Pauli Matrices and orthogonal projections

[clumsy9irl]

Ok, I'm working with the Pauli Matrices, and I've already gone through showing a few bits of information. I've got a good idea how to keep going, but I'm not exactly sure about this one--

say M= 1/2(alphaI + a*sigma)

where alpha E C, a=(ax, ay, az) a complex vector, a*sigma=ax sigmax+ay sigmay+ az sigmz, and I is the identity matrix.


So, an orthogonal projection means that for a matrix P, P^2 and P dagger are both equal to P, right?

Supposedly alpha and a can beonstrained so that M is an orthogonal projection.


How would I go about doing that?


Thanks muchly!

 

[florianb]

Some hints

Hey there !

Here are some hints on a was to do it :

You will only use the fact that M^2 = M...

1) Compute M^2 :

M^2 = \cfrac{1}{4}\left(\alpha I + \sum_i a_i \sigma_i \right)^2 = ...

You will find a term like \sum_i \sum_j a_i a_j \sigma_i \sigma

In order to reduce this term, use some common identities :

\sigma_i \sigma_j = I \delta_{ij} + i \epsilon_{ijk} \sigma_k

and

\sigma_i \sigma_j + \sigma_j \sigma_i = 2 \delta_{ij} I

If you applied them correctly, you should get something quite simple. You then use the fact that M^2 = M and you will find by identification :
\alpha = 1 \text{ and } ||a|| = 1, which is your final answer (Hopefully, I didn't mess up).

If you're not confident with the use of the Levi-Civita symbol, another (more tedious) way is to write all in matrix notation (2x2), compute M^2 and put this equal to M... You will get 4 equations, which will reduce to the answer given above.

Hope this helps,
Cheers,
Florian

 

[clumsy9irl]

i must've screwed something up, for I'm getting a sqrt(2) for my alpha?

maybe i screwed up something with the identities...

 

[clumsy9irl]

oh, and i almost forgot, THANK YOU VERY MUCH!

 

[mayriluseeya]

can we solve the particle in a box problem using schrodingers equation?

 

[quantumdude]

can we solve the particle in a box problem using schrodingers equation?

Sure, it's easy; they do it in Halliday and Resnick.

(But not with the Pauli spin matrices.)