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Suppose there is a spin 1/2 particle in a state $\chi = {1 \over \sqrt{5}} \begin{bmatrix} 1\\ 2\\ \end{bmatrix} $. To determine the probability of finding the particle in a spin up($\hbar/2$) state, we simply multiply the state of the particle by the adjoint of the eigenspinor matrix representing spin up, and square the result. Therefore we get:

$P_+ = {1 \over 5}$

But what if we want the spin up state when we measure $S_x$ and $S_z$?

EDIT: (Example from Griffiths )

The problem I have has been marked by red box. Where from we have the factor $(3+i)$in probability measurement for $S_x$ ? Can you please elaborate that? enter image description here

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closed as off-topic by ACuriousMind, Danu, yuggib, Ryan Unger, DanielSank Sep 9 '15 at 23:26

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  • $\begingroup$ This question has a few problems. First of all, we do not like screenshots of textbooks. If there's something relevant to the question in a book then type that content into the question your self. There are several reasons for this 1) It's easier to read, 2) It leads to more focused questions, which means that 3) It makes it much more likely for you to figure out the answer to your own question. The other problem is that the question is vague. You ask if we can "elaborate". That's not a question. Please find a specific question and ask that :-) $\endgroup$ – DanielSank Sep 9 '15 at 23:26
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Let $\chi$ be the spinor defined as follows:-

$$\chi=\begin{pmatrix} a\\b\end{pmatrix}$$

then for measuring $S_x$ we need to find the eigenspinors of $S_x$ which are

$$\chi_{+}^x= \frac{1}{\sqrt{2}}\begin{pmatrix} 1\\ 1 \end{pmatrix} ,\hspace{1cm} \chi_{-}^x= \frac{1}{\sqrt{2}}\begin{pmatrix} 1\\ -1 \end{pmatrix}$$

Now the spinor $\chi$ can be written as a linear combination of the above two as shown in Griffith eq[4.152]

$$ \chi=\left(\frac{a+b}{\sqrt{2}}\right)\chi_{+}^x + \left(\frac{a-b}{\sqrt{2}}\right)\chi_{-}^x $$

So the probability for $S_x$ is $(1/2)|a+b|^2$ for $+\hbar/2$ and $(1/2)|a-b|^2$ for $-\hbar/2$.

Similarly you can show that for $S_y$ it is $(1/2)|a-ib|^2$ for $+\hbar/2$ and $(1/2)|a+ib|^2$ for $-\hbar/2$.

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  • $\begingroup$ Crustal clear :) $\endgroup$ – Numerical Person Sep 6 '15 at 12:40
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The adjoint eigenspinor you multiplied by was the unit length eigenvector of $\sigma_z$ with positive eigenvalue.

If you you want a spin up result for the direction $(n_x,n_y,n_z)$ find a unit length eigenvector of $n_x\hat\sigma_x+n_y\hat\sigma_y+n_z\hat\sigma_z$ with positive eigenvalue. And use that instead.

If you wanted to do an interaction in the x direction and follow by an interaction in the z direction. Then you need to project onto the two eigenspaces for of $1\hat\sigma_x+0\hat\sigma_y+0\hat\sigma_z$ and then take each result and project them onto the two eigenspaces for of $0\hat\sigma_x+0\hat\sigma_y+1\hat\sigma_z.$ Where I wrote it in an overly complicated way so that you can do any type of directions, not just $\hat x$ and $\hat z.$

To be clear, if you choose the z basis (as you did) then the reason you multiplied by $[1,0]$ is because it was the adjoint of the eigenvector of $\sigma_z$ with positive eigenvalue. Do the exact same thing with $\sigma_x.$

If I don't know what physics concept you are asking about I can't explain the concept more clearly. Pick a direction, get a matrix, find an eigenvector, normalize it, take its adjoint, multiply by your vector, take the magnitude of the result, then square that. Done, that's the probability. Repeat for each eigenvector of the matrix.

If you are doing repeated measurements actually project onto the eigenspaces of the matrices. And take the square of the magnitudes of the projections.

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  • $\begingroup$ Can you please see the example from Griffiths I have attached with the question ? $\endgroup$ – Numerical Person Sep 6 '15 at 11:01
  • $\begingroup$ @SabbirHasan If I don't know what physics concept you are asking about I can't explain the concept more clearly. Pick a direction, get a matrix, find an eigenvector, normalize it, take its adjoint, multiply by your vector, take the magnitude of the result, then square that. Done. Repeat for each eigenvector of the matrix. $\endgroup$ – Timaeus Sep 6 '15 at 11:13

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