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Does half-integer angular momenta mean that the particle will always be found spinning? For example, if a particle is in a $l=\frac{1}{2}$ state, this means $ m=\pm\frac{1}{2}$ and since $L_z=\hbar m f_l ^m$ the particle will always be spinning in the z-component.

Am I understanding this wrong? How can a particle be always spinning? Is this perpetual motion? I searched around a bit and I keep seeing that particles can actually only have integer angular momentum. Apparently this has something to do with normalisation. What's going on?

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  • $\begingroup$ If you're talking about quantum mechanics (I'm pretty sure you are), then spin has (as far as we know) nothing to do with rotation. $\endgroup$ – QuantumBrick Jul 6 '16 at 4:35
  • $\begingroup$ Check out "the story of spin" by tomonaga. $\endgroup$ – user122066 Jul 6 '16 at 13:22
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Particles do have half-integer spin! All fermions do. It is the bosons which have to have integral spin.

Spin is an intrinsic angular momentum of a particle. You can try to imagine that the particle is spinning, if this helps you sleep better, but it isn't really what is happening.

You might have heard that the angular momentum is conserved because of the symmetry under rotations. In quantum mechanics this is made even more extreme -- angular momentum just characterizes how a quantum state transforms under rotations. In this sense saying that an electron has spin 1/2 means that electron is not rotationally-symmetric (but it doesn't mean its an ellipsoid or something!), but rather transforms (quantum-mechanically!) under rotations in the simplest non-trivial way possible. It turns out that this means that it also has intrinsic angular momentum, by virtue of quantum-mechanical laws. The only way to really understand it is to carefully study quantum mechanics from very basics.

By the way, $m$ that you use in the question is just a projection of angular momentum on an axis. Even if you have integral angular momentum, you are still always "spinning" in some direction with that full angular momentum, although projection of your "rotation" on orthogonal axis can be zero.

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Assume, without the loss of generality, that the angular momentum operator has only one component, $\hat{L_z}$. It can be shown that the following eigenvalue equation holds true.

$$ L_{z} |l, m \rangle = m \hbar |l, m \rangle \quad \text{such that} \quad m \in \mathbb{Z}. $$

This is to say that the eigenvalues of the orbital angular momentum operator are integer multiples of $\hbar$. You can show this is two ways:

In the co-ordinate (polar) representation, $\hat{L_z} = -i \hbar \frac{\partial}{\partial \phi}$. The eigenvalue equation,

$$ \hat{L_z} |l, m \rangle = l_z |l, m\rangle, $$

implies that the (unnormalized) eigenfunction of $\hat{L_z}$ is:

$$ e^{\frac{il_z\phi}{\hbar}}. $$

We can prove the desired result by:

  1. Imposing the condition that the aforementioned eigenfunction has to be single-valued; and
  2. $\hat{L_z}$ is a hermitian operator. That is: $$ \langle \psi_1 | \hat{L_z} | \psi_2 \rangle = \langle \psi_2 | \hat{L_z} | \psi_1 \rangle ^{*} $$

We can also have angular momentum whose eigenvalues are half integer multiplies of $\hbar$. This is referred to as spin angular momentum. One can show that in the matrix mechanical formulation of quantum mechanics, the eigenvalues of angular momentum can in general be both half or full integer multiples of $\hbar$.

One shouldn't think of the spin angular momentum as having a classical counterpart. Let me give you an example:

Apply the rotation operator (whatever its functional form) on a quantum state with spin pointing in the positive $z$ direction. The result is the following:

$$ \hat{R}(2 \pi) | + z \rangle = -| + z \rangle, $$

where I have chosen to rotate the spin state by 360 degrees. Defying classical intution, the state of, say electron, isn't what we expect: classically, we expect the result of the application of the rotation operator to be the same as the result of our applying the identity operator, which'll yield the initial state as the final state. It is only when one rotates the electrons by $4 \pi$ radians that the final and initial states are the same. The result is both counter-intuitive and against classical intuition.

Note: To be pedantic, the quantum states on both the right and left sides are equivalent/same. (why?) So please choose to ignore the terminology, in its correct QM interpretation, since in the previous paragraph what I'm tying to convert is that the expressions on the right and the left hand sides don't match.

Therefore, one shouldn't be making a correspondence between the spin angular momentum and classical intuition/results. For example, the spin of an electron can be fully specified on a Bloch Sphere and one usually uses the term that one can perform "rotations" on the Bloch sphere, which can be thought of (for example) as a result of time evolution on a Bloch sphere. Yes, we can but one shouldn't think of these "rotations" as operations that have a classical counterpart in all cases, as for example in the previous example. Therefore, the quantum operator of interest in this case is usually not a rotation operator bur rather a unitary operator, which can be thought of as generalization of a rotations operator in either a complex vector space or a Hilbert space.

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  • $\begingroup$ It is misleading to say that the spin "flipped" after a 360 rotation. You do say later that the states are equivalent, but "flipped" can be mistaken by someone for an actual change in direction of the spin. Just saying this because I often encounter this misconception:) $\endgroup$ – Peter Kravchuk Jul 6 '16 at 5:50
  • $\begingroup$ Yes, I have edited the post to remove the aforementioned ambiguity. $\endgroup$ – Junaid Aftab Jul 6 '16 at 5:54
  • $\begingroup$ Nitpicking: there is still a part saying "... that the spin would remain the same". Personally, I do not think that any classical intuition is violated, since there is no classical intuition for state vectors; only for the state which, being a ray (a vector defined up to a phase), behaves as one would expect classically. This is the reason why we allow projective representations in the first place. In any case, it doesn't seem very relevant to the question at hand, which is mostly concerned with the idea of intrinsic angular momentum. $\endgroup$ – Peter Kravchuk Jul 6 '16 at 6:01
  • $\begingroup$ Yes, true. Although I don't have a very deep understanding of projective representations, even in the context of SO(3) and SU(2), I was initially contemplating whether or not to add a comment along these lines in the answer but then I chose not to do so since the scope of the question is clearly different. $\endgroup$ – Junaid Aftab Jul 6 '16 at 6:04

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