States of a spin 1/2 particle in a magnetic field

In summary, the question is about finding the normalized energy eigenstates and eigenvalues for a spin 1/2 particle with eigenstates of Sz, and finding the normalized eigenstates and eigenvalues of Sx in terms of the eigenstates of Sz. This is done by calculating the Hamiltonian and using the eigenvalue equation to find the eigenvalues and eigenvectors. The vectors should have a magnitude of 1 and the eigenvalues should be +/- hbar/2. The eigenvectors are already in terms of the eigenstates of Sz, as the matrix representation of Sx is with respect to the basis of the Sz eigenstates.
  • #1
ulcrid
3
0
Hi, for my quantum mechanics course I'm confronted with the following question:

Homework Statement


A spin 1/2 particle has the following eigenstates of Sz: |+> = (1 0) and |-> = (0 1). A magnetic field is pointing in the z direction, B = (0,0,B). The Hamiltonian is H = -B * n, with n = -(e/mc)S and S = h/4pi * s (with B, n, S, and s vectors and H of course an operator).

The questions are a) to find the normalized energy eigenstates and eigenvalues and b) to find the normalized eigenstates and eigenvalues of Sx in terms of the eigenstates of Sz.

Homework Equations


The Pauli spin matrices are most important here, for a) only the z component is relevant as the magnetic field only has a z component and the Hamiltonian is defined as -B * n. For b) I think the x component is also relevant.
sz=
(1 0)
(0 -1)
sx=
(1 0)
(0 1)

The Attempt at a Solution


I think I got the most far on question a). I calculated H and found it to be H = (ehB/4pi*mc)*
(1 0)
(0 -1)
Then, using the determinant of (H - lambda * I) I calculated the eigenvalues of H: lambda = +/- (ehB/4pi*mc). After that, I used the eigenvalue equation Av = lambda*v to find the eigenvectors (1 0) and (0 1). I don't think that should be surprising though because somewhere in the book (Introductory Quantum Mechanics by Liboff) it says that Sz and H have the same eigenfunctions.
But are these normalized? And are these values/vectors correct?

On question b) I don't really know what to do - if I try finding out the eigenstates of Sx I get stuck on the vectors. I get equations like h/4pi (b a) = (a b) which doesn't have a solution except for a = b = 0 which obviously is incorrect. How do I find the correct eigenstates? And how do I then find out how to write them as combinations of eigenstates of Sz? Hope someone can help!
 
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  • #2
ulcrid said:

The Attempt at a Solution


I think I got the most far on question a). I calculated H and found it to be H = (ehB/4pi*mc)*
(1 0)
(0 -1)
Then, using the determinant of (H - lambda * I) I calculated the eigenvalues of H: lambda = +/- (ehB/4pi*mc). After that, I used the eigenvalue equation Av = lambda*v to find the eigenvectors (1 0) and (0 1). I don't think that should be surprising though because somewhere in the book (Introductory Quantum Mechanics by Liboff) it says that Sz and H have the same eigenfunctions.
But are these normalized? And are these values/vectors correct?
If the vectors are normalized, they'd have a magnitude of 1. What are the magnitude of your vectors?
On question b) I don't really know what to do - if I try finding out the eigenstates of Sx I get stuck on the vectors. I get equations like h/4pi (b a) = (a b) which doesn't have a solution except for a = b = 0 which obviously is incorrect. How do I find the correct eigenstates? And how do I then find out how to write them as combinations of eigenstates of Sz? Hope someone can help!
You need to show more details of your work. You might just be dropping a constant somewhere.

The matrix representation of Sx you're using is with respect to basis of the Sz eigenstates, so the eigenvectors you find will already be representations in terms of the eigenstates of Sz.
 
  • #3
Well, the magnitude of the vectors seems to be 1. After all, |(1 0)|2 is 1. So that would be okay?

For the second question I used the Pauli-spinmatrix of Sx:
Code:
[U]h[/U]   (0 1)
4pi (1 0)

Then, I can use det(Sx-lambda*I) = 0 to get lambda2-1 = 0 so lambda is either 1 or -1.
Using that in the eigenvalue equation I get Sx*v = +/- v
Which gets me to
Code:
[U]h[/U]   (0 1)(a) = +/- (a)
4pi (1 0)(b)       (b)
Which is equal to
Code:
[U]h[/U]   (b) = +/- (a)
4pi (a)       (b)
Which in my view is only solvable for a = b = 0. ?
 
  • #4
Your eigenvalues should be [itex]\pm\hbar/2[/itex]. Can you see the mistake you made in calculating them? If you fix those, the eigenvectors should work out for you.
 
  • #5
Ah, yes, I see where I went wrong. And because I used the Pauli-spinmatrix for Sx, which apparently already is in the Sz/S2 representation, these vectors and eigenvectors are already in terms of the eigenstates of Sz?
 

What is a spin 1/2 particle?

A spin 1/2 particle is a type of elementary particle, such as an electron or proton, that has an intrinsic angular momentum or "spin" of 1/2. This spin value is a fundamental property of the particle and is not related to its physical rotation.

What is a magnetic field?

A magnetic field is a region in space where a magnetic force can be detected. It is created by moving electric charges, such as the movement of electrons in a wire or the orbiting of electrons around an atom's nucleus.

How does a spin 1/2 particle behave in a magnetic field?

A spin 1/2 particle in a magnetic field will experience a torque, or turning force, due to its intrinsic magnetic moment. This can cause the particle to precess, or rotate, around the direction of the magnetic field.

What are the possible states of a spin 1/2 particle in a magnetic field?

The possible states of a spin 1/2 particle in a magnetic field are the spin-up state, where the particle's spin is aligned with the direction of the magnetic field, and the spin-down state, where the particle's spin is opposite to the direction of the magnetic field.

What is the energy of a spin 1/2 particle in a magnetic field?

The energy of a spin 1/2 particle in a magnetic field depends on the orientation of its spin relative to the magnetic field. The spin-up state has a lower energy than the spin-down state, resulting in a splitting of energy levels known as the Zeeman effect.

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