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The simplest non-trivial system is a two-level system. Classically, it is a system which can be in one state labelled $H$ or another state labelled $T$.

There is no necessary reference to any particular physical system when thinking abstractly about this system and its observables, yet once we have, we can use it to describe any two-level physical system (I think), e.g. a coin that can be heads or tails. Is this the case in quantum mechanics?

Let's say we use the orthonormal $\{|H\rangle, |T\rangle\}$ basis where $\sigma_3 |H\rangle = |H\rangle$ and $\sigma_3 |T\rangle = -|T\rangle$. For a two-state system, the Hermitian operators that can represent observables are real linear combinations of the Pauli operators $\sigma_1$, $\sigma_2$, $\sigma_3$ and the identity operator $I_2$. However, unlike the classical case, it seems that when applying this abstraction to a physical system, there are not always candidate observables for all of them apart from $a \sigma_3+ b I_2$, i.e. the observables that are diagonal for the choice of basis.

I ask this question because I'm interested in whether it is possible to abstractly describe the difference between classical and quantum systems without reference to particular examples such as spin or polarisation.

3 Answers3

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Starting from the Hilbert space is the wrong way to conceptualize a quantum system - the Hilbert space isn't what makes a quantum system, its observables are. It is perfectly common to have cases where not every self-adjoint operator is an observable, usually in the case of superselection effects, see this answer of mine.

For more on the idea that the observables are the fundamental part of a quantum system, formally expressed by the view that quantum mechanics is about representations of $C^\ast$-algebras of observables rather than fixed Hilbert spaces, see this answer of mine.

ACuriousMind
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  • To expand on the second answer you link to, will the algebra of observables also contain additional Hermitian operators that are still not observables? For example, if $\sigma_z$ and $\sigma_x$ are observables, does that mean $\sigma_y$ has to be an observable too? – tomdodd4598 Aug 13 '22 at 13:24
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There is no necessary reference to any particular physical system when thinking abstractly about this system and its observables, yet once we have, we can use it to describe any two-level physical system (I think), e.g. a coin that can be heads or tails. Is this the case in quantum mechanics?

Yes, you can think abstractly about a system with no necessary reference to any particular physical system. But it is not necessarily a helpful way to proceed. The two state system is a model system that might or might not provide a useful approximate model for a real physical system.

As illustrated below, you can proceed in complete generality with the two state system and its observables, if you would like.

Let's say we use the orthonormal $\{|H\rangle, |T\rangle\}$ basis where $\sigma_3 |H\rangle = |H\rangle$ and $\sigma_3 |T\rangle = -|T\rangle$. For a two-state system, the Hermitian operators that can represent observables are real linear combinations of the Pauli operators $\sigma_1$, $\sigma_2$, $\sigma_3$ and the identity operator $I_2$. However, unlike the classical case, it seems that when applying this abstraction to a physical system, there are not always candidate observables for all of them apart from $a \sigma_3+ b I_2$, i.e. the observables that are diagonal for the choice of basis.

In general an observable is a Hermitian operator. (The converse is not true, as we discuss further in the update section below.) The eigenvalues of the observable represent the possible outcomes of a measurement of the observable.

For your two state systems, the most general Hermitian operator can be represented in matrix form as: $$ M = \begin{bmatrix}a & b+ic\\b-ic & d\end{bmatrix}\;, $$ where $a$, $b$, $c$, and $d$ are real.

The eigenvalues of $M$ are: $$ \lambda_M = \frac{a+d}{2}\pm\frac{\sqrt{(a-d)^2+4|b+ic|^2}}{2} $$

The most general state has the form: $$ |\psi> = \alpha|H> + \beta|T>\;, $$ where $\alpha$ and $\beta$ are complex numbers. And we also impose the normalization condition $$|\alpha|^2 + |\beta|^2 = 1 $$

The expected value of the above-mentioned general Hermitian operator in the above-mentioned general state is: $$ <\psi|M|\psi> = a|\alpha|^2 + 2\Re(\alpha^*\beta(b+ic))+ d|\beta|^2\;, $$ subject to the normalization condition.

This is all completely general, and in my opinion, all pretty opaque.

But. it is still useful to have these formulas since you can apply them to any specific cases as well. E.g., if the you are interested in a Hamiltonian of a coupled two level system like: $$ H = \begin{bmatrix}-\epsilon/2 & v\\ v & \epsilon/2\end{bmatrix}\;, $$ where the lower level is at energy $-\epsilon/2$, the upper level at energy $+\epsilon/2$ and the coupling is a real number $v$. And you can read off the observable energy levels immediately from the above formula as: $$ E_{\pm} = \pm \frac{\sqrt{\epsilon^2+4v^2}}{2}\;. $$ In practice you might, say, tune these parameters to model some energy splitting from an experiment.


Update:

I didn't notice there was a question in the title... which reads:

"For a generic two-state quantum system, are there interpretations for the observables corresponding to all Hermitian operators?" [Emphasis added.]

We note that "are there interpretations for..." is really asking. Are there interpretations for the following equation: $$ \lambda_M = \frac{a+d}{2}\pm\frac{\sqrt{(a-d)^2+4|b+ic|^2}}{2} $$

Or, rather, can we come up with some "other" meaning beside that of a pure function for the following function of four real variables: $$ \lambda_M(a, b, c, d) = \frac{a+d}{2}\pm\frac{\sqrt{(a-d)^2+4|b+ic|^2}}{2} $$

We provided one possible meaning for $$ \lambda_M(\epsilon/2, v, 0, -\epsilon/2) $$ as energy levels of a coupled two-state system.

There are clearly other interpretations. This is up to the imagination of the reader.

Now, to the word "observable" and its relation to a Hermitian operator. Any Hermitian operator whose eigenvectors span the space (in this case a two-dimensional space) is an observable. (See, for example, Messiah's Quantum Mechanics textbook, volume 1, chapter VII, second 9 titled "Eigenvalue problems and Observables.")

Finally, the issue of super-selection rules, where the Hilbert space is a direct sum of spaces, is not very relevant in our present analysis (the analysis of a two-state system). This is because we have already agreed that the one-state system is trivial; the only possible direct-sum to consider would be a direct sum of trivial systems.

hft
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  • My "this" refers to the precending sentence, i.e. the final sentence of the previous paragraph. Perhaps a rearranging of the paragraphs would make it clearer? – tomdodd4598 Aug 12 '22 at 20:46
  • Your "this" refers to "There is no necessary reference to any particular physical system when thinking abstractly about this system and its observables, yet once we have, we can use it to describe any two-level physical system (I think), e.g. a coin that can be heads or tails."? That's even less clear then... – hft Aug 12 '22 at 20:47
  • I mean, yeah, rearrange it if you think it will make the question clearer... – hft Aug 12 '22 at 20:48
  • I don't think your answer and ACuriousMind's are incompatible, but I suppose apart from a Hamiltonian perhaps, it's difficult to associate any old Hermitian operator with an observable. – tomdodd4598 Aug 13 '22 at 12:41
  • Yeah, they aren't incompatible. Mine is kind of a non-answer since I actually didn't realize the main question was in the title of the post. I was just reading the body of the post and working from what was there. My bad, I should pay more attention. Anyways, I might update my answer today if there's time. – hft Aug 13 '22 at 17:31
  • I updated the answer... – hft Aug 13 '22 at 18:12
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Yes. Quantum optics realizes single qubits in terms of polarization. Here the Pauli basis together with the identity has a clear interpretation as the 4 components of the Stokes vector, and any hermitian operator represented in this basis is a linear combination, with an associate interpretation. Full details can be found in my essay https://www.physicsforums.com/insights/a-classical-view-of-the-qubit/