Find Min B to Flip Spin-Up in Magnetic Field

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Homework Help Overview

The discussion revolves around a quantum mechanics problem involving a spin-1/2 particle placed in a time-dependent magnetic field along the z-direction. The initial state of the spin is defined as spin-up along the x-direction, and the goal is to determine the minimum magnetic field strength required to flip the spin state.

Discussion Character

  • Conceptual clarification, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the Hamiltonian of the system and the evolution of the spin state over time. There is an exploration of the probability of measuring the spin-down state and the conditions under which the spin flips. One participant questions whether the problem statement is complete, while another confirms the context of the exercise.

Discussion Status

The discussion is ongoing, with participants sharing their interpretations and attempts at solving the problem. Some guidance has been provided regarding the formulation of the Hamiltonian and the probability expression, but there is no explicit consensus on the correctness of the approaches taken.

Contextual Notes

One participant notes that the problem is an exercise given by an instructor rather than a formal homework assignment, which may influence the expectations for completeness in the problem statement.

KFC
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Homework Statement


Assuming a spin 1/2 is put in a magnetic filed along z direction B_z = B*cos(w_0 t). At the beginning, the spin is in spin-up along x direction, i.e.

\psi(0) = \frac{1}{\sqrt{2}}\left(<br /> \begin{matrix}<br /> 1 \\ 1<br /> \end{matrix}\right)

Try to find out the minimum B such that S_x is flip.

2. The attempt at a solution
First of all, I write than the Hamiltonian of the system

H \propto<br /> \left(<br /> \begin{matrix}<br /> B\cos w_0t &amp; 0\\<br /> 0 &amp; -B\cos w_0t <br /> \end{matrix}<br /> \right)<br />

From that, in any time t>0, the state will evolute as \psi(t) (I already solved that). From the time-dependent solution, I can figure out the probability to find the spin-down when measuring S_x, which is of the following form

P = \sin^2(\gamma \sin(w_0 t))

where \gamma is a constant containing B. Hence, to make the system flip, I have to let P=1, i.e.

\gamma \sin(w_0 t)=\pi/2

and solve for B gives

B = \frac{k}{\sin w_0 t}

where k is another constant. For finding the minimum B, I just take sin w_0t =1. I don't know if my solution is correct or not. Any comment?
 
Last edited:
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KFC said:

Homework Statement


Assuming a spin 1/2 is put in a magnetic filed along z direction B_z = B*cos(w_0 t). At the beginning, the spin is in spin-up along x direction, i.e.

\psi(0) = \frac{1}{\sqrt{2}}\left(<br /> \begin{matrix}<br /> 1 \\ 1<br /> \end{matrix}\right)

Try to find out the minimum B such that S_x is flip.

2. The attempt at a solution
First of all, I write than the Hamiltonian of the system

H \propto<br /> \left(<br /> \begin{matrix}<br /> B\cos w_0t &amp; 0\\<br /> 0 &amp; -B\cos w_0t <br /> \end{matrix}<br /> \right)<br />

From that, in any time t>0, the state will evolute as \psi(t) (I already solved that). From the time-dependent solution, I can figure out the probability to find the spin-down when measuring S_x, which is of the following form

P = \sin^2(\gamma \sin(w_0 t))

where \gamma is a constant containing B. Hence, to make the system flip, I have to let P=1, i.e.

\gamma \sin(w_0 t)=\pi/2

and solve for B gives

B = \frac{k}{\sin w_0 t}

where k is another constant. For finding the minimum B, I just take sin w_0t =1. I don't know if my solution is correct or not. Any comment?

it seems like something may be missing from the question... is that the exact statement of the question on the homework?
 
olgranpappy said:
it seems like something may be missing from the question... is that the exact statement of the question on the homework?

Thanks for reply. This is not actually an hw. This is an exercise given by my instructor and that's all statement :)
 
KFC said:
Thanks for reply. This is not actually an hw. This is an exercise given by my instructor and that's all statement :)

then what you have looks good to me.
 

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