# Homework Help: Expectation Value of Composite System

1. Oct 21, 2014

### Sci

1. The problem statement, all variables and given/known data
System of 2 particles with spin 1/2. Let
$$\vert + \rangle = \begin{pmatrix} 0 \\ 1 \end{pmatrix} \\ \vert - \rangle = \begin{pmatrix} 1 \\ 0 \end{pmatrix}$$

singlet state $$\vert \Phi \rangle = \frac{1}{\sqrt{2}} \Big( \vert + \rangle \otimes \vert - \rangle - \vert - \rangle \otimes \vert + \rangle \Big)$$
observables:
$$(2 \vec{a} \vec{S}^1) \otimes 1 \\ (2 \vec{a} \vec{S}^1) \otimes (2 \vec{b} \vec{S}^2)$$
for arbitraty a,b
2. Relevant equations
$$S_x^i= \begin{pmatrix} 0 & 1\\ 1 & 0 \end{pmatrix}$$
and similar for S_y and S_z
3. The attempt at a solution
I have to calculate
$$\langle \Phi \vert(2 \vec{a} \vec{S}^1) \otimes 1 \vert \Phi \rangle$$
in the first task. Does the tensor product notation of Phi means that particle A is in state + and particle B is in state - or the other way round?
Does the 1 in the observable means that the state of B is simply ignored in the meaurement?
So does the first case simplify to
$$\langle + \vert a_1 \hat S_x +a_2 \hat S_y +a_3 \hat S_z \vert + \rangle - \langle - \vert a_1 \hat S_x +a_2 \hat S_y +a_3 \hat S_z \vert - \rangle$$
and the expectation value becomes zero, as expected for the singlet state

2. Oct 21, 2014

### Orodruin

Staff Emeritus
The state $|+\rangle \otimes |-\rangle$ is the state with the first particle in the + state and the second particle in the - state. The singlet state $|\Phi\rangle$ that you have is a linear combination of $|+\rangle \otimes |-\rangle$ and $|-\rangle \otimes |+\rangle$.

In general:
$$(\hat A\otimes \hat B)(|a\rangle \otimes |b\rangle) = (\hat A |a\rangle) \otimes (\hat B |b\rangle).$$

3. Oct 22, 2014

### Sci

Thank you!
I am still confused about the basic rules
$$(\langle + \vert \otimes \langle - \vert )( \vec{S} \vec{a} \otimes 1 ) (\vert + \rangle \otimes \vert - \rangle ) \\ -(\langle + \vert \otimes \langle - \vert )( \vec{S} \vec{a} \otimes 1 ) (\vert - \rangle \otimes \vert + \rangle )\\ -(\langle - \vert \otimes \langle + \vert )( \vec{S} \vec{a} \otimes 1 ) (\vert + \rangle \otimes \vert - \rangle )\\ +(\langle - \vert \otimes \langle + \vert )( \vec{S} \vec{a} \otimes 1 ) (\vert - \rangle \otimes \vert + \rangle )$$
$$(\langle + \vert \otimes \langle - \vert )( \vec{S} \vec{a}\vert + \rangle \otimes 1\vert - \rangle ) + others$$
can I do the following step;
$$(\langle + \vert \vec{S} \vec{a}\vert + \rangle \otimes \langle - \vert 1 \vert - \rangle) + others$$

the tensor product doesn't mke sense here...

Last edited: Oct 22, 2014
4. Oct 22, 2014

### Orodruin

Staff Emeritus
When taking the inner product between two states, you simply get the product of the inner products on the separate spaces:
$$(\langle a'|\otimes\langle b'|)(|a\rangle\otimes |b\rangle) = \langle a'|a\rangle \langle b'|b\rangle$$
There is no need to keep the tensor product and as you say it makes little sense to do so.

Edit: that said, you are otherwise on the right track.