Quantum mechanics question.

  • #1
I was recently given a problem that asked me to show that the classical velocity of a particle is given by:

[tex] v = \frac{d \langle x\rangle}{dt} = \frac{1}{m} \langle({\bf p}-q{\bf A})\rangle[/tex]

The expectation value of the time derivative of x is given by:

[tex]\langle v\rangle = \int{\Psi ^{*} (x \frac{d}{dt}) \Psi[/tex]

So then I just work this out, and what do I do after that? How do I get from here to the form the problem is asking for?
 
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Answers and Replies

  • #2
HallsofIvy
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I assume p and q are the "generalized coordinates" (p is "x-position" and q is momentum) but what is A?

Also you are using "v" to mean two different things. In the first equation, v is the "classical" velocity while in the second it is the velocity "operator".
In other words, the v in the first equation IS the <v> of the second.
 
  • #3
HallsofIvy said:
I assume p and q are the "generalized coordinates" (p is "x-position" and q is momentum) but what is A?

Also you are using "v" to mean two different things. In the first equation, v is the "classical" velocity while in the second it is the velocity "operator".
In other words, the v in the first equation IS the <v> of the second.
I'm sorry, this material is new to me and I'm not sure what is "standard" notation and what isn't. The question was asked in the context of gauge invariance in electrodynamics(that's what we're discussing in my class right now), so I believe the whole expression "p-qA" is the substitution:

[tex]\bf{p} = \frac{\hbar}{i} \bigtriangledown \to \bf{p} - q\bf{A}[/tex]

so p is momentum, q is charge, and A is the vector potential. We also have the substitution:

[tex]i\hbar\frac{\partial}{\partial t} \rightarrow i\hbar\frac{\partial}{\partial t} - q\phi[/tex]

Where q is again the charge and phi is the scaler potential.

So the schrodinger equation should look like:

[tex]i\hbar\frac{\partial\Psi}{\partial t}= \left[\frac{1}{2m} \left(\frac{\hbar}{i}\bigtriangledown + qA\right)^2+q\phi\right]\Psi[/tex]
 
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  • #4
Dr Transport
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Start with the commutator [tex] [H,x] [/tex], from there you should be able to work out the solution.

dt
 
  • #5
Dr Transport said:
Start with the commutator [tex] [H,x] [/tex], from there you should be able to work out the solution.

dt

So the hamiltonian in this case is
[tex]\left[\frac{1}{2m} \left(\frac{\hbar}{i}\bigtriangledown + qA\right)^2+q\phi\right][/tex]
right?

So if I take the commutator of that with x:
[tex]\left[\left[\frac{1}{2m} \left(\frac{\hbar}{i}\bigtriangledown + qA\right)^2+q\phi\right],x\right][/tex]

I should be able to figure out the time derivative of <x> by

[tex]\frac{d\langle x \rangle}{dt}=\frac{i}{\hbar}\left\langle\left[\left[\frac{1}{2m} \left(\frac{\hbar}{i}\bigtriangledown + qA\right)^2+q\phi\right],x\right]\right\rangle+\frac{\partial \langle x \rangle}{\partial t}[/tex]

Right? I just want to make sure I've got the idea right before I try to go and do the math. Thanks for the help guys.
 

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