What is the proof for the expectation value of a quantum system?

[Warr]

I am trying to show that

\frac{d}{dt}<x^2>=\frac{1}{m}(<xp>+<px>)....(1)

With the wavefunction \Psi being both normalized to unity and square integrable

Here is what I tried...

<xp> = \int_{-\infty}^{\infty}{\Psi}^*xp{\Psi}dx


<px> = \int_{-\infty}^{\infty}{\Psi}^*px{\Psi}dx=\frac{\hbar}{i}\int_{-\infty}^{\infty}{\Psi}^*\frac{\partial}{{\partial}x}(x{\Psi})dx=\frac{\hbar}{i}\int_{-\infty}^{\infty}{\Psi}^*({\Psi}+x\frac{{\partial}{\Psi}}{{\partial}x})dx=\frac{\hbar}{i}\int_{-\infty}^{\infty}{\Psi}^*{\Psi}dx+\frac{\hbar}{i}\int_{-\infty}^{\infty}{\Psi}^*x\frac{{\partial}{\Psi}}{{\partial}x}dx

from here the first part in the addition reduces to \frac{\hbar}{i} since the integral reduces to 1 for a normalized wavefunction. The second part is just the integral for <xp>, so I'll just reduce it to that for now...

&lt;px&gt;=\frac{\hbar}{i}\int_{-\infty}^{\infty}{\Psi}^*{\Psi}dx+\frac{\hbar}{i}\int_{-\infty}^{\infty}{\Psi}^*x\frac{{\partial}{\Psi}}{{\partial}x}dx =\frac{\hbar}{i}+&lt;xp&gt;

so now we have

\frac{1}{m}(&lt;xp&gt;+&lt;px&gt;) = \frac{\hbar}{im}+\frac{2}{m}&lt;xp&gt;...(2)

Looking at the left side of of equation (1) now

\frac{\partial}{{\partial}t}&lt;x^2&gt;=\int_{-\infty}^{\infty}\frac{{\partial}}{{\partial}t}(\Psi^*x^2{\Psi})dx=\int_{-\infty}^{\infty}x^2\frac{{\partial}}{{\partial}t}(\Psi^*{\Psi})dx+\int_{-\infty}^{\infty}\Psi^*\frac{d}{dt}(x^2){\Psi}dx

now.. I don't know if I can even do this: \frac{d}{dt}x^2=2x\frac{dx}{dt}=\frac{2}{m}xp

assuming I could, that reduces the second integral to \int_{-\infty}^{\infty}\Psi^*\frac{d}{dt}(x^2){\Psi}dx=\frac{2}{m}\int_{-\infty}^{\infty}\Psi^*xp{\Psi}dx=\frac{2}{m}&lt;xp&gt;

subbing into the LS and RS of equation (1) gives

\int_{-\infty}^{\infty}x^2\frac{{\partial}}{{\partial}t}(\Psi^*{\Psi})dx + \frac{2}{m}&lt;xp&gt; = \frac{\hbar}{im} + \frac{2}{m}&lt;xp&gt;

the same terms on each side cancel leaving me to show

\int_{-\infty}^{\infty}x^2\frac{{\partial}}{{\partial}t}(\Psi^*{\Psi})dx = \frac{\hbar}{im}

The inner derivative of this equation can be expanded to

\frac{{\partial}}{{\partial}t}(\Psi^*{\Psi})=\Psi^*\frac{{\partial\Psi}}{{\partial}t}+\frac{{\partial}\Psi^*}{{\partial}t}\Psi

From Shroedinger's eq'n we know

\frac{{\partial}\Psi}{{\partial}t}=\frac{i\hbar}{2m}\frac{{\partial}^2\Psi}{{\partial}x^2}-\frac{i}{\hbar}V\Psi

\frac{{\partial}\Psi^*}{{\partial}t}=-(\frac{i\hbar}{2m}\frac{{\partial}^2\Psi^*}{{\partial}x^2}-\frac{i}{\hbar}V\Psi^*)

we then find that

\Psi^*\frac{{\partial\Psi}}{{\partial}t}+\frac{{\partial}\Psi^*}{{\partial}t}\Psi=\frac{i\hbar}{2m}(\Psi^*\frac{{\partial^2\Psi}}{{\partial}x^2}-\frac{{\partial^2}\Psi^*}{{\partial}x^2}\Psi)=\frac{\partial}{{\partial}x}(\frac{i\hbar}{2m}(\Psi^*\frac{{\partial\Psi}}{{\partial}x}-\frac{{\partial}\Psi^*}{{\partial}x}\Psi))

From here I don't know where I'm going..seems like kind of a dead end..

 

[dextercioby]

I'll give you a hint. Work in the Heisenberg representation. No integrations, just operator equalities (surely, assuming a common dense domain for the operators and a specific form of the Hamiltonian).

 

[Warr]

I don't really know anything about 'Heisenberg representation'...this is homework too, so it would probably be pointless to figure out since we aren't expected to know it.

 

[Avodyne]

now.. I don't know if I can even do this: \frac{d}{dt}x^2=2x\frac{dx}{dt}=\frac{2}{m}xp

No, you can't; x doesn't depend on t, it's just an integration variable.

Start with
{\partial\over\partial t}\langle x^2\rangle=\int_{-\infty}^{\infty}\left[{\partial\Psi^*\over\partial t}x^2\Psi+\Psi^*x^2{\partial\Psi\over\partial t}\right]dx
Then use the Schrodinger equation and it's complex conjugate. Then do some integrations by parts to try to get it to look like your integral expression for \langle(xp+px)\rangle

Dexter is right that this problem is *much* easier in the Heisenberg representation.

 

[Warr]

Thanks, I got it to work out now.