Heisenberg equations for spin

In summary, the conversation discusses the Heisenberg equations for spin operators and obtaining decoupled second-order differential equations for the expectation values of these operators in a particle at rest in a B-field. The equations are solved for a particle initially in a state of ##S_x = \frac{\hbar}{2}##, and the resulting expectation values are ##\langle \hat S_y \rangle_{\psi} = - \frac{\hbar}{2} \sin (\gamma B_0 t)## and ##\langle \hat S_x \rangle_{\psi} = \frac{\hbar}{2} \cos (\gamma B_0 t)##.
  • #1
BOAS
552
19

Homework Statement


Consider a spin ##\frac{1}{2}## particle at rest in a B-field ##\vec B = B_0\vec e_z##.

(a) Find the Heisenberg equations for ##\hat S_x## and ##\hat S_y##.

(b) Obtain from the Heisenberg equations two decoupled second-order differential equations for ##\langle \hat S_x \rangle_{\psi}## and ##\langle \hat S_y \rangle_{\psi}## for a general state ##| \psi \rangle## of the particle.

(c) Solve the equations for a particle that initially (at t = 0) is in a state of ##S_x = \frac{\hbar}{2}##

Homework Equations

The Attempt at a Solution



For part (a) I have done the following;

##\langle \psi (t) | \hat B_S | \psi (t) \rangle = \langle e^{-\frac{i \hat H t}{\hbar}} \psi (0) | \hat B_S | e^{-\frac{i \hat H t}{\hbar}} \psi (0) \rangle = \langle \psi (0) | e^{\frac{i \hat H t}{\hbar}} \hat B_S e^{-\frac{i \hat H t}{\hbar}} | \psi (0) \rangle##

##B_H = e^{\frac{i \hat H t}{\hbar}} \hat B_S e^{-\frac{i \hat H t}{\hbar}}##

where the H and S subscripts correspond to the Heisenberg and Schrodinger pictures respectively.

The hamiltonian that I have is ##\hat H = - \gamma B_0 \hat S_z##

which leads me to the conclusion that my spin operators are unchanged in the Heisenberg picture.

for part (b) I use the Heisenberg equation of motion on both the ##\hat S_x## and ##\hat S_y## operators

##\frac{d}{dt} \hat S_x = \frac{i}{\hbar} [\hat H, \hat S_x] = \gamma B_0 \hat S_y##

##\frac{d}{dt} \hat S_y = \frac{i}{\hbar} [\hat H, \hat S_y] = - \gamma B_0 \hat S_x##

I then differentiate both again, and substitute the original equations into the result to find

##\frac{d^2}{dt^2} \hat S_y = -(\gamma B_0)^2 \hat S_y##

##\frac{d^2}{dt^2} \hat S_x = -(\gamma B_0)^2 \hat S_x##

These two equations are SHM equations with solutions of the form ##y = A \cos \omega t + B \sin \omega t ##

solving these I find that ##\hat S_y (t) = \hat S_y (0) \cos (\gamma B_0 t) - \hat S_x (0) \sin (\gamma B_0 t)## and ##\hat S_x (t) = \hat S_x (0) \cos (\gamma B_0 t) + \hat S_y (0) \sin (\gamma B_0 t)##

I don't really understand what the question is trying to get me to do here. Any guidance would be much appreciated.
 
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  • #2
Looks all ok, I think the only thing you are assumed still to do is to form the expectations values with the psi(0) specified.
 
  • #3
DrDu said:
Looks all ok, I think the only thing you are assumed still to do is to form the expectations values with the psi(0) specified.

I'm still having trouble solving this. I can't see how to find ##\hat S_y (0)##
 
  • #4
At t=0, the operators in the Schroedinger and Heisenberg picture coincide.
 
  • #5
DrDu said:
At t=0, the operators in the Schroedinger and Heisenberg picture coincide.

Thanks, I see.

Ok, so I know that ##\psi (s_x = \hbar / 2) = \frac{1}{\sqrt{2}} \begin{pmatrix} 1\\ 1 \end{pmatrix}##. Does "solving" these equations mean taking the expectation value for my state?
 
  • #6
I think so, what do you get?
 
  • #7
DrDu said:
I think so, what do you get?

I find that ##\langle \hat S_y \rangle_{\psi} = - \frac{\hbar}{2} \sin (\gamma B_0 t)## and ##\langle \hat S_x \rangle_{\psi} = \frac{\hbar}{2} \cos (\gamma B_0 t)##
 
  • #8
BOAS said:
I find that ##\langle \hat S_y \rangle_{\psi} = - \frac{\hbar}{2} \sin (\gamma B_0 t)## and ##\langle \hat S_x \rangle_{\psi} = \frac{\hbar}{2} \cos (\gamma B_0 t)##
Looks good!
 
  • #9
DrDu said:
Looks good!

Thank you for your help
 

1. What are the Heisenberg equations for spin?

The Heisenberg equations for spin, also known as the Heisenberg equations of motion, are a set of equations developed by physicist Werner Heisenberg to describe the time evolution of a quantum mechanical system. They relate the rate of change of a system's observables, such as spin, to the system's Hamiltonian.

2. How do the Heisenberg equations for spin differ from the Schrödinger equation?

The Heisenberg equations for spin are an alternative formulation of quantum mechanics that describes the time evolution of a system's observables, while the Schrödinger equation describes the time evolution of a system's wavefunction. In the Heisenberg picture, the operators are time-dependent, while in the Schrödinger picture, the wavefunction is time-dependent.

3. Can the Heisenberg equations for spin be used to predict the exact outcome of a measurement?

No, the Heisenberg equations for spin cannot predict the exact outcome of a measurement. They can only give the probabilities for different outcomes. This is due to the inherent uncertainty in quantum mechanics, described by the Heisenberg uncertainty principle.

4. What is the physical significance of the Heisenberg equations for spin?

The Heisenberg equations for spin have significant physical significance as they allow scientists to accurately describe and predict the behaviors of quantum mechanical systems. They have been used to make important discoveries in fields such as quantum computing and have helped to advance our understanding of the fundamental principles of the universe.

5. How are the Heisenberg equations for spin derived?

The Heisenberg equations for spin are derived using the principles of quantum mechanics and the concept of operator algebra. They are based on the commutation relations between operators, which describe how the operators of different observables interact with each other. From these commutation relations, the equations of motion for a system's observables can be derived.

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