Expectation Value of Momentum for Wavepacket

torq123
Messages
8
Reaction score
0

Homework Statement



What is the average momentum for a packet corresponding to this normalizable wavefunction?

\Psi(x) = C \phi(x) exp(ikx)

C is a normalization constant and \phi(x) is a real function.

Homework Equations


\hat{p}\rightarrow -i\hbar\frac{d}{dx}

The Attempt at a Solution



\int\Psi(x)^{*}\Psi(x)dx = \int C^2 \phi(x)^{2}dx= 1

Plugging in the momentum operator and using the chain rule:

<\hat{p}> = \hbar k \int C^2 \phi(x)^2 dx - i \hbar \int C^2 \phi^{'}\phi dx

The second term is always imaginary since \phi(x) is real, so I said the momentum is \hbar k which I think might be right, but for the wrong reasons? I didn't think Hermetian operators could give imaginary expectation values...
 
Physics news on Phys.org
torq123 said:
<\hat{p}> = \hbar k \int C^2 \phi(x)^2 dx - i \hbar \int C^2 \phi^{'}\phi dx

The second term is always imaginary since \phi(x) is real, so I said the momentum is \hbar k which I think might be right, but for the wrong reasons? I didn't think Hermetian operators could give imaginary expectation values...
Try working on the 2nd term a bit more. Hint: use integration by parts.
 
Are you saying that the second term must be zero since \phi vanishes at ±∞ and the integral evaluates to \phi^2(x)/2? That makes sense to me.
 
torq123 said:
Are you saying that the second term must be zero since \phi vanishes at ±∞ and the integral evaluates to \phi^2(x)/2?
That's the idea.
 
Awesome. Thanks for the help.
 
##|\Psi|^2=\frac{1}{\sqrt{\pi b^2}}\exp(\frac{-(x-x_0)^2}{b^2}).## ##\braket{x}=\frac{1}{\sqrt{\pi b^2}}\int_{-\infty}^{\infty}dx\,x\exp(-\frac{(x-x_0)^2}{b^2}).## ##y=x-x_0 \quad x=y+x_0 \quad dy=dx.## The boundaries remain infinite, I believe. ##\frac{1}{\sqrt{\pi b^2}}\int_{-\infty}^{\infty}dy(y+x_0)\exp(\frac{-y^2}{b^2}).## ##\frac{2}{\sqrt{\pi b^2}}\int_0^{\infty}dy\,y\exp(\frac{-y^2}{b^2})+\frac{2x_0}{\sqrt{\pi b^2}}\int_0^{\infty}dy\,\exp(-\frac{y^2}{b^2}).## I then resolved the two...
Hello everyone, I’m considering a point charge q that oscillates harmonically about the origin along the z-axis, e.g. $$z_{q}(t)= A\sin(wt)$$ In a strongly simplified / quasi-instantaneous approximation I ignore retardation and take the electric field at the position ##r=(x,y,z)## simply to be the “Coulomb field at the charge’s instantaneous position”: $$E(r,t)=\frac{q}{4\pi\varepsilon_{0}}\frac{r-r_{q}(t)}{||r-r_{q}(t)||^{3}}$$ with $$r_{q}(t)=(0,0,z_{q}(t))$$ (I’m aware this isn’t...
Hi, I had an exam and I completely messed up a problem. Especially one part which was necessary for the rest of the problem. Basically, I have a wormhole metric: $$(ds)^2 = -(dt)^2 + (dr)^2 + (r^2 + b^2)( (d\theta)^2 + sin^2 \theta (d\phi)^2 )$$ Where ##b=1## with an orbit only in the equatorial plane. We also know from the question that the orbit must satisfy this relationship: $$\varepsilon = \frac{1}{2} (\frac{dr}{d\tau})^2 + V_{eff}(r)$$ Ultimately, I was tasked to find the initial...

Similar threads

Back
Top