# Two spin half particles

Viona
Homework Statement:
the matrices representing the operators S^2 and S_z
Relevant Equations:
the matrices representing the operators S^2 and S_z
I have this homework: consider the case of two spin half particles. Use the basis: |++>, |+->, |-+>, |--> to find the matrices representing the operators S^2 and S_z.
My idea for the solution for S_z is: S_z=S_z(1)+S_z(2) where S_z(1) is the operator for the first particle ... etc
So I will first find the S_z(1) matrix. The first element in the matrix will be: <++|S_z(1)|++>=(hbar/2) the second element will be <++|S_z(1)|+->=0 ... etc Where I finally get a diagonal matrix. Is this procedure correct?

## Answers and Replies

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Homework Statement:: the matrices representing the operators S^2 and S_z
Relevant Equations:: the matrices representing the operators S^2 and S_z

I have this homework: consider the case of two spin half particles. Use the basis: |++>, |+->, |-+>, |--> to find the matrices representing the operators S^2 and S_z.
My idea for the solution for S_z is: S_z=S_z(1)+S_z(2) where S_z(1) is the operator for the first particle ... etc
So I will first find the S_z(1) matrix. The first element in the matrix will be: <++|S_z(1)|++>=(hbar/2) the second element will be <++|S_z(1)|+->=0 ... etc Where I finally get a diagonal matrix. Is this procedure correct?
What do you mean by ##S_z = S_z(1) + S_z(2)##?

What dimension of matrices are you looking for?

Viona
I am looking for 4x4 matrices.

PeroK
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What do you mean by ##S_z = S_z(1) + S_z(2)##?
I think he means ##S_z=s_{z1}+s_{z2}##. Then $$(s_{z1}+s_{z2})|+-\rangle=\frac{\hbar}{2}|+-\rangle+\left(-\frac{\hbar}{2}\right)|+-\rangle.$$

Viona
Viona
I think he means ##S_z=s_{z1}+s_{z2}##. Then $$(s_{z1}+s_{z2})|+-\rangle=\frac{\hbar}{2}|+-\rangle+\left(-\frac{\hbar}{2}\right)|+-\rangle.$$
Yes this what i mean.

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I am looking for 4x4 matrices.
What do you know about the properties of ##S_z## and ##S^2## on the space of two-particle spin states? Do you know how they operate on the spin states you are given? Or, do you have to work that out from first principles?

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Yes this what i mean.
Technically that means that $$S'_z = S_z \otimes I + I \otimes S_z$$Where ##S'_z## is the z-spin operator on the two-particle space and ##I## is the identity operator. I think it helps sometimes to know this so that you don't go wrong oversimplifying the product of operators.

Last edited:
George Jones and vanhees71
Viona
Technically that means that $$S'_z = S_z \otimes I + I \otimes S_z$$Where ##S'_z## is the z-spin operator on the two-particle space and ##I## is the identity operator. I think it helps sometimes to know this so that you don't wrong oversimplifying the product of operators.
The problem is I am not familiar with this type of product using this symbol: \otimes ! So will it be wrong if I used the way I described above a got the correct matrices?

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The problem is I am not familiar with this type of product using this symbol: \otimes ! So will it be wrong if I used the way I described above a got the correct matrices?
You haven't said what you're going to do about ##S^2## - that may be the tricky one. What do you get for ##S_z##?

PS I don't understand why, in your original post, you appear to be using only ##S_z(1)##.

Viona
For Sz: I added the two 4x4 matrices together: Sz1 + Sz2 and I got 4x4 diagonal matrix with diagonal elements: hbar, 0, 0, hbar.

Viona
for S^2 I will say: S^2=(S1+S2)^2= (S1)^2 + (S2)^2 +2S1.S2 where S1.S2=Sx1Sx2+Sy1Sy2+Sz1Sz2
and add all these 4x4 matrices together to find S^2

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The problem is I am not familiar with this type of product using this symbol: \otimes ! So will it be wrong if I used the way I described above a got the correct matrices?
The formal notation for the vectors is, for example:$$|+-\rangle = |+\rangle \otimes |-\rangle$$ In this case, then:$$S'_z|+-\rangle = (S_z \otimes I + I \otimes S_z)(|+\rangle \otimes |-\rangle) = (S_z|+\rangle \otimes I |-\rangle) + (I|+\rangle \otimes S_z |-\rangle)$$$$= (\frac \hbar 2 |+\rangle \otimes |-\rangle) + (|+\rangle \otimes \frac \hbar 2 |-\rangle) = \hbar (|+\rangle \otimes |-\rangle)$$And we see that ##|+-\rangle## is an eigenstate of ##S'_z## with eigenvalue ##\hbar##.

You can drop the prime - I was just using it to make it clear that it's technically a different operator on the two-particle space.

vanhees71 and Viona
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for S^2 I will say: S^2=(S1+S2)^2= (S1)^2 + (S2)^2 +2S1.S2 where S1.S2=Sx1Sx2+Sy1Sy2+Sz1Sz2
and add all these 4x4 matrices together to find S^2
That's not going to work: $$S^2 = S_x^2 + S_y^2 + S_z^2 = \vec S \cdot \vec S$$

docnet
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PS Calculating ##S^2## from first principles is more work than you might think! That's why I asked if you already know how the operator ##S^2## acts on the two-particle space.

This is tricky until you get the hang of it, so don't worry if you are confused at this stage.

Viona
Viona
What do you know about the properties of ##S_z## and ##S^2## on the space of two-particle spin states? Do you know how they operate on the spin states you are given? Or, do you have to work that out from first principles?
I only studied two spin half particles from the book Introduction to Quantum Mechanics by David J. Griffiths. I do not know if this enough For example see the attached picture.

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I only studied two spin half particles from the book Introduction to Quantum Mechanics by David J. Griffiths. I do not know if this enough For example see the attached picture.
View attachment 289739
I know he uses that shorthand notation, but that was something I didn't like. I learned QM from that book but I took some time to find out about the tensor product of hilbert spaces ##\otimes##. As, otherwise, I couldn't figure out what was going on with those composite operations.

If you've got that book, then look at the triplet and singlet states on page 185 (2nd edition). That should give a big clue of how ##S^2## works.

Also, Griffiths does some of the calculations for ##S^2## on page 186: note the decomposition of ##S^2## as I indicated above:

That's not going to work: $$S^2 = S_x^2 + S_y^2 + S_z^2 = \vec S \cdot \vec S$$

Viona
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For Sz: I added the two 4x4 matrices together: Sz1 + Sz2 and I got 4x4 diagonal matrix with diagonal elements: hbar, 0, 0, hbar.
It might just be a typo, but there's a negative sign missing on one of the ##\hbar##s.

Viona and PeroK
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It might just be a typo, but there's a negative sign missing on one of the ##\hbar##s.
Where? In post #15?

Viona
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In the post I quoted, #10.

Viona
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In the post I quoted, #10.
Ah yes. That matrix should be traceless. Thanks.

Viona