Consider the state-space with a base formed by the eigenstates of the operator $\hat{S}_z$. For the state $|\phi\rangle=\frac{1}{\sqrt2}|+\rangle_z-\frac{1}{\sqrt2}|-\rangle_z$, what is the value of $\langle\hat{S}_x\rangle$?

I have absolutely no idea of how to do this. I don't even understand quite well the expression of $|\phi\rangle$ itself. How could this be done?

EDIT: Thanks to Asaf's answer I've been able to understand the subject better. For simplicity, I'll write just $|+\rangle$ instead of $|+\rangle_z$ and $|-\rangle$ instead of $|-\rangle_z$. So I did the following:

$$\langle\phi|\hat{S}_x|\phi\rangle=\left(\frac{1}{\sqrt2}\langle+|-\frac{1}{\sqrt2}\langle-|\right)\left(\frac{1}{\sqrt2}\hat{S}_x|+\rangle-\frac{1}{\sqrt2}\hat{S}_x|-\rangle\right) =\left(\frac{1}{\sqrt2}\langle+|-\frac{1}{\sqrt2}\langle-|\right) \left(\frac{1}{\sqrt2}\frac{\hbar}{2}|-\rangle-\frac{1}{\sqrt2}\frac{\hbar}{2}|+\rangle\right) =\frac12\frac{\hbar}{2}\langle+|-\rangle-\frac12\frac{\hbar}{2}\langle+|+\rangle-\frac12\frac{\hbar}{2}\langle-|-\rangle+\frac12\frac{\hbar}{2}\langle-|+\rangle=-\frac12\frac{\hbar}{2}-\frac12\frac{\hbar}{2}=-\frac{\hbar}{2}$$

But I don't know if this is right. I would have expected that, as $|\phi\rangle$ only has spin components in $z$, the spin in another axis would be $0$. Did I do anything wrong in the calculations or is this right but I'm getting the concept wrong?


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  • 1
    $\begingroup$ Hint: Try writing $|+\rangle_z$ and $|-\rangle_z$ as linear combinations of $|+\rangle_x$ and $|-\rangle_x$. $\endgroup$ – WillO May 1 '16 at 6:18
  • $\begingroup$ @WillO Should I use the Pauli matrixes to change the basis of the vectors? I haven't found any exercise of this type so I'm pretty lost $\endgroup$ – Tendero May 1 '16 at 13:30
  • 1
    $\begingroup$ You have the right answer, I will expand mine to help you interpret the result. $\endgroup$ – Asaf May 1 '16 at 19:33

First, let's clarify the expression of $|\phi\rangle$.

The kets $|+\rangle_z$ and $|-\rangle_z$ are eigenvectors of $\hat{S}_z$ such that $$ \hat{S}_z |+\rangle_z = +\frac{\hbar}{2}|+\rangle_z \\ \hat{S}_z |-\rangle_z = -\frac{\hbar}{2}|-\rangle_z $$ This means that in the $\{|+\rangle_z = \left(\substack{1\\0}\right),|-\rangle_z=\left(\substack{0\\1}\right)\}$ basis the $\hat{S}_z$ has the matrix representation $$ \hat{S}_z = \frac{\hbar}{2}\left( \begin{array}{cc} +1 & 0\\ 0 & -1 \end{array} \right) = \frac{\hbar}{2}\left(|+\rangle_z\langle+|_z-|-\rangle_z\langle-|_z\right). $$

Now, the $\hat{S}_x$ has the following matrix representation $$ \hat{S}_x = \frac{\hbar}{2}\left( \begin{array}{cc} 0 & 1\\ 1 & 0 \end{array} \right) = \frac{\hbar}{2}\left(|+\rangle_z\langle-|_z+|-\rangle_z\langle+|_z\right) $$

To calculate $\langle\phi|\hat{S}_x|\phi\rangle$ you can now substitute everything and find it.


To interpret the result, think about it like this: An eigenstate of $\hat{S}_z$ has a well defined $z$ component of the angular momentum $\vec{S}$ but you don't know the values of the $x$ and $y$ components. In fact, you cannot know because there is an uncertainty principle that prevents it.

It works like in this picture. If the state is $|+\rangle$, you know that $\vec{S}$ is somewhere in the upper cone but you can't know exactly where. Same goes for $|-\rangle$ and the lower cone.

enter image description here

Now, if you take a look you will see that your state $|\phi\rangle$ is actually an eigenvector of the operator $\hat{S}_x$. It is in fact the $|-\rangle_x$ state so you can think about it as a cone pointing in the $-x$ direction.

  • $\begingroup$ I'm not familiar with this spin notation thing yet. Aren't the eigenvalues supposed to be $\frac{\hbar}{2}$ and $-\frac{\hbar}{2}$? Or am I wrong? $\endgroup$ – Tendero May 1 '16 at 17:14
  • $\begingroup$ Sorry, you are right. I was leaving that out for simplicity. $\endgroup$ – Asaf May 1 '16 at 17:21
  • $\begingroup$ No problem, I was just asking, thank you. Another thing, am I trying to calculate $\langle+|\hat{S}_x|+\rangle$ or $\langle\phi|\hat{S}_x|\phi\rangle$? $\endgroup$ – Tendero May 1 '16 at 17:28
  • $\begingroup$ You want $\langle\phi|\hat{S}_x|\phi\rangle$. I didn't write the full answer here so you can work it out. I will edit for clarity. $\endgroup$ – Asaf May 1 '16 at 17:56
  • $\begingroup$ I will edit my question with the progress I've made so far. Please correct me if you see anything wrong $\endgroup$ – Tendero May 1 '16 at 18:41

I wonder why we so seldom mention, when discussing these things, that you cannot answer this question without adopting a human convention with respect to the combination of spin states. If we take the +/- z direction to be the north and south poles, then any state with equal amplitude-squared in both components will correspond to a spinor pointing somewhere towards the equator. Could be the x direction, could be the y direction, but somewhere in the equator. To say exactly what direction you have to adopt some human convention with respect to the relative complex ratio of the two amplitudes. There is no intrinsically correct answer to this question based on pure physics.

  • $\begingroup$ I guess in this case it's because the convention is very well established and widely used. If we stick to use the standard Pauli matrices then it's all set. $\endgroup$ – Asaf May 1 '16 at 20:01

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