How Does Ehrenfest Theorem Apply to Particle Position in Quantum Mechanics?

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Homework Help Overview

The discussion revolves around the application of the Ehrenfest theorem in quantum mechanics, specifically regarding the expectation value of the position of a particle moving in three dimensions under a given Hamiltonian. Participants are examining how to demonstrate that the time derivative of the expectation value of position relates to momentum and mass.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants are attempting to manipulate the Ehrenfest theorem's equation and explore the implications of the commutator between position and Hamiltonian. There are questions about the correctness of LaTeX formatting and clarity in mathematical expressions.

Discussion Status

Some participants are providing feedback on the clarity of mathematical notation and suggesting ways to improve the presentation of equations. There is an ongoing exploration of the relationships between the variables involved, but no consensus has been reached on the final outcome or method.

Contextual Notes

Participants are discussing the use of specific mathematical symbols and notation in quantum mechanics, indicating a focus on precision in communication. There is also a mention of potential confusion regarding the representation of the potential energy function.

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Homework Statement


Use ehrenfest theorem (i*\hbar*d<Q>/dt=(\varphi(t),[Q,H],\varphi(t)) to show that the expectation value of the position of a particlee that moves in 3 dimensions with the Hamiltonian H=p^2/2m+V(r) satisfies d<r>/dt=<p>/m

Homework Equations

(i*\hbar*d<Q>/dt=(\varphi(t),[Q,H],\varphi(t))

or d<Q>/dt=<-i[Q,H]/(\hbar)

The Attempt at a Solution



[Q,H]=QH-HQ=Q((-i*\hbar*d/dx)^2/2m+V(r))-((-i*\hbar*d/dx)^2/2m+V(r))(Q)=Q*(\hbar)^2 d^2/dx^2*1/2m +QV(r)-(\hbar)^2 d^2Q/dx^2*1/2m+V(r)Q=QV(r)-(\hbar)^2 d^2Q/dx^2*1/2m+V(r)Q not sure how to continue this problem

Perhaps i should say: i*\hbar*d<r>/dt=[\varphi, [r,H]\varphi]
 
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anybody find a hard time reading the latex code or all of the country
 
noblegas said:
Perhaps i should say: i*\hbar*d<r>/dt=[\varphi, [r,H]\varphi]

That looks like the best starting point to me...what do you get when you do that?
 
i*\hbar*d/dt=[\varphi, [r,H]\varphi]=[\varphi, [r,p^2/2m+V(x,t)]\varphi]=[\varphi, (r*p^2/2m+V(x,t)-p^2/2m+V(x,t)*r)\varphi]=1/(i*\hbar*2*m)*<[x,p]*d(p^2)/dp>=(<i*\hbar*2*p>)/(i*\hbar*2*m)=<p>/m? Please take a look at my latex code because I don't think latex displayed all of my solution
 
Your \LaTeX is terrible!

Click on the image below to see how to generate something more legible:

i\hbar\frac{d\langle r\rangle}{dt}=(\varphi, [r,H]\varphi)=(\varphi, [r,p^2/2m+V(r)]\varphi)

You need to be careful to only use square brackets to represent commutators,and round brackets otherwise. There is also no need to use the * symbol to represent multiplication, it just makes things look messy. And you should use \frac when appropriate. Also, your potential is given to you as a function of r...why would you write it as a function of x and t?

continue from here...
 
<br /> i\hbar\frac{d\langle r\rangle}{dt}=(\varphi, [r,H]\varphi)=(\varphi, [r,p^2/2m+V(r)]\varphi)=1/(2*\hbar*m*i)*(&lt;[x,p]d(p^2)/dp&gt;)=1/(2*\hbar*m*i)*(&lt;[i*\hbar*2p&gt;)=&lt;p&gt;/m<br /> hope this is better
 

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