Ehrenfest theorem and Hamiltonian operator

You can show it by using the commutator identities or using the fact that if a function f(x) commutes with x then it must be of the form f(x)=ax+b. Convince yourself that this is true and use it to prove the commutation relation. In summary, the generalized Ehrenfest theorem can be used to show that any free particle with the one-dimensional Hamiltonian operator H=p^2/2m follows the equation d^2<x^2> / dt^2 = (2/m)<p^2>, by using the commutation relations and the commutator identity. The commutator [x^2,p^2] can be evaluated using the identity [A,BC]=[A
  • #1
cleggy
29
0

Homework Statement



Use the generalized Ehrenfest theorem to show that any free particle with the one-dimensional Hamiltonian operator

H= p^2/2m obeys

d^2<x^2> / dt^2 = (2/m)<p^2>,



Homework Equations



The commutation relation xp - px = ih(bar)



The Attempt at a Solution




d^2<x^2> / dt^2 = (1/m)(d<p^2>/dt)


H = p^2/2m + V(x)

then [x^2,H] = x^2[(p^2/2m) + V(x)) - ((p^2/2m) + V(x))x^2 = (1/2m)[x^2,p^2]


[x^2,p^2] = xxpp - ppxx = 2ih(bar)(xp + px)

I'm stuck at this point?

am i on the right track?

the < > brackets represent the expectation value
 
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  • #2
You're not using the generalized Ehrenfest theorem from the looks of it. T

The theorem you want to use is:
[tex]
\frac{d \langle A \rangle}{dt}=\frac{1}{i \hbar}\langle [A,H] \rangle+\left\langle \frac{dA}{dt} \right\rangle
[/tex]

With A an operator in the Schrodinger picture. Now use A=x and A=p.
 
  • #3
According to my textbook the generalized Ehrenfest theorem is

what you have put but without the derivative term of A at the end
 
  • #4
Don't forget the tex bracket. Well you could write it like that, because the last term in my expression is usually zero for operators in the Schrodinger picture.

It wasn't clear from your original post that you were actually using that one however.

That said can you evaluate the equation for x and p?
 
  • #5
Which x and p ?

the term (xp + px) ?

I really thought i was on the right track.

I'm not sure where to go now. Pointers?
 
  • #6
If

[x^2,p^2] = xxpp - ppxx = 2ih(bar)(xp + px)

Then how do I find the term (xp + px) ?
 
  • #7
Write the commutator as [itex][x^2,pp][/itex] and use the commutator identity [itex][A,BC]=[A,B]C+B[A,C][/itex] then do the same for the x^2.
 
  • #8
I'm still not following

I'm given the commutation relations in my text as


[x^2,p^2] = xxpp - ppxx = 2ih(bar)(xp + px)

[xp,p^2] = xxpp - ppxx = 2ih(bar)(p^2)

[px,p^2] = xxpp - ppxx = 2ih(bar)(p^2)
 
  • #9
You can let a test function work on it. Evaluate (xp+px)f(x).
 
  • #10
This makes no sense to me. I think I'm going to call it a day. Thanks for all your help Cyosis
 
  • #11
Do you know how you can show that [itex][x,p]=i \hbar[/itex] in the first place?
 

What is the Ehrenfest theorem?

The Ehrenfest theorem is a fundamental principle in quantum mechanics that describes how the expectation values of physical quantities evolve over time. It states that the time derivative of the expectation value of any observable is equal to the expectation value of the commutator between that observable and the Hamiltonian operator.

What is the Hamiltonian operator?

The Hamiltonian operator is a mathematical representation of the total energy of a quantum system. It takes into account the potential and kinetic energies of the system's particles and is used to calculate the time evolution of the system.

What is the significance of the Ehrenfest theorem?

The Ehrenfest theorem allows us to relate the classical concept of motion to the quantum mechanical behavior of particles. It also provides a way to calculate the time evolution of expectation values, which are important for understanding the behavior of physical systems.

How is the Ehrenfest theorem derived?

The Ehrenfest theorem can be derived mathematically using the Schrödinger equation and the definition of the expectation value of an operator. It is a consequence of the Heisenberg uncertainty principle and the correspondence principle, which states that classical mechanics is a limiting case of quantum mechanics.

What are the limitations of the Ehrenfest theorem?

The Ehrenfest theorem is only valid for systems with time-independent Hamiltonians and does not take into account the effects of quantum fluctuations. It also does not apply to systems in which the Hamiltonian is not Hermitian, such as systems with time-dependent potentials.

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