# Does the mean value change?

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:

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

Then

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

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 $$\psi(x+x_0)$$?

• 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. – 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 $$\frac{\partial}{\partial x} = \frac{\partial}{\partial (x+x_0)}$$.