Expectation Value of Composite System

[Sci]

Homework Statement

System of 2 particles with spin 1/2. Let
$$\vert + \rangle =<br /> \begin{pmatrix}<br /> 0 \\<br /> 1<br /> \end{pmatrix} \\<br /> \vert - \rangle =<br /> \begin{pmatrix}<br /> 1 \\<br /> 0<br /> \end{pmatrix}$$

singlet state $$\vert \Phi \rangle = \frac{1}{\sqrt{2}} \Big( \vert + \rangle \otimes \vert - \rangle - \vert - \rangle \otimes \vert + \rangle \Big)$$
observables:
$$(2 \vec{a} \vec{S}^1) \otimes 1 \\<br /> (2 \vec{a} \vec{S}^1) \otimes (2 \vec{b} \vec{S}^2)$$
for arbitraty a,b

Homework Equations

$$S_x^i=<br /> \begin{pmatrix}<br /> 0 & 1\\<br /> 1 & 0<br /> \end{pmatrix}$$
and similar for S_y and S_z

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
$$\langle + \vert a_1 \hat S_x +a_2 \hat S_y +a_3 \hat S_z \vert + \rangle -<br /> \langle - \vert a_1 \hat S_x +a_2 \hat S_y +a_3 \hat S_z \vert - \rangle$$
and the expectation value becomes zero, as expected for the singlet state

 

[Orodruin]

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?

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 $|-\rangle \otimes |+\rangle$.

Does the 1 in the observable means that the state of B is simply ignored in the meaurement?

In general:
$$
(\hat A\otimes \hat B)(|a\rangle \otimes |b\rangle) = (\hat A |a\rangle) \otimes (\hat B |b\rangle).
$$
This should help you solve your problem.

 

[Sci]

Thank you!
I am still confused about the basic rules
$$(\langle + \vert \otimes \langle - \vert )( \vec{S} \vec{a} \otimes 1 ) (\vert + \rangle \otimes \vert - \rangle ) \\<br /> -(\langle + \vert \otimes \langle - \vert )( \vec{S} \vec{a} \otimes 1 ) (\vert - \rangle \otimes \vert + \rangle )\\<br /> -(\langle - \vert \otimes \langle + \vert )( \vec{S} \vec{a} \otimes 1 ) (\vert + \rangle \otimes \vert - \rangle )\\<br /> +(\langle - \vert \otimes \langle + \vert )( \vec{S} \vec{a} \otimes 1 ) (\vert - \rangle \otimes \vert + \rangle )$$
using your rule
$$(\langle + \vert \otimes \langle - \vert )( \vec{S} \vec{a}\vert + \rangle \otimes 1\vert - \rangle ) + others$$
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...

 

[Orodruin]

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.