We know that $x_0$ and $p_0$ are the mean values for the position and momentum of a particle in the normalized state characterized by the function $\psi (x)$, ( that is, $x_0=\langle x \rangle_\psi$ and $p_0=\langle p \rangle_\psi$).

Does the mean value of $x$ change for the function $\psi(x+x_0)$ ?

If the mean value for x for the funtion $\psi(x)$ is:

$$\langle x \rangle_\psi=\int_{-\infty}^\infty \psi^\ast (x)\; x\; \psi(x) dx=x_0$$

The mean value for x for the function $\psi(x+x_0)$ is:

$$\langle x \rangle_{\psi' }=\int_{-\infty}^\infty \psi^\ast(x+x_0)\; x\; \psi(x+x_0) dx$$

If we center the function in $x=x_0$, we have:

$$\langle x \rangle_{\psi' }=\int_{-\infty}^\infty \psi^\ast(x)\; (x-x_0)\; \psi(x) dx$$


$$\langle x \rangle_{\psi' }=\int_{-\infty}^\infty \psi^\ast(x)\; x\;\psi(x)dx -\int_{-\infty}^\infty \psi^\ast(x)\;x_0\; \psi(x) dx$$

Where the first integral is $\langle x \rangle_\psi$ and the second one, we know that

$$\int_{-\infty}^\infty \psi^\ast(x)\psi(x) dx=1$$

Because it is normalized, so:

$$\langle x \rangle_{\psi' }= \langle x \rangle_\psi -x_0$$

Where $\langle x \rangle_\psi=x_0$, then:

$$\langle x \rangle_{\psi' }=0$$

Is that correct?

If so, what will be the mean value for $p$ with $\psi(x+x_0)$?

  • $\begingroup$ Try this question again with the proper definition of the integral over all space: $\int_{-\infty}^{\infty}$ is, properly, defined as $\lim_{a\to -\infty}\lim_{b\to\infty}\int_a^b$. When you "center the function at $x=x_0$," something also happens to $a$ and $b$, so by ignoring that change, you might be doing a different integral than the one you started with. $\endgroup$ – probably_someone Oct 25 '18 at 11:03

Your result is correct. You have a function $\psi(x)$ whose mean is $x_0$. So the function $\psi(x+x_0)$ is the original function shifted by an amount $|x_0|$ towards $x=0$. But then this means (pun always intended) that your new mean has to be at $x=0$. As others have pointed out, this only works for an integral from $-\infty$ to $\infty$. In general if you are finding the average of something over a finite interval, you would have to shift your interval as well (there are some contrived exceptions to this, but I won't go into them here).

As for the average momentum, I would think nothing changes as long as your system has translational invariance, but @kryomaxim seems to think it does matter in general. I don't think the correct momentum operator would involve $\frac{\partial}{\partial(x+x_0)}$, because you are still working in the original position basis. Many arguments in QM text books I have seen exploit the fact that we can center the wavefunction so that $\langle X\rangle=0$ without changing the mean momentum. So I believe shifting the wavefunction to a new position will not change the mean momentum if there is translational invariance in the system.


If you have an integral over the interval $[-\infty, \infty]$, yes, that's correct.

The wave function $\psi'$ is simply displaced by $-x_0$, so the Position expectation value does Change exactly this amount. For finite intervals, note that also the Integration endpoints Change.

For computing the expectation value $p$, use the Definition of the Momentum Operator and use the fact that $\frac{\partial}{\partial x} = \frac{\partial}{\partial (x+x_0)}$.


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