2}\int_{-\infty}^{+\infty}dp~e^{iqp/2} \left| q\right>\\ =&(2\sqrt{\pi})^{-1} \int_{-\infty}^{+\infty}dp~e^{iqp/2}e^{it\hat{q}/2} \left| q\right>\\ =&(2\sqrt{\pi})^{-1} \int_{-\infty}^{+\infty}dp~e^{iqp/2} \sum_{k=0}^{\infty}\frac{1}{k!}\left(\frac{it}{2} \right)^{k} \hat{q}^k\left| q\right>\\ =&(2\sqrt{\pi})^{-1} \int_{-\infty}^{+\infty}dp~e^{iqp/2} \sum_{k=0}^{\infty}\frac{1}{k!}\left(\frac{it}{2} \right)^{k} q^k\left| q\right>\\ =&(2\sqrt{\pi})^{-1} \int_{-\infty}^{+\infty}dp~e^{iqp/2}e^{itq/2} \left| q\right>\\ =&(2\sqrt{\pi})^{-1} \int_{-\infty}^{+\infty}dp~e^{iq(p + t)/2} \left| q\right>\\ =&\left| p + t\right>. \end{aligned} $$

Now my question: I want to apply, the $Z(t)$ operator to an arbitrary state $\left|\psi\right>$ expanded in the momentum basis, and, on the other hand, I want to apply the $X(s)$ operator to the same state $\left|\psi\right>$ but expanded in the position basis, that is,

$$ Z(t)\left|\psi(q)\right>=e^{it\hat{q}/2} \int_{-\infty}^{+\infty}dp \psi(p) \left|p\right>, \tag{9} $$ $$ X(s)\left|\psi(p)\right>=e^{-is\hat{p}/2} \int_{-\infty}^{+\infty}dq \psi(q) \left|q\right>, \tag{10} $$

So, what is the result that I should get?

if I follow the same steps for the proofs of $X(s)\left| q\right>$ and $Z(t)\left| p\right>$ I think that the result should be

$$ Z(t)\left|\psi(q)\right>= \int_{-\infty}^{+\infty}dp \psi(p) \left|p + t\right>, \tag{11} $$ $$ X(s)\left|\psi(p)\right>= \int_{-\infty}^{+\infty}dp \psi(q) \left|q + s\right>, \tag{12} $$

However I am not sure if I should also shift the arguments of the $\psi$ functions inside the integral, that is,

$$ Z(t)\left|\psi(q)\right>= \int_{-\infty}^{+\infty}dp \psi(p+t) \left|p + t\right>, \tag{13} $$ $$ X(s)\left|\psi(p)\right>= \int_{-\infty}^{+\infty}dp \psi(q+s) \left|q + s\right>. \tag{14} $$

So, could someone give me a hint to solve my question?

  • $\begingroup$ You may be getting confused about states and representations. Any state $|\psi\rangle$ can be written in both the position and momentum bases as $|\psi\rangle =\int dp \psi(p)|p\rangle=\int dq \psi(q)|q\rangle$, where sometimes people use different symbols for the wavefunctions to make things more clear [e.g., $|\psi\rangle =\int dp \psi(p)|p\rangle=\int dq \tilde{\psi}(q)|q\rangle$]. $\endgroup$ Sep 4, 2021 at 1:12

1 Answer 1


Let's expand: \begin{aligned} Z(t)|\psi\rangle=Z(t)\int_{-\infty}^\infty dp\, \psi(p)|p\rangle=\int_{-\infty}^\infty dp \,\psi(p)|p+t\rangle=\int_{-\infty}^\infty dp \,\psi(p-t)|p\rangle. \end{aligned} This could have also been written in the position basis as \begin{aligned} Z(t)|\psi\rangle=Z(t)\int_{-\infty}^\infty dq \,\tilde{\psi}(q)|q\rangle=\int_{-\infty}^\infty dq\, e^{itq/2}\tilde{\psi}(q)|q\rangle, \end{aligned} where I distinguish the position-space wavefunction using a tilde: $\tilde{\psi}(q)$. Similarly, \begin{aligned} X(s)|\psi\rangle=X(s)\int_{-\infty}^\infty dq\, \tilde{\psi}(q)|q\rangle=\int_{-\infty}^\infty dq\, \tilde{\psi}(q)|q+s\rangle=\int_{-\infty}^\infty dq\, \tilde{\psi}(q-s)|q\rangle \end{aligned} and \begin{aligned} X(s)|\psi\rangle=X(s)\int_{-\infty}^\infty dq \,{\psi}(p)|p\rangle=\int_{-\infty}^\infty dq\, e^{-isp/2}{\psi}(p)|p\rangle. \end{aligned} So all of your calculations were initially correct, but then the shift inside the integral was incorrect (it is just a change of variables, like defining $k\equiv p+s$, $dk=dp$, using substitution, etc.). The crux is to not write the argument of the wavefunction in the ket, because the ket exists regardless of the basis in which you expand it, so we can always write things like $$|\psi\rangle =\int_{-\infty}^\infty dp\, \psi(p)|p\rangle=\int_{-\infty}^\infty dq\, \tilde{\psi}(q)|q\rangle.$$

  • $\begingroup$ Thanks a lot @Quantum Mechanic $\endgroup$ Sep 5, 2021 at 0:02
  • $\begingroup$ @JulioAbrahamMendozaFierro glad to help! $\endgroup$ Sep 5, 2021 at 16:53

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