Expectation Value of Composite System

[Sci]

Homework Statement

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

Homework Equations

$$S_x^i= \begin{pmatrix} 0 & 1\\ 1 & 0 \end{pmatrix}$$
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
$$\langle + \vert a_1 \hat S_x +a_2 \hat S_y +a_3 \hat S_z \vert + \rangle - \langle - \vert a_1 \hat S_x +a_2 \hat S_y +a_3 \hat S_z \vert - \rangle$$
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
$$(\langle + \vert \otimes \langle - \vert )( \vec{S} \vec{a} \otimes 1 ) (\vert + \rangle \otimes \vert - \rangle ) \\ -(\langle + \vert \otimes \langle - \vert )( \vec{S} \vec{a} \otimes 1 ) (\vert - \rangle \otimes \vert + \rangle )\\ -(\langle - \vert \otimes \langle + \vert )( \vec{S} \vec{a} \otimes 1 ) (\vert + \rangle \otimes \vert - \rangle )\\ +(\langle - \vert \otimes \langle + \vert )( \vec{S} \vec{a} \otimes 1 ) (\vert - \rangle \otimes \vert + \rangle )$$
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.

 

[paranormal]

?

I would first clarify the notation used in the problem. The tensor product notation in this case means that the two particles are in separate states, with the first particle in state + and the second particle in state -. The 1 in the observable means that the second particle's state is not involved in the measurement, so the expectation value simplifies to just the first particle's state.

Based on this, the expectation value calculation becomes:

\langle \Phi \vert (2 \vec{a} \vec{S}^1) \otimes 1 \vert \Phi \rangle = \langle + \vert a_1 \hat{S}_x + a_2 \hat{S}_y + a_3 \hat{S}_z \vert + \rangle = 0

This is expected for the singlet state, as it represents a state where the two particles are entangled and their individual states cannot be determined. The second observable, (2 \vec{a} \vec{S}^1) \otimes (2 \vec{b} \vec{S}^2), will also result in an expectation value of 0.

In summary, the expectation value of the composite system in the singlet state is 0 for both observables, as expected. This shows that the two particles are entangled and their individual states cannot be measured independently.