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.