Assuming the eigenvalue of position operator $\hat x$ equal to $k$, can I not write:

$$\begin{align} \langle\psi_n|x|\psi_m\rangle &= \langle x\psi_n|\psi_m\rangle \\ &=\langle k\psi_n|\psi_m\rangle \\ &=k\langle\psi_n|\psi_m\rangle \\ &=k\delta_{nm} \end{align}$$

But I know that $\langle x \rangle =0$ in case of even potentials (I don't know how that happens) and what I have written above is wrong, at least in case of even potentials.

Taking the example of infinite 1D square well, the states are : $$ \psi_{n} \left(x\right)=A\sin\left(\frac{nx\pi}{L}\right)dx $$ then, $$ \langle\psi_n|x|\psi_m\rangle =A\int_{-\infty}^{\infty}\sin\left(\frac{mx\pi}{L}\right)x\sin\left(\frac{nx\pi}{L}\right)dx $$ If m=n=1, $$ <x>=A\int_{-\infty}^{\infty}x\sin^{2}\left(\frac{x\pi}{L}\right)dx $$ if we apply an even potential then the equation gets reduced to$$ <x>=\frac{1}{L}\int_{-L}^{L}x\sin^{2}\left(\frac{x\pi}{L}\right)dx=0 $$ while in case of a potential(neither even nor odd), the equation leads to $$<x>=\frac{2}{L}\int_{0}^{L}x\sin^{2}\left(\frac{x\pi}{L}\right)dx=L/2 ~? $$ Here $n=m=1$ but $ <x>=0 $ for even potentials, which is confusing me! It should be $k$ right?


The $|\psi_n\rangle$ you are using are eigenstates of the Hamiltonian, $\hat{H}$, not eigenstates of the position operator $\hat{X}$. There's no sense in which you may write $\langle\psi_m|\hat{X}|\psi_n\rangle=(\text{some constant constant})*\langle\psi_m|\psi_n\rangle$, because it is NOT true that $\hat{X}|\psi_m\rangle=k|\psi_m\rangle$.


First, If you are looking at the 1D infinite potential well, then the states you mention $\langle x|\psi _m \rangle=\psi_m(x)=A\sin \left (\frac{n\pi x}{L}\right )$ are not eigenstates of the position operator. So, $$\hat X|\psi_m\rangle\neq x|\psi_m\rangle$$

Second, if you modify your 1D potential to not be even, then your wave functions change. You cannot just change the limits of integration. If you want the well to have length $L$ and the potential to be even, then your ground state function is actually a cosine function $\psi_1(x)=A\cos \left (\frac{n\pi x}{L}\right )$ and you would want to integrate between $x=-L/2$ and $x=L/2$ for the integral to calculate the average position of this state.

Then, if you want a 1D infinite potential well where the function is defined from $x=0$ to $x=L$ so that the potential function is not even, then you would perform the final integral in your question from $0$ to $L$ to get the average position in this state.

You can reason to the average position of a particle in the ground state of the infinite potential well by just thinking about the shape of the wave function. It is a symmetric function that is $0$ at the ends of the well and hits a maximum in the middle. Therefore, it should be no surprise that the average position is always found in the middle of the well ($0$ for the even potential case and $L/2$ when it is between $0$ and $L$).

  • $\begingroup$ I wasn't expecting k to pop out of my integral. It's just that I was expecting <x> to be a non-zero quantity for even potentials as well because my argument (which is invalid now ) led me to believe that since m=n=1,DELTA function must be equal to 1,so k*DELTA must be something non-zero. $\endgroup$ – Jalaj Chaturvedi Sep 1 '18 at 23:36
  • $\begingroup$ @JalajChaturvedi ok I get it now. I'll edit my answer accordingly. $\endgroup$ – BioPhysicist Sep 1 '18 at 23:41

In the case of even potentials, the wavefunction $[f(x)]$ for a bound state is either even or odd under parity.

case 1:
Even wavefunction- The integrand $f(x) \cdot x \cdot f(x)$ is odd and hence while integrative over all of space vanishes.

case 2:
$f(-x) \cdot f(-x) = [-f(x)] \cdot [-f(x)] = f(x) \cdot f(x) \implies$ The product of 2 odd functions is even.
Hence: The integrand again is odd and the integral over all space vanishes.

Ps: By the way, have used the fact the wavefunctions for bound states can be chosen to be real upto a phase.

  • $\begingroup$ Hi and welcome to Phys.SE. Please use MathJax to write equations. $\endgroup$ – Abhay Hegde Apr 26 '19 at 12:59

$\langle x \rangle$ is not always neccesarily 0, even in the case of even-potentials. Do not learn the rule as a holy truth, but think about it.

$\langle x \rangle $ is $0$, with even wavefunctions, as long as the interval is symmetrical. If you do the integral

$$\int_{-\infty}^{+\infty} \psi_{even}^*\cdot x \cdot \psi_{even} dx $$

you'll get $0$, because of symmetry. The same happens if $\psi$ is odd, but you can't tell if $\psi$ isn't even or odd.

HOWEVER in the case of the infinite well, you usually do the integral FROM $0$ TO $L$. Of course that is not a symmetrical interval, so the fact that $\psi$ is odd or even will not be relevant.

In fact, it is correct that the mean value is $L/2$. There's no reason to think that it can be $\langle x\rangle$=0. Check that, for the particle to have the mean position in the origin, it would need to be exactly there, as it's the left top of the well. That's against Heisenberg's principle.

However, you'd actually find $\langle x \rangle = 0$ if you defined the well to be from $-L/2$ to $+L/2$.

So it depends on what your system is, and on the coordinates you choose!

The only thing you can say is that the integral vanishes if you're integrating a globally odd function along a symmetric interval, but not otherwise.

  • $\begingroup$ Let me know if you don't understand. $\endgroup$ – FGSUZ Sep 1 '18 at 22:54
  • $\begingroup$ Okay, I didn't say that they're not position's eigenfunctions, but I think I solved the second part of the quesiton, and I explained that one sentence he says is wrong (And he didn't know why). SO I don't think my answer is wrong either. Is it? Please, explain your downvotes, that'll be helpful. $\endgroup$ – FGSUZ Sep 1 '18 at 23:13
  • $\begingroup$ So basically I can't write that DIRAC NOTATION stuff which I wrote using eigenstates of the Hamiltonian as JAHAN CLAES has mentioned, but I have to calculate <x> using the whole wavefunction as you have done. But then isn't the wavefunction always an eigenstate of a Hamiltonian? $\endgroup$ – Jalaj Chaturvedi Sep 1 '18 at 23:17
  • $\begingroup$ After all, the whole Schrodinger Equation is written as(time independent) : $$H\psi=E\psi ? $$ $\endgroup$ – Jalaj Chaturvedi Sep 1 '18 at 23:26
  • 1
    $\begingroup$ When you soulve a problem, you want to find the eigenstates of the hamiltonian. Those are the useful eigenstates. Those are the ones you try to find. Nevertheless, the mean value must be calculated using the wavefunction of the system. You have to know what wavefunction is describing your system. If your particle is in a defined energy state, then it's fine; but if your system is a linear conbination of eigenstates, you have to use the whole wavefunction for $\langle x \rangle$. $\endgroup$ – FGSUZ Sep 1 '18 at 23:30

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