Is the Spin Ket an Eigenket for Zero Magnetic Field or Interaction Strength?

Gregg
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Homework Statement

Hamiltonian for an electron positron system in a uniform magnetic field B (in the z direction)

## \hat{H} = AS^{(-)} \cdot S^{(+)} +\frac{eB}{mc} (S_z^{(-)} +S_z^{(+)} )##

##A \in \Re ##

We have a spin ket given by

## |\psi \rangle = | \uparrow \rangle^{(+)} | \uparrow \rangle^{(-)} ##

For A = 0 is ##| \psi \rangle ## an eigenket?

For B = 0 is it an eigenket?

Homework Equations


The Attempt at a Solution



I believe that the (+) denotes the positron and the (-), an electron. This is a two spin-1/2 particle system.

In lectures we have been given eigenvalues of various spin operators

The second term in the Hamiltonian is

## \frac{eB}{mc}(S^{(+)}_z + S^{(-)}_z ) = \frac{eB}{mc} (S^{(+)+(-)})_z ## which has eigenvalue ## m\hbar = (m_1+m_2)\hbar ##

The first term in the Hamiltonian is something I find confusing

## A S^{(-)} \cdot S^{(+)} ##

My attempt is

## S^{(+)+(-)} = S^{(+)} + S^{(-)} ##

## S^{(-)} \cdot S^{(+)} = ( S^{(+)+(-)}-S^{(+)} )( S^{(+)+(-)}-S^{(-)}) ##

##= S^{(+)+(-)} \cdot S^{(+)+(-)} + S^{(+)} \cdot S^{(-)} - S^{(+)} \cdot S^{(+)+(-)} - S^{(+)+(-)} \cdot S^{(-)} ##

The first term gives ##s(s+1) \hbar^2## as an eigenvalue. I can't work out the rest and I do not know how to apply it to the spin ket

## |\psi \rangle = | \uparrow \rangle^{(-)} | \uparrow \rangle ^{(+)} ##

I don't get the notation impled by this ket. Does it just mean that it is the state with both spin +1/2. meaning that we have s=1/2+1/2=1?
 
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My problem is for finding expectation values of these operators...

## \langle S_1 \cdot S_2 \rangle = \frac{\hbar^2}{2}(s(s+1) -s_1(s_1+1)-s_2(s_2+1)) ## isn't it?and

##| \uparrow \rangle | \uparrow \rangle = | \uparrow \rangle \bigotimes | \uparrow \rangle ## ?The thing I do not understand is how to convert the

## | \uparrow \rangle^{(+)} | \uparrow \rangle^{(-)} \to | m_1 m_2 ; j_1 j_2 \rangle = | j m ; j_1 j_2 \rangle ##

It is ## |\frac{1}{2}\frac{1}{2};\frac{1}{2}\frac{1}{2} \rangle ##

i don't know what ## j_1## and ##j_2## denote.

##m_1## and ##m_2## denote spin up or spin down i.e. ## m_1 = \pm \frac{1}{2} ## and ##m=m_1+m_2 \in {-1,0,1} ## because we have ## \uparrow \uparrow, \uparrow \downarrow, \downarrow \uparrow, \downarrow \downarrow ## making ## {-1,0,1} ## what indicates that ##j = 1## here?

I have seen that ##j = |l\pm s| = |\pm s| = |s_1 +s_2| ## which certainly indicate some basis ##\{ |11\rangle,|10\rangle ,|1-1\rangle ,|00\rangle \}## is it my job to find the basis in terms of the ## \{ |\uparrow \uparrow\rangle, |\uparrow \downarrow\rangle, |\uparrow \downarrow\rangle, |\downarrow \downarrow\rangle \} ##?
 
Last edited:
Gregg said:

Homework Statement




Hamiltonian for an electron positron system in a uniform magnetic field B (in the z direction)

## \hat{H} = AS^{(-)} \cdot S^{(+)} +\frac{eB}{mc} (S_z^{(-)} +S_z^{(+)} )##

##A \in \Re ##

We have a spin ket given by

## |\psi \rangle = | \uparrow \rangle^{(+)} | \uparrow \rangle^{(-)} ##

For A = 0 is ##| \psi \rangle ## an eigenket?

For B = 0 is it an eigenket?

Homework Equations





The Attempt at a Solution



I believe that the (+) denotes the positron and the (-), an electron. This is a two spin-1/2 particle system.

In lectures we have been given eigenvalues of various spin operators

The second term in the Hamiltonian is

## \frac{eB}{mc}(S^{(+)}_z + S^{(-)}_z ) = \frac{eB}{mc} (S^{(+)+(-)})_z ## which has eigenvalue ## m\hbar = (m_1+m_2)\hbar ##

The first term in the Hamiltonian is something I find confusing

## A S^{(-)} \cdot S^{(+)} ##

My attempt is

## S^{(+)+(-)} = S^{(+)} + S^{(-)} ##

## S^{(-)} \cdot S^{(+)} = ( S^{(+)+(-)}-S^{(+)} )( S^{(+)+(-)}-S^{(-)}) ##

##= S^{(+)+(-)} \cdot S^{(+)+(-)} + S^{(+)} \cdot S^{(-)} - S^{(+)} \cdot S^{(+)+(-)} - S^{(+)+(-)} \cdot S^{(-)} ##

The first term gives ##s(s+1) \hbar^2## as an eigenvalue. I can't work out the rest and I do not know how to apply it to the spin ket
I'm going to use 1 and 2 instead of + and - because it's easier to read and type.

It would be more straightforward to start with ##\vec{S} = \vec{S}_1 + \vec{S}_2##, square it, and solve for ##\vec{S}_1\cdot\vec{S}_2##.

## |\psi \rangle = | \uparrow \rangle^{(-)} | \uparrow \rangle ^{(+)} ##

I don't get the notation impled by this ket. Does it just mean that it is the state with both spin +1/2. meaning that we have s=1/2+1/2=1?
What you've said is correct, but you might be making a bit of a conceptual error here. It means both particles are in the spin-up state so that ##m_{s_1} = 1/2## and ##m_{s_2} = 1/2##. This means that ##m = m_{s_1}+m_{s_2} = 1##, which, in turn, implies that s=1.

Gregg said:
My problem is for finding expectation values of these operators...

## \langle S_1 \cdot S_2 \rangle = \frac{\hbar^2}{2}(s(s+1) -s_1(s_1+1)-s_2(s_2+1)) ## isn't it?
No, expectation values have nothing to do with this problem. You simply want to check if the given state is an eigenstate of the operator. In other words, is it true that
$$(\vec{S}_1 \cdot \vec{S}_2) |\psi\rangle = \lambda|\psi\rangle$$ for some ##\lambda##?

The thing I do not understand is how to convert the

## | \uparrow \rangle^{(+)} | \uparrow \rangle^{(-)} \to | m_1 m_2 ; j_1 j_2 \rangle = | j m ; j_1 j_2 \rangle ##

It is ## |\frac{1}{2}\frac{1}{2};\frac{1}{2}\frac{1}{2} \rangle ##

i don't know what ## j_1## and ##j_2## denote.
This problem is about the addition of angular momentum, i.e. ##\vec{J} = \vec{J}_1 + \vec{J}_2##. The letter J is a generic letter associated with angular momentum. When you talk about a particle's intrinsic spin, it's customary to use S. In other words, in this problem, j1=s1 and j2=s2.

##m_1## and ##m_2## denote spin up or spin down i.e. ## m_1 = \pm \frac{1}{2} ## and ##m=m_1+m_2 \in \{-1,0,1\} ## because we have ## \uparrow \uparrow, \uparrow \downarrow, \downarrow \uparrow, \downarrow \downarrow ## making ## \{-1,0,1\} ## what indicates that ##j = 1## here?

I have seen that ##j = |l\pm s| = |\pm s| = |s_1 +s_2| ## which certainly indicate some basis ##\{ |11\rangle,|10\rangle ,|1-1\rangle ,|00\rangle \}## is it my job to find the basis in terms of the ## \{ |\uparrow \uparrow\rangle, |\uparrow \downarrow\rangle, |\uparrow \downarrow\rangle, |\downarrow \downarrow\rangle \} ##?
Yes, that's essentially what the problem boils down to. Your textbook should cover the addition of angular momentum. I think you're getting confused because you're seeing bits and pieces. Reviewing the topic now that you have an idea of what's going on might prove helpful at this point.
 
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