# Prove Quantum Spin Problem: θ, Su11, Su22, Sz1, Sz2, Sx2

• Cracker Jack
For example, if σx=-1 then |↓↑>=-(1,0) (and vice versa for σx=+1). In summary, the homework statement is that -ħ2/4 * u1⋅u2 = |00>.f

## Homework Statement

It was noted that <00|Su11Su22|00> = -ħ2/4 * u1⋅u2.

where u1 and us are unit vectors along which the two spin operators are measured and θ is the angle between. |00> is the singlet state that the two electrons are entangled in (corresponding to total spin values ). Prove this relationship using

Su11=Sz1

and

Su22=cosθ*Sz2+sinθ*Sx2

Also use |00>=1/√(2) * (|↓↑>-|↑↓>)

and that Su11 and Su22 act on particle 1 and 2, respectively.

## Homework Equations

Su22=cosθ*Sz2+sinθ*Sx2

Also use |00>=1/√(2) * (|↓↑>-|↑↓>)

## The Attempt at a Solution

I tried to use the matrix representations of these states (i.e. |↑>=(1,0) and |↓>=(0,1)) then writing out

|↓↑>=|↓>*|↑> = (0,1)*(1,0)

and trying to use the respective spin matrices on each particle but I keep getting zero each time I try.

i.e. when I try the first multiplication <00|Su11Su22|

I get -ħ2/4√(2)* {(0,1)*σz*[(1,0)*σzcosθ+sinθ*(1,0)*σx]-(1,0)*σz*[(0,1)*σz*cosθ+sinθ*(0,1)*σx]}

where the σ are the pauli spin matricies.

my answer for <00|Su11Su22| is then just zero.

writing out

|↓↑>=|↓>*|↑> = (0,1)*(1,0)
That's not the way you translate a composite system into the matrix representation. Two or more systems paired together to form a new vector space becomes a single vector space and all vectors in this space must admit the usual column matrix notation. Here you are pairing two two-dimensional vector spaces, the result is a 4 dimensional vector space and the state vector in there will be a 4 elements column matrix.

I get -ħ2/4√(2)* {(0,1)*σz*[(1,0)*σzcosθ+sinθ*(1,0)*σx]-(1,0)*σz*[(0,1)*σz*cosθ+sinθ*(0,1)*σx]}

I think this is OK, but the notation is a bit intimidating. I don't think you should get zero.

For example, what do you get when you simplify this part: (0,1)*σz*[(1,0)*σzcosθ]

This will not cancel out with the other cosθ term: -(1,0)*σz*[(0,1)*σz*cosθ]

I think this is OK, but the notation is a bit intimidating. I don't think you should get zero.

For example, what do you get when you simplify this part: (0,1)*σz*[(1,0)*σzcosθ]

This will not cancel out with the other cosθ term: -(1,0)*σz*[(0,1)*σz*cosθ]
I get zero for both terms, with σz=[(1,0),(0,-1)] (the 2x2 form)

(0,1)*[(1,0),(0,-1)] =(0,-1)

then (0,-1)*(1,0)=0

That's not the way you translate a composite system into the matrix representation.

I believe Cracker Jack is just using direct product notation: |↓↑>=|↓>*|↑> means |↓↑>=|↓>⊗|↑> = (0,1)T⊗(1,0)T where Cracker Jack did not bother to write the transpose symbols to indicate column vectors.

with σz=[(1,0),(0,-1)] (the 2x2 form)

(0,1)*[(1,0),(0,-1)] =(0,-1)
OK

then (0,-1)*(1,0)=0
No. You cannot multiply these two vectors like this. The two "factors" correspond to different particles.
The first factor is (0, -1) = -(0, 1) = - <↓| and refers to particle 1. The second factor is (1, 0) = <↑| and refers to particle 2.
So, (0,-1)*(1,0) = - <↓|⊗<↑| = - <↓↑|.

The OP seems to be not quite familiar with the direct product, which is necessary if he/she wants to work in the matrix representation. In fact, the OP does not need to resort to the matrix representation if he knows how to proceed with the braket notation (this is what he has been doing actually only that the bras and kets are written in matrix form which can turn out to be misleading).

The OP seems to be not quite familiar with the direct product, which is necessary if he/she wants to work in the matrix representation. In fact, the OP does not need to resort to the matrix representation if he knows how to proceed with the braket notation (this is what he has been doing actually only that the bras and kets are written in matrix form which can turn out to be misleading).
I agree. It's easier to stay with the arrow notation as long as the student knows the effect of σx on |↑> or |↓>.