Expectation values and probabilities for spinors in a well

[Cameron Roberts]

Working on a homework at the moment involving spinors. The algebra isn't hard at all, I just want to make sure my understanding is right and I'm not doing this incorrectly.

1. Homework Statement
An electron in a one-dimensional infinite well in the region 0≤x≤a is described by the spinor ψ(x) =A (3*sin(πx/a) , 4*sin(2πx/a)).

Find A .

What is the probability that S_z = ħ/2?

Homework Equations

[/B]
1 = ∫ψ*ψ dx

P(χ⁺) = <χ⁺|ψ>^2 = ∫(ψχ)²dx

The Attempt at a Solution

[/B]
So, for normalization, I handled the problem as if it were a simple wave function rather than a spinor, with the only real difference being that ψ* is ψ^†, so that the two matrices reduce to a single term. This gives me A² = 2/(25a).

For the second problem, I'm a bit less confident. My options seem to be either 1. simply do matrix multiplication with the transpose of my spinor and the χ⁺ and get 3A*sin(πx/a) (ħ/2). This doesn't seem in line with other expectation values since it has some x dependence still, so I don't think this is it, though it seems to be how we've handled simple spins in the pas. Or, 2., integrate this term from 0 to a and get a numerical answer, rather than a function of x.

I intend to try the second option, and press on using a similar method for the rest of the problem, but if someone could confirm this is the correct way to do it or guide me towards the correct method, I'd be greatly appreciative!

Edit: Forgot a squared sign. I've got 9/25 as an answer now, after squaring the inner product of the eigenspinor and wavefunction and integrating. Seems like a reasonable answer, as doing the -ħ/2 should yield 16/25.

 

[TSny]

I've got 9/25 as an answer now, after squaring the inner product of the eigenspinor and wavefunction and integrating. Seems like a reasonable answer, as doing the -ħ/2 should yield 16/25.

Looks good. This assumes that you are using up and down along the z-axis as basis vectors when writing the spinor ψ(x).

 

[Cameron Roberts]

Looks good.

Thank you! Pressing on then.