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.