Commutation Relations and Ehrenfest

In summary, the expectation value for <x^2> in the state \psi(\vec{r},t) is a quadratic function of t. The constants b and c relate to the expectation values of operators at t=0.
  • #1
keniwas
59
1

Homework Statement


Let [tex]\psi(\vec{r},t)[/tex] be the wavefunction for a free particle of mass m obeying Schrodinger equation with V=0 in 3 dimensions. At t=0, the particle is in a known initial state [tex]\psi_0(\vec{r})[/tex]. Using Ehrenfest's theorem, show that the expectation value [tex]<x^2>[/tex] in the state [tex]\psi(\vec{r},t)[/tex] is a quadratic function of t:
[tex]<x^2>_t=a+bt+ct^2[/tex]
where a, b and c are constants. Relate the constants b and c to the expectation values of operators at t=0. Show that these lead to the expression for [tex]<(\delta x)^2>_t[/tex] quoted in class.


Homework Equations


we have that [tex]<(\delta x)^2>=<x^2>-<x>^2=<(\delta x)^2>_0+\frac{1}{m}(<xp_x+p_xx>_0-2<x>_0<p_x>_0)+\frac{<(\delta p_x)^2>_0}{m^2}[/tex]
where
[tex]<x>=<x>_0+\frac{<px>_0}{m}t[/tex]


The Attempt at a Solution


[tex]\frac{d}{dt}<x^2>=\frac{1}{i\hbar}<[x^2,H]>=\frac{1}{m}<xp_x+p_xx>[/tex]

[tex]\frac{d}{dt}\frac{1}{m}<xp_x+p_xx>=\frac{1}{i\hbar m}<[xp_x,H]+[p_xx,H]>=\frac{1}{i\hbar m}<x[p_x,H]+[x,H]p_x+p_x[x,H]+[p_x,H]x>=\frac{1}{i\hbar m}<\frac{i\hbar}{m}p_xp_x+p_x\frac{i\hbar}{m}p_x>=\frac{i\hbar}{i\hbar m^2}<2p_{x}^2>[/tex]

[tex]\frac{d}{dt}\frac{1}{m^2}<p_{x}^2>=0[/tex]

Taking three integrals of the last equation from 0 to t and substituting the values we got in the previous equations we get the equation for <x^2> as

[tex]<x^2>=<x^2>_0+\frac{1}{m}<xp_x+p_xx>_0t+\frac{2<p_{x}^2>_0}{m^2}t^2[/tex]

The problem is if I try to subtract [tex]<x>^2[/tex] from this, the 2 in the last term ([tex]\frac{2<p_{x}^2>_0}{m^2}t^2[/tex]) causes problems. I feel there should have been a 1/2 pop out somewhere in here that will cancel with that 2 to make this work. But i can't see where it is. Does anyone see where I am going wrong that is giving me this extra factor of 2?

Thanks for any input.
 
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  • #2
You're gunna kick yourself:

[tex]\int_0^t \frac{2\langle p_x^2\rangle_0}{m^2}t'dt'=\frac{\langle p_x^2\rangle_0}{m^2}t^2[/tex]

:wink:
 
  • #3
HAHAHA oh... that hurts...

Thank you =)
 

Related to Commutation Relations and Ehrenfest

1. What are commutation relations?

Commutation relations are mathematical equations that describe the relationships between different operators in quantum mechanics. These equations are used to calculate the probability of obtaining a certain measurement or observable in a quantum system.

2. Why are commutation relations important in quantum mechanics?

Commutation relations are important because they allow us to predict the behavior and properties of quantum systems. By understanding the relationships between different operators, we can make accurate predictions about the behavior of particles in a quantum system.

3. What is the Ehrenfest theorem?

The Ehrenfest theorem is a mathematical principle that relates the expectation values of different operators to the rate of change of their corresponding observables. It states that the time derivative of the expectation value of an operator is equal to the expectation value of the commutator of the operator with the Hamiltonian.

4. How is the Ehrenfest theorem used in quantum mechanics?

The Ehrenfest theorem is used to relate the classical and quantum descriptions of a system. It allows us to make predictions about the behavior of a quantum system by using classical analogies. This theorem is also used to derive equations of motion for quantum systems and to prove the conservation of energy in quantum mechanics.

5. Can commutation relations and the Ehrenfest theorem be applied to all quantum systems?

Yes, commutation relations and the Ehrenfest theorem can be applied to all quantum systems, regardless of their complexity. These principles are fundamental to the understanding of quantum mechanics and are applicable to a wide range of systems, from individual particles to large-scale systems.

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