Show that if the wave function $\langle x|\psi\rangle$ is modified by a position-dependent phase $\langle x|\psi\rangle \to e^{\frac{ip_ox}{\hbar}}\langle x|\psi\rangle$ then $\langle x\rangle \to \langle x \rangle$ and $\langle p_x\rangle \to \langle p_x\rangle +p_o.$

In my earlier question here I made the mistake where I commuted the translation operator with the position observable, which you can't do. But, here you can, which confuses me.

Expanding in Taylor series for $\exp\left(i\frac{p_{0}}{\hbar}\hat{x}\right)$:

$$ \exp\left(i\frac{p_{0}}{\hbar}\hat{x}\right) \approx 1+\frac{ip_ox}{\hbar} $$

Expanding in Taylor series for my earlier question $\exp\left(-i\frac{p_{x}}{\hbar}\delta x\right)$: $$ \exp\left(-i\frac{p_{x}}{\hbar}\delta x\right) \approx 1-\frac{ip_x\delta x}{\hbar}. $$

Since they are both translation operators why does one commute with the position observable and the other does not?

  • 1
    $\begingroup$ Well in one $\delta x$ is an infinitesimal scalar, and in the other $\hat{x}$ is an operator with finite eigenvalues. So the first expansion is incorrect. $\endgroup$
    – user12029
    Commented Apr 4, 2015 at 20:36

1 Answer 1


In this case $p_0$ is just a number. It's the amount of phase that you add to the wave function. $\text{e}^{ip_0 x/\hbar}$ is not a translation operator. It's just multiplying the wave function by a complex number of norm $1$.

In you previous question you had $\hat{p}$ which is an operator. It acts on the wave function by differentiating it.

  • $\begingroup$ I see, thank you very much! Once I reach 15 reps, I will upvote your answer. $\endgroup$
    – Luffy
    Commented Apr 4, 2015 at 21:18

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