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I'm trying to get an idea of what spin in quantum mechanics means. I have the following questions regarding spin:

The eigenvalues of the operator corresponding to the z component of spin is $\hat{S}_{z}$ with eigenvalues $\frac{\hbar}{2}$ and $-\frac{\hbar}{2}$ respectively. We then have two corresponding eigenvectors $|\frac{1}{2}(\frac{1}{2}) \rangle$ and $|\frac{1}{2}(-\frac{1}{2}) \rangle$. As I understand from the postulates of QM, any quantum state can be expressed as a linear combination of these eigenvectors. What I don't understand is how we know that these eigenvectors can be expressed as $\chi_+ = \left( \begin{array}{c} 1\\ 0\\ \end{array} \right) \text{ and } \chi_- = \left( \begin{array}{c} 0\\ 1\\ \end{array} \right)$?

What exactly does it mean for a state to be in z spin up state $\chi_+ = \left( \begin{array}{c} 1\\ 0\\ \end{array} \right)$ where it has eigenvalue $\frac{\hbar}{2}$? Is the following the correct interpretation: Does it simply mean that the spin around the z axis can take only two values $\frac{\hbar}{2}$ and $-\frac{\hbar}{2}$ which corresponds to counter clockwise spin around the $\hat{z}$ axis (spin up) so that the vector points up the z axis and respectively counter clockwise spin around the $- \hat{z}$ axis (spin down) so the vector points down the z axis?

If this is the correct interpretation then how does the form of the eigenvectors $\left( \begin{array}{c} 1\\ 0\\ \end{array} \right) \text{ and } \left( \begin{array}{c} 0\\ 1\\ \end{array} \right)$ correspond to a vector up the z axis and one down the z axis?

Thanks for any assistance.

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  • $\begingroup$ The $z$-axis isn't the spatial $z$-direction that you may think of: it is one of the three components of the spin and has no connection at all with the spatial dimensions. There is no (counter)-clockwise rotation anywhere and the state being a composition of the spin eigenstates work like any other state being the composition of eigenstates of any other operator (under the assumption that the operators spectra span the entire state space). $\endgroup$
    – gented
    Nov 22, 2016 at 19:04
  • $\begingroup$ @GennaroTedesco Okay thanks, if you have a chance please see the question in my post. $\endgroup$
    – Alex
    Nov 22, 2016 at 19:09
  • $\begingroup$ @GennaroTedesco What do these components of spin of a particle refer to? $\endgroup$
    – Alex
    Nov 22, 2016 at 19:10
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    $\begingroup$ Spin is a set of three observables that any particle possesses that happen to have the same commutation relations as the orbital angular momentum; that is why in the literature their components are referred to as $(x,y,z)$. In general you can consider it as any set of observables $A_i$ that you use to describe a quantum system. $\endgroup$
    – gented
    Nov 22, 2016 at 19:15
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    $\begingroup$ @gented "The $z-$axis isn't the spatial $z-$direction that you may think of: it is one of the three components of the spin and has no connection at all with the spatial dimensions." This is precisely false. Spin has everything to do with spatial dimensions and only with them. The $S_z$ operator is a generator of rotations in the 3-dimensional space--not in some abstract space. Spin has exactly as much to do with rotations as orbital angular momentum has to do with rotations. They are both components of the same quantity, i.e., angular momentum. $\endgroup$
    – user87745
    May 4, 2020 at 19:29

4 Answers 4

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Understanding a quantum quantity such as the spin of a particle is an impossible task. Many physicists told us so, and trying to explain the spin of a particle using your "rotation picture" is doomed to failure. Nevertheless, personally, I feel that trying to use pictures is still fruitful, because it helps to memorise the quantum effects. However, we have to keep in mind that quantum mechanics is a mathematical description of nature. This fact is key to answer your questions.

1. What I don't understand is how we know that these eigenvectors can be expressed as $\chi^+$ and $\chi^-$?

Linear algebra tells us that we can represent a vector in many different ways. The key of this representation are the base vectors. Both bases {$\frac{1}{\sqrt 2}|1, 1\rangle$, $\frac{1}{\sqrt 2}|1, -1\rangle$}, and {$|1, 0\rangle$, $|0,1\rangle$}, can be used to represent any vector in the 2-dimensional space. Note that base vectors neither need to be orthogonal, not normalised. However, often these two properties are convenient.

2. What exactly does it mean for a state to be in z spin up state 𝜒+ where it has eigenvalue ℏ/2?

It means that, if we measure the spin of this particle in $z$-direction, we know the outcome before the measurement is performed. As was mentioned before, you could read about the Stern-Gerlach experiment, which proofed, that if we remeasure a particle spin-state multiple times in the same spatial direction, it does not change the result. However, if we first measure the spin in the $z$-direction, then in the $x$-direction, and finally in the $z$-direction again, the final measurement yields random results -- the measurement in $x$-direction randomises the spin-component along the $z$-direction.

Your interpretation

 Does it simply mean that the spin around the z axis can take only two
 values ℏ/2 and −ℏ/2?

is also correct. If the spin state of a spin-1/2 particle is measures, we obtain only one of the two results: Either spin-up, or spin-down. In quantum mechanic slang we say that the possible measurement results are the eigenvalues of the observable. Therefore, after the measurement the spin-1/2 particle will be in the eigenstate of the observable, which is consistent with the measurement result.

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Good question---that's awesome you are studying quantum mechanics---keep it up!

I think it may help to understand what $|\frac{1}{2} \frac{1}{2}\rangle$ and $|\frac{1}{2} \frac{-1}{2}\rangle$ mean. For example, I have a state $|\Psi \rangle = |\frac{1}{2} \frac{1}{2}\rangle$ . What is the inner product of $\Psi$ with respect to $|\frac{1}{2}\frac{1}{2}\rangle$ (i.e., $\langle\frac{1}{2}\frac{1}{2}|\Psi\rangle$? What is the inner product of $\Psi$ with respect to $|\frac{1}{2} \frac{-1}{2}\rangle$ (i.e., $\langle\frac{1}{2}\frac{-1}{2}|\Psi\rangle$)? (If you cannot answer these questions, let me know and I will alter my teaching method.)

Now, think about a putatively different example. Suppose I am given a vector $\vec{r} = \hat{x} + \hat{y}$. How would you represent this vector in Cartesian coordinate? How about the vector $\vec{r} = \hat{x}$?

Now try to find a relationship between the two different examples above. Did you answer your own question?

Once you understand this mathematical representation of quantum states, you can begin to answer your second question: The quantum state, $\Psi = | \frac{1}{2}\frac{1}{2}\rangle$ is an eigenstate of the angular momentum, spin-z component with eigenvalue equal to $\frac{\hbar}{2}$. Thus, in this quantum state, when the spin-z angular momentum component is measured, you will obtain the value $\frac{\hbar}{2}$.

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  • $\begingroup$ I think I understand your point thanks. The $(1 , 0)$ just corresponds to the component in a given basis which is not explicitly shown in the notation. In this case the basis would be $| \frac{1}{2}(\frac{1}{2}) \rangle$. $\endgroup$
    – Alex
    Nov 22, 2016 at 19:49
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It's the convention.

The spin for the electron was detected experimentally - the math came later.

A non-homogeneous magnetic field (or a magnetic filter) picks out the z direction - which has 2 eigenvalues, namely, $+\hbar/2$ and $-\hbar/2$.

Simply having one eigenvector pointing up and the other one pointing down along the same axis is odd.

But if you haven't tried to predict the outcomes in the Stern-Gerlach experiment by blocking one of the beams or rotating the magnetic, things get weirder.

See the Stern-Gerlach experiments in the Feynman lecture series chapter 5 and chapter 6.

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This confused me a lot when I was studying quantum mechanics, too - it seems like $\binom{1}{0}$ should be perpendicular, not opposite to $\binom{0}{1}$. As other posters have already said, the key is to not think about it visually - the two eigenvectors are not physical vectors that point in the same direction as the spin, they are just abstract mathematical objects which represent one of hte two possible spin states.

Remember, when we work with continuous variables like position or momentum, our operators have infinite eigenvectors (corresponding to the infinite possible values that these operators can take on - they are not quantized). These eigenvectors don't somehow point in the direction of the position or momentum - again, their meaning is more abstract.

I hope all of this helps!

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