In a typical quantum mechanics course/textbook for physisists, and even in Hall's Quantum Theory for Mathematicians, states of a quantum mechanical system are said to be vectors in some Hilbert space $\cal H$. I have found some other seemingly more precise sources, for example Takhtajan's Quantum Mechanics for Mathematicians, that say that states are in fact trace class bounded operators on $\cal H$. Particular instances of the latter are 1-dimensional projections onto vectors in $\cal H$, so this second view encompasses the first.

I am unsure whether or not there is something wrong with the first model and this is the reason for introducing the second one, or is the second one simply a mathematically more challenging extension of the first one and this is the reason why textbooks deal only with the simpler one? Is this something similar as in classical mechanics, where you might first study only Newton's equations and only later introduce Hamilton's approach?

  • $\begingroup$ The first model came first. Don't underestimate the power of inertia! $\endgroup$ – user5174 Sep 19 '16 at 14:47
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    $\begingroup$ @Hurkyl It's not just inertia. Schrödinger's equation is most cleanly formulated in terms of vectors, and while you can rephrase it as the quantum Liouville equation (which you do do when you need to), you take a significant hit in conceptual clarity, ease of conceptual manipulation, and numerical efficiency when you do it. You keep the simpler picture where it works because it makes for better science more easily. $\endgroup$ – Emilio Pisanty Sep 19 '16 at 17:41

The second view is strictly broader than the first, and it is used to describe mixed states as well as the pure states described by the first view.

There are several different situations that require the use of mixed states, but the simplest to understand is that of a (classically) probabilistic source for the states. In this paradigm, you have a preparation procedure which normally results in the pure state $|\psi⟩\in\mathcal H$, but, for whatever reason, occasionally bugs out and produces something else. Thus, with some (say) 5% probability, the preparation procedure will actually give you some other $|\psi'⟩\in \mathcal H$, and for whatever reason you cannot identify these cases before they are used or post-select them out afterwards.

The way to describe these situations is using the formalism of density matrices. Any and all quantum mechanical predictions for experimentally measurable quantities will essentially involve a matrix element of the form $$ A=⟨\psi|\hat A|\psi⟩. $$ The density matrix formalism works by a clever re-phrasing of this expectation value into the form $$ A=\mathrm{Tr}\left(\hat A|\psi⟩⟨\psi|\right), $$ which can easily be seen to give the same quantity (either by direct computation in a basis, or by choosing a basis with one member along $|\psi⟩$).

In our situation, however, we're contemplating a preparation procedure that might produce a bunch of different states, $|\psi_j⟩$, with probabilities $p_j$. Each of these will produce an expectation value $A_j=\mathrm{Tr}\left(\hat A|\psi_j⟩⟨\psi_j|\right)$ for our quantity, and we then need to average those results, weighed with the corresponding probabilities $p_j$. Because of the linearity of the trace, and the clever choice of our representation, we can encapsulate all of the preparation procedure into a single part of the expression: $$ A=\sum_jA_j=\mathrm{Tr}\left(\hat A\sum_jp_j|\psi_j⟩⟨\psi_j|\right). $$ This means that the true descriptor of the system is the density matrix $\rho=\sum_jp_j|\psi_j⟩⟨\psi_j|$, and it cleanly gives us the expectation value of any operator $\hat A$ via $A=\mathrm{Tr}(\hat A\rho)$.

Now, the existence of a completely classical probabilistic source of quantum states is not entirely without challenge, but either way you're stuck with density matrices: the alternative is to have a state that is entangled with its environment, and when you decide to ignore the environment (because it takes no further part in the measurement) what you're left with is also a mixed quantum state for the system.

Past a certain point, then, mixed states are a necessity, and whenever those are around you need to use the trace-class operator (the density matrix) as your descriptor of the state of the system.

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  • $\begingroup$ For any quantum state (pure or mixed) there is always a representation (by the GNS construction) where the state is (the projection in) a vector of the Hilbert space (it is a reducible representation for mixed states, but still...). Due to this fact, the description of mixed states as non-vector states and pure states as vector states is a bit misleading, and in some sense simplistic since it does not take into account which representation of the algebra of observables we are dealing with. $\endgroup$ – yuggib Sep 19 '16 at 16:18
  • $\begingroup$ @yuggib Sure, but the bigger vector space is not particularly unique and most damningly the construction comes with no guarantee that the resulting pure state is physically meaningful, (I do think in the end traced-out entanglement is essentially the cause of all mixed states, but you don't necessarily get the correct description of said entanglement), so in that sense that representation is also misleading. (Also: what if you get a genuinely, fundamentally mixed state? Then you're barking at completely the wrong tree.) $\endgroup$ – Emilio Pisanty Sep 19 '16 at 17:01
  • $\begingroup$ It's a pretty pointless debate ultimately, though - there's reasons to go either way, and this posts documents the reasons why you'd go with $\rho$, with as much of a nod to the existence of debate as I felt still fit within the flow. $\endgroup$ – Emilio Pisanty Sep 19 '16 at 17:01
  • $\begingroup$ I am just saying that "purity" of a quantum state is not related to the fact it is a vector state or not in some representation. It is a more fundamental property: a pure state is one with "maximal information" on the system (it is extremal wrt a suitable partial ordering), while a mixed state has non-maximal information. This depends only on the algebra of observables considered, and not on eventual entanglement with an environment. Of course a pure state on a composite system may become a mixed state when restricted to one subsystem, but that is another story. $\endgroup$ – yuggib Sep 19 '16 at 17:21
  • $\begingroup$ With all due respect, while that is interesting and true, I feel it is overpowered w.r.t. the OP as I read it ("why even use operators instead of vectors?") - and I'm happy for you to take the credit if the OP is indeed more interested in those aspects of the difference. $\endgroup$ – Emilio Pisanty Sep 19 '16 at 17:37

States of quantum mechanical systems are in general functionals of the W* algebra of observables of the system, that are positivity preserving and of norm one.

All quantum states cannot be represented at the same time as vectors on a given Hilbert space. All the so-called normal states are representable either with the orthogonal projection on a single vector, or with a positive trace class operator with trace one. There exist however also non-normal states (at least for algebras of observables irreducibly represented on infinite dimensional Hilbert spaces). These non-normal states are, however, often considered "unphysical" (I think mostly because nobody ever observed a phenomenon that was explainable only with the system being in a non-normal state).

Let us restrict to normal states only. There is an important distinction between pure and non-pure states that is in some sense related to the distinction between vector-states and trace-class states. The pure states are the ones that carry maximal information on the system, while the mixed ones have an incomplete information, and may be thought as a "statistical mixture" of pure states. The pure states are the ones that yield an irreducible representation of the algebra of observables (via the GNS construction). In a given irreducible representation, all the pure states are given by rank one projections on vectors of the Hilbert space, while the mixed states are given by trace class operators that are not rank one.

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