Vanishing Wavefunction: Show Expectation Values of x and p Vanish

[misterpickle]

Homework Statement

If <x> and <p> are the expectation values of x and p formed with the wave-function of a one-dimensional system, show that the expectation value of x and p formed with the wave-function vanishes. The wavefunction is:

\phi(x)=exp(-\frac{i}{h}\langle p\rangle x)\psi(x+\langle x\rangle)Basically I don't know how to start this problem. Do I plug in the integral forms of the expectation values into the exponential and distribute the psi over x and \langle x\rangle?

 

[fzero]

You left out some important text, but I found the problem set with a google search. \langle x\rangle, \langle p\rangle are computed with the wavefunction \psi(x), i.e.

\langle x\rangle = \int x \psi^*(x) \psi(x) \, dx, \ldots

You are asked to compute expectation values with the wavefunction \phi(x):

\langle x\rangle_\phi = \int x \phi^*(x) \phi(x) \, dx, \ldots

These expressions can be reduced to expectation values for certain quantities in the wavefunction \psi(x). You don't have to do any integrals explicity, just some algebra and derivatives.

 

[misterpickle]

I've set it up the way you suggested, and the exponential terms computing \langle x \rangle cancel. However there is a \langle x \rangle term in \psi(x+\langle x \rangle). Is this not a discrepancy to have the value you are trying to calculate inside the equation for calculating its value?

Specifically,

\int{ x \psi^{*}(x+\langle x \rangle) \psi(x+\langle x \rangle)\,dx}

 

[fzero]

I've set it up the way you suggested, and the exponential terms computing \langle x \rangle cancel. However there is a \langle x \rangle term in \psi(x+\langle x \rangle). Is this not a discrepancy to have the value you are trying to calculate inside the equation for calculating its value?

Specifically,

\int{ x \psi^{*}(x+\langle x \rangle) \psi(x+\langle x \rangle)\,dx}

Well \langle x \rangle is just a number (it came out of another integral over x, so it doesn't depend on x). There's still something you have to do to this integral though.

 

[misterpickle]

If \psi(x+\langle x \rangle) is distributable (meaning f(x+y)=f(x)+f(y)) then I come up with the following:

\int{ x \psi^{*}(x+\langle x \rangle) \psi(x+\langle x \rangle)\,dx}
\int{ x[ (\psi^{*}x+\psi^{*}\langle x \rangle)( \psi x+\psi\langle x \rangle)\,dx}
\int{ (\psi^{*}\psi x^{3}+2\psi^{*}\psi x^{2}\langle x \rangle+\psi^{*}\psi x\langle x \rangle^{2})\,dx}
\langle x^{3}\rangle +2\langle x^{2}\langle x \rangle\rangle + \langle x\langle x\rangle^{2}\rangle

...which doesn't make any sense to me. Also the "distributable" assumption I made does not work for exponentials, since f(x+y)=f(x)f(y) for exponential functions.

 

[fzero]

Yeah wavefunctions will almost never have that property since they're always related to exponential functions. Think about how you can convert this to an expectation value in \psi(x) by making a substitution in the integration variable.