Composite Spin 1/2 System Probability Question

[CDL]

Homework Statement

Two spin $\frac 12$ particles form a composite system. Spin A is in the eigenstate $s_z = + \frac 12$ and Spin B is in the eigenstate $s_x = + \frac 12$ What is the probability that a measurement of the total spin will give the value $0$?

Homework Equations

I know particle A is in the state $| \psi_A \rangle = (1,0)$ and particle B is in the state $| \psi_B \rangle = (\frac{1}{\sqrt{2}} , \frac{1}{\sqrt{2}} ) = \frac{1}{\sqrt{2}} (1,0) + \frac{1}{\sqrt{2}} (0,1)$

The basis for the composite system is $B = \{ \ |\uparrow \uparrow \rangle, |\downarrow \uparrow \rangle, |\uparrow \downarrow \rangle, |\downarrow \downarrow \rangle \}$

The Attempt at a Solution

My interpretation is that question is essentially asking us to find the probability that upon measurement, the particles will be measured in the singlet state, or the triplet state with $m = 0$. Is this correct? In the case of 1 dimension, we measure projections of total spin. So is it true that we want states with $m = 0$?

My strategy is to express the given state of the particles in terms of the basis $B$, and then calculate the probability as follows: $$\mathbb{P}(\text{Particles measured to have total spin} \ 0) = |\langle 0 \ 0 | \psi \rangle|^2 + |\langle 1 \ 0 | \psi \rangle|^2$$ Where $| \psi \rangle$ is the initial composite state.

I do not know how to find the initial composite state though! How do I form a composite state from some given single particle spin states? Once I have this, and verified my question about measurements of projections of spin, I should be able to solve it.

 

[DrClaude]

My interpretation is that question is essentially asking us to find the probability that upon measurement, the particles will be measured in the singlet state, or the triplet state with $m = 0$. Is this correct? In the case of 1 dimension, we measure projections of total spin. So is it true that we want states with $m = 0$?

That's not correct. A total spin of zero means that the particles are in the singlet state. The value of the projection is irrelevant.

I do not know how to find the initial composite state though! How do I form a composite state from some given single particle spin states? Once I have this, and verified my question about measurements of projections of spin, I should be able to solve it.

It would be easier if you used a more consistent notation. The Dirac notation is well suited here, so
$$
| \uparrow \downarrow \rangle = | \uparrow \rangle_A \otimes | \downarrow \rangle_B
$$
and so on.

 

[CDL]

It would be easier if you used a more consistent notation. The Dirac notation is well suited here, so

|↑↓⟩=|↑⟩A⊗|↓⟩B|↑↓⟩=|↑⟩A⊗|↓⟩B​

| \uparrow \downarrow \rangle = | \uparrow \rangle_A \otimes | \downarrow \rangle_B
and so on.

Is the $\otimes$ symbol denoting a tensor product? I am finding this concept tough to understand and find good resources on. Do you have any recommendations?

Using that notation, does this working produce the correct composite state?

$\begin{align*} | \psi_a \rangle \otimes | \psi_b \rangle &= | \uparrow \rangle \otimes \frac{1}{\sqrt{2}} \left( | \uparrow \rangle + | \downarrow \rangle \right) \\ &= \frac{1}{\sqrt{2}} \left(| \uparrow \uparrow \rangle + | \uparrow \downarrow \rangle \right) \end{align*}$

 

[DrClaude]

Is the $\otimes$ symbol denoting a tensor product? I am finding this concept tough to understand and find good resources on. Do you have any recommendations?

Greiner's Quantum Mechanics - An Introduction has a chapter on the mathematics of QM. I think that Sakurai and Napolitano also covers this. Online, I was able to find this document: http://web.mst.edu/~parris/QuantumOne/Class_Notes/ManyParticleSystems.pdf

Using that notation, does this working produce the correct composite state?

$\begin{align*} | \psi_a \rangle \otimes | \psi_b \rangle &= | \uparrow \rangle \otimes \frac{1}{\sqrt{2}} \left( | \uparrow \rangle + | \downarrow \rangle \right) \\ &= \frac{1}{\sqrt{2}} \left(| \uparrow \uparrow \rangle + | \uparrow \downarrow \rangle \right) \end{align*}$

That's correct!