Superposition of different spins

[Isaac0427]

TL;DR Summary

Is it theoretically possible to have a particle in a superposition of states with different $S^2$ eigenvalues?

Hi all, it has been quite a while since I've been on here.

I am curious, as the TLDR summary says, if it theoretically possible to have a particle in a superposition of states with different $S^2$ eigenvalues. For example, a particle being in the state $\frac{1}{\sqrt{2}}\left(|s=\frac{1}{2}, m_s=\frac{1}{2}\rangle + |s=1, m_s=0\rangle\right)$. This would make the particle be neither a boson nor a fermion, and have some very weird behavior under exchange of two identical particles. I was thinking that strange behavior could lead to some explanation of why these particles (to my knowledge) don't exist.

So, is there a theoretical reason this type of particle could not exist? Or is it possible in theory and they are simply not observed?

 

[Hyperfine]

A particle has a single, unique value of S². Thus the proposed superposition cannot exist for a single particle as specified.

 

[strangerep]

So, is there a theoretical reason this type of particle could not exist?

Yes. The $S^2$ operator is a Casimir operator for the relevant physical group, in this case the group of spatial rotations $SO(3)$. The point of a Casimir operator is that it is invariant under those group operations, and thus characterizes physically distinct feature(s) of particles.

Moreover, when one analyzes the structure of the (unitary) group representations for any given eigenvalue of the Casimir operator, one finds that the Hilbert spaces have different dimensions. E.g., for a spin-$\frac12$ particle, the Hilbert space is 2-dimensional (corresponding to the 2 distinct orientations of spin). For a spin-1 particle the Hilbert space is 3-dimensional, etc, etc. It makes no mathematical sense to try and form a superposition between such incompatible spaces. (Ballentine ch7 is recommended reading on these points.)

Such situations that exclude particular superpositions are often called "superselection rules".

Another way to look at it is that the basis states for any given $S^2$ (i.e., the different "m" states corresponding to different spin orientations) can always be transformed among themselves by the group operations. E.g., spin-up can be rotated into spin-down, etc. There exists no such similar transformation for different total spins -- you can't rotate a spin-1 particle into spin-2, etc.

Bottom line: it's all about the intricate structure of unitary irreducible representations of physically relevant groups. Casimirs and representation theory are powerful tools for investigating theoretical aspects of the real world.

 

[vanhees71]

The superselection rule, that excludes the superposition of angular-momentum states with both half-integer and integer representations, is simple to explain.

We define angular-momentum eigenstates $|J,M \rangle$ by
$$\hat{\vec{J}}^2 |J,M \rangle=\hbar^2 J(J+1), \quad \hat{J}_z |J,M \rangle=M \hbar.$$
Here $J \in \{0,1/2,1,\ldots \}$ and for a given $J$, $M \in \{-J,-J+1,\ldots,J-1,J \}$.

Now for a given $J$ the rotation around the $z$-axis with angle $(2 \pi)$ changes such an angular-momentum eigenstate to
$$\exp(-2 \pi \mathrm{i} \hat{J}_Z /\hbar)|J,M \rangle=\exp(-2 \pi \mathrm{i} M) |J,M \rangle.$$
For $J \in \mathbb{N}$ also $M \in \mathbb{N}$, which means that the exponential (phase) factor is $+1$. For $J \in (2 \mathbb{N}+1)/2$ also $M \in (2 \mathbb{N}+1)/2$, and thus the phase factor is $-1$.

Of course a rotation with angle $(2 \pi)$ shouldn't change your state, and that's the case as long as all vectors multiply by the same (!) phase factor. That's the case of any superposition of only (!) integer or only (!) half-integer-$J$ states, while for superpositions of an integer and a half-integer $J$ state, that's no longer the case, i.e., the rotation around $(2 \pi)$ leads not anymore simply to a phase factor but to a different state. That's a contradiction in the interpretation of the representation of rotations within the quantum-theoretical formalism, and thus we must forbid superpositions of integer- and half-integer-$J$ states. That's the angular-momentum superselection rule.

 

[Isaac0427]

Moreover, when one analyzes the structure of the (unitary) group representations for any given eigenvalue of the Casimir operator, one finds that the Hilbert spaces have different dimensions. E.g., for a spin-12 particle, the Hilbert space is 2-dimensional (corresponding to the 2 distinct orientations of spin). For a spin-1 particle the Hilbert space is 3-dimensional, etc, etc. It makes no mathematical sense to try and form a superposition between such incompatible spaces. (Ballentine ch7 is recommended reading on these points.)

Thank you for the answer. It leaves me with this question though: wouldn't the same logic prohibit, say, the state $\frac{1}{\sqrt{2}}(|210\rangle + |220\rangle)$ (expressing states as $|nlm\rangle$? My understanding is that a state like that would be allowed. Additionally, I checked out Ballentine section 7.6 as you suggested, where it says

The superselection rule generated by R(2π), which separates states of integer angular momentum from states of half odd-integer angular momentum, is the only superselection rule that occurs in the quantum mechanics of stable particles.

Is there still something different about the spin case that is not true with orbital or total angular momentum?

 

[vanhees71]

No, that's the only superselection rule due to angular momentum.

 

[Hyperfine]

Is it theoretically possible to have a particle in a superposition of states with different $S^2$ eigenvalues?

Emphasis added.

Is there still something different about the spin case that is not true with orbital or total angular momentum?

I think you need to be careful to recognize that spin is an intrinsic physical property of a given particle. Change S and you change the particle.

By comparison, the quantum numbers $n, \ell, m$, and $j$ characterize a state of the system. All of those quantum number are allowed to take on a restricted range of values corresponding to distinct states of the system. States of the system can mix.

 

[Isaac0427]

I think you need to be careful to recognize that spin is an intrinsic physical property of a given particle. Change S and you change the particle.

Right, I do know this. My question here was why that is the case, specifically if it was a theoretical necessity or an empirical observation. What I have gathered so far is that the superposition of integer-spin and half-integer spin is forbidden by theoretical necessity. The superposition of spins in general, regardless of whether they are integer or half-integer spins, is simply not observed. Is this correct? Or is there something in particle physics that places the further limitation on spins?

 

[Hyperfine]

The superposition of spins in general, regardless of whether they are integer or half-integer spins, is simply not observed. Is this correct? Or is there something in particle physics that places the further limitation on spins?

That depends on the situation under consideration. For a single particle, as initially stated, then yes. The limitation is the identity of the single particle. S is not just a quantum number, it is an intrinsic physical property.

For an ensemble of particles, then not necessarily. Allow me to provide an example where a superposition of electron spins is most certainly allowed. Consider an ensemble of hydrogen atoms with random electron spin states as they undergo a chemical reaction to form molecular hydrogen.
$$H\cdot + H\cdot \rightarrow H_{2}$$The electron spin system of the pair of hydrogen atoms can be described in terms of a superposition resulting in four possible states where $\mid \alpha >$ or $\mid \beta >$ represents the spin state of one of the electrons:
$$T_{+}=\mid \alpha \alpha > $$$$T_{-}=\mid \beta \beta >$$$$T_{0}=\frac{1}{\sqrt{2}}\left ( \mid \alpha \beta > +\mid \beta \alpha >\right )$$$$S=\frac{1}{\sqrt{2}}\left ( \mid \alpha \beta > -\mid \beta \alpha >\right )$$The singlet state $S$ will mix with the triplet $T_{0}$ state. That mixing is driven by the hyperfine interactions. Only the singlet state will form a stable molecular bond.

Note that the system is now described in terms of the pair of spins, not the individual spins which always have $S=1/2$. I know this is a far cry from your initial question, but I offer it as a real world example of mixing of spin states in composite systems.

 

[Isaac0427]

That depends on the situation under consideration. For a single particle, as initially stated, then yes. The limitation is the identity of the single particle. S is not just a quantum number, it is an intrinsic physical property.

Ok, this is making much more sense. I think the last follow-up question is this:

I can see that if we somehow discovered a particle that was a mix of spin-0 and spin-1/2, this would force us to fundamentally reconsider our understanding of quantum mechanics as a whole. If we discovered a particle that was a mix of spin-0 and spin-1, or spin-1/2 and spin-3/2, how much of quantum and/or particle theory would we need to "fix"?

 

[Hyperfine]

Ok, this is making much more sense. I think the last follow-up question is this:

I can see that if we somehow discovered a particle that was a mix of spin-0 and spin-1/2, this would force us to fundamentally reconsider our understanding of quantum mechanics as a whole. If we discovered a particle that was a mix of spin-0 and spin-1, or spin-1/2 and spin-3/2, how much of quantum and/or particle theory would we need to "fix"?

I would say a great deal of it would need to be "fixed". Including the Standard Model of particles.
I doubt you need worry about that happening however.

 

[Isaac0427]

I doubt you need worry about that happening however.

Thank you! I was not worried; I just curious as to how fundamental this empirical fact is to our theoretical understanding of particles.

 

[Hyperfine]

My pleasure.

 

[PeterDonis]

I can see that if we somehow discovered a particle that was a mix of spin-0 and spin-1/2, this would force us to fundamentally reconsider our understanding of quantum mechanics as a whole. If we discovered a particle that was a mix of spin-0 and spin-1, or spin-1/2 and spin-3/2, how much of quantum and/or particle theory would we need to "fix"?

I would say, about the same. As @strangerep pointed out in an earlier post, the issue with different spins is that each value of spin (0, 1/2, 1, 3/2, etc.) has its own distinct Hilbert space, and each of those Hilbert spaces is separate from the position/momentum Hilbert space. So if we want to describe, say, a single spin-1/2 particle, the total Hilbert space is a tensor product of the position/momentum Hilbert space with the spin-1/2 (qubit) Hilbert space. If we want to describe, say, a system with one spin-0 particle and one spin-1/2 particle, the total Hilbert space would be, heuristically,

<spin-0 particle position/momentum> tensor <spin-0 spin> tensor <spin-1/2 particle position/momentum> tensor <spin-1/2 spin>

And the two "position/momentum-spin" pairs would be connected to each other, so that, for example, the "total angular momentum of the spin-0 particle" operator would operate only on the first two factors in the tensor product above.

And even if we want to describe, say, a system with one spin-0 particle and one spin-1 particle (so both are bosons and have the same statistics), the Hilbert space would still look like the tensor product I described above (just substitute "spin-1" for "spin-1/2"), and the "total angular momentum of the spin-0 particle" operator would still work as I described above.

If, however, it turned out that a particle could be in a superposition of different spins (not just superposition of boson and fermion but superposition of any two different spins, such as spin-0 and spin-1), then the whole structure of the Hilbert spaces we use would have to change. It wouldn't matter whether we were "mixing" spin-0 and spin-1/2 or spin-0 and spin-1. The change would be just as drastic either way.

 

[strangerep]

If we discovered a particle that was a mix of spin-0 and spin-1, or spin-1/2 and spin-3/2, how much of quantum and/or particle theory would we need to "fix"?

If such a newly discovered particle were truly elementary it would be in conflict with special relativity. Physicists around the world would fall off their collective perches.

Have you studied anything yet about Wigner's theory of unitary irreducible representations of the Poincare group? The key point is that all elementary particle types (or perhaps I should say "field types") must correspond to one of these representations. The latter are characterized by 2 Casimirs: the mass$^2$ and spin$^2$ (actually the square of the Pauli-Lubanski vector).

 

[vanhees71]

That depends on the situation under consideration. For a single particle, as initially stated, then yes. The limitation is the identity of the single particle. S is not just a quantum number, it is an intrinsic physical property.

For an ensemble of particles, then not necessarily. Allow me to provide an example where a superposition of electron spins is most certainly allowed. Consider an ensemble of hydrogen atoms with random electron spin states as they undergo a chemical reaction to form molecular hydrogen.
$$H\cdot + H\cdot \rightarrow H_{2}$$The electron spin system of the pair of hydrogen atoms can be described in terms of a superposition resulting in four possible states where $\mid \alpha >$ or $\mid \beta >$ represents the spin state of one of the electrons:
$$T_{+}=\mid \alpha \alpha > $$$$T_{-}=\mid \beta \beta >$$$$T_{0}=\frac{1}{\sqrt{2}}\left ( \mid \alpha \beta > +\mid \beta \alpha >\right )$$$$S=\frac{1}{\sqrt{2}}\left ( \mid \alpha \beta > -\mid \beta \alpha >\right )$$The singlet state $S$ will mix with the triplet $T_{0}$ state. That mixing is driven by the hyperfine interactions. Only the singlet state will form a stable molecular bond.

Note that the system is now described in terms of the pair of spins, not the individual spins which always have $S=1/2$. I know this is a far cry from your initial question, but I offer it as a real world example of mixing of spin states in composite systems.

Note, however, that also here the superselection rule is fulfilled, i.e., you have only states with integer angular momentum here, i.e., your two-electron system can be either in a spin state with $S=0$ or $S=1$.