Mixed states II: decoherence; important measures of purity and entropy

Decoherence

At the end of this section, you should be able to:

• appreciate why the density matrix is really a great way of expressing states, even for pure states (they uniquely determine the dynamics of the system, without any "overall phase", etc.)
• develop an intuition for measurement, even "inadvertent" measurement
• understand on a somewhat high level how classical physics arises as a limit of quantum physics
• hang out with Wigner's friend
• admit that complex phases matter in quantum mechanics and link them to interference

Let's talk about measurement.

Suppose we have a system that we wish to measure it under an operator whose eigenvectors are $|0\rangle_A$ and $|1\rangle_B$. The idea is that we have some measurement apparatus, and their original combined state evolves from something like:

$$|\psi\rangle_{AB}=(\lambda|0\rangle_A+\mu|1\rangle_B)\otimes|0\rangle_B$$
To the entangled state:

$$|\psi\rangle_{AB} = \lambda|0\rangle_A\otimes|0\rangle_B+\mu|1\rangle_A\otimes|1\rangle_B$$
Then observing the apparatus is sufficient to observe the system. The idea is that ultimately, the observer himself (or his "knowledge") are the apparatus, and the he entangles with the system to measure it.

Well, we know that often, we end up seeing things we didn't really want to. After all, physics does not care about your wants and preferences. In fact, in pretty much any situation, information about the system will leak out into the surroundings in some specific way. For example, Schrodinger's cat leaks information about the life of the cat by making the environment smelly, i.e. the state evolves from:

$$|\psi\rangle_{AB}=(\lambda|\mathrm{alive}\rangle+\mu|\mathrm{dead}\rangle)\otimes|\mathrm{clean}\rangle$$
To the entangled state:

$$|\psi\rangle_{AB}=\lambda|\mathrm{alive}\rangle\otimes|\mathrm{clean}\rangle+\mu|\mathrm{dead}\rangle\otimes|\mathrm{smelly}\rangle$$
What this means is that the density matrix of the cat evolves as:

$$\left[ {\begin{array}{*{20}{c}}{{{\left| \lambda  \right|}^2}}&{\lambda \bar \mu }\\{\mu \bar \lambda }&{{{\left| \mu  \right|}^2}}\end{array}} \right] \mapsto \left[ {\begin{array}{*{20}{c}}{{{\left| \lambda  \right|}^2}}&0\\0&{{{\left| \mu  \right|}^2}}\end{array}} \right]$$
(Check that I got the right transpose.) OK, what happened here?

Recall that the probabilities of collapsing to $|0\rangle$ and $|1\rangle$ are determined purely by the elements on the diagonal -- the off-diagonal elements, or the coherences, are only relevant for collapsing on to some combination of $|0\rangle$ and $|1\rangle$. What's going on here is that when the environment entangles with the system, it has "kinda" already observed it -- like your Wigner's friend. It "knows" that the system isn't in $|0\rangle+|1\rangle$, and even though you haven't observed the environment yet (you haven't smelled it), you know how the combined state has evolved, and the probability has become a classical probability, because the quantum stuff has already been observed -- by the environment.

The idea behind decoherence is the same idea that ensures that the Wigner's friend scenario is consistent.

"Eventually", "all" the information about the system will leak into the environment -- i.e. in principle, we should be able to determine anything about the system from measuring the environment, and our uncertainty about the system arises entirely from our completely classical uncertainty about the environment -- so the density matrix becomes a classical one, i.e. a diagonal one (the off-diagonal terms go to zero).

What basis is it diagonal in? In the basis corresponding to the states of the environment -- i.e. if the environment can be in states $|0\rangle_B$ and $|1\rangle_B$, then the states of the system that precisely induce these states of the environment form the preferred basis. These are often called the "environmentally selected basis".

This process is called decoherence. You may also hear the terms pointer states (for the preferred basis), einselection (environmentally induced selection of the preferred basis), or Quantum Darwinism (what the heck?) -- but they're really synonymous. We'll just use the fancy words when they're grammatically useful.

Well, the following may not be completely clear, but you should at least be able to appreciate that it is true: the off-diagonal terms approach zero, rather than hit it. Why? Although the system leaks information into the surroundings, we aren't really certain about what we're inferring about the system from the environment -- a live cat may be smelly too, etc. So the pointer states are not exactly orthogonal, either.

The precise behavior of decoherence depends on the Hamiltonian of the system -- e.g. predicting the generation of the smelliness of the air from the state of the cat based on what's going on microscopically is something that could be done in principle by solving a really complicated Schrodinger equation. You can, given a Hamiltonian, at least make order-of-magnitude estimates of at how much time and at how macroscopic a scale (i.e. with how many degrees of freedom) does the system begin to behave in a way that can be described as classical.

Decoherence does not remove the need for wavefunction collapse -- one still needs the observer to note an observation, collapsing the system.

TBC: purity, entropy, correlation functions

Time evolution, Schrodinger and Heisenberg pictures, Noether's theorem

So far, we have discussed quantum mechanics without any reference to changes across time. You might think we could just upgrade $\psi(x)$ to $\psi(x,t)$ and e.g. an observable $Q_t$ measuring a value $q$ at time $t$ would have eigenvectors whose cross-section at $t$ are of the form $\delta(x-a)$. But this would mean the entire $\psi(x,t)$ is the state of the object, rather than there being a state at each value of $t$, and time would be an observable. This is clearly not what we want (right now -- although to be consistent with special relativity we will need to treat space and time on an equal footing later in this series).

Instead, a more appropriate approach is to say that the state is a function of time $|\psi(t)\rangle$ and the evolution of the state is given by some operation $|\psi(t)\rangle=U[|\psi(0)\rangle]$.

How do we know that $U$ is a linear operator? What does it mean for $U$ to be a linear operator anyway? The only sense in which such a linearity can be tested is by looking at a state in a superposition. So suppose $|\psi(0)\rangle=|\psi_1(0)\rangle+|\psi_2(0)\rangle$. Now $|\psi(t)\rangle=U[|\psi_1(0)\rangle+|\psi_2(0)\rangle]$ -- this is from the perspective of some observer Alice.

But if another observer Bob had previously observed and collapsed the system to $|\psi_1(0)\rangle$ at time 0, then according to him, the state should evolve to $U[|\psi_1(0)\rangle]$, and if he had observed the system in $|\psi_2(0)\rangle$, his knowledge of the system would evolve to $U[|\psi_2(0)\rangle]$.

So according to Alice, who doesn't know what Bob has observed (she has not observed him), her knowledge of the system can also be written as $U[|\psi_1(0)\rangle]+U[|\psi_2(0)\rangle]$. Thus

$$U[|\psi_1(0)\rangle+|\psi_2(0)\rangle]=U[|\psi_1(0)\rangle]+U[|\psi_2(0)\rangle]$$
I.e. $U$ is linear, so we can write it as a linear operator as in $U|\psi(0)\rangle$. (The above scenario is called Wigner's friend)

$U$ is also clearly a unitary operator, as it must preserve all lengths.

We can consider infinitesimal time evolutions $U_t(dt)$ representing evolution of the state from $t$ to $t+dt$. Then:

$$|\psi(t)\rangle=U_0(dt)\dots U_{t-dt}(dt)|\psi(0)\rangle$$
This product integral can be written alternatively as:

$$|\psi(t)\rangle=\mathcal{T}\left\{e^{\int \ln U_t(dt)}|\psi(0)\rangle\right\}$$

$\mathcal{T}$ is the time-ordering operator which orders a product like $H(t_1)H(t_2)$ in order of ascending $t$ in an expansion. Can you see why this is necessary (hint: $e^{AB}\ne e^{A+B}$ for noncommuting $A,B$).

Oh, and it's not actually an operator -- not even in the math sense, it's a "formal operation", one that takes a form or sentence (rather than its value) -- in this case the $\exp$ Taylor expansion -- and changes it some way.

$\ln U_t(dt)$ is an infinitesimal, and it's easy to see that it is equal to $U_{t}'(0)dt$ -- a member of the Lie algebra. We know, of course, that the Lie Algebra of the unitary group is comprised of anti-Hermitian operators (this can be checked without Lie Algebra, of course), and so $iU_{t}'(0)$ is Hermitian. From Lie Algebra, we can tell that this represents a generator of time translations -- and from a little experience of classical mechanics, we want this to represent energy. So for dimensional consistency with energy, we write:

$$H(t)=i\hbar U_{t}'(0)$$
(Why $\hbar$ and not $h$? Because $U_{t}'(0)$ is basically already in "radians per second".) This is called the Hamiltonian operator. It determines $U_t$, and thus describes the time evolution of a state. How exactly? Since:

$$|\psi(t+dt)\rangle=U_t(dt)|\psi(t)\rangle$$
We can write re-arranging:

$$\frac{\partial |\psi(t)\rangle}{\partial t}=-\frac{i}{\hbar}H(t)|\psi(t)\rangle$$
This is the most general form of the Schrodinger equation. Note that the earlier exponential equation is the "general solution" to this equation -- obviously not very useful, rewritten as what is known as the Dyson series:

$$|\psi(t)\rangle=\mathcal{T}\left\{e^{-i/\hbar\int H(t) dt}|\psi(0)\rangle\right\}$$
It's easy to show from this that the evolution of a density matrix $\rho(t)$ is similarly:

$$\frac{\partial\rho(t)}{\partial t}=-\frac{i}\hbar [H, \rho]$$
Which is the von Neumann equation, whose solution is given by:

$$\rho(t)=\mathcal{T}\left\{e^{-i/\hbar\int H(t) dt}\rho(0)e^{i/\hbar \int H(t) dt}\right\}$$
These should all appear as obvious special cases of Lie theoretic results.

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$H(t)$ is not the same as $i\hbar\partial/\partial t$. $H(t)$ is a Hermitian operator, i.e. an observable, while $\partial/\partial t$ does not act on the Hilbert space at all. One could also see what could wrong by equating the two in the "solution to the Schrodinger equation" above. The Schrodinger equation does not say that $H$ and the time-derivative are equal in general -- rather, it says that they are the same on a valid state vector $|\psi(t)\rangle$ -- you cannot just "factor this out".

So the Hamiltonian is fundamentally what determines the dynamics of a quantum system. Give me a Hamiltonian, and you've given me a theory. The Schrodinger equation (or equivalently the von Neumann equation) above is just an axiom of quantum mechanics/of any quantum mechanical theory.

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Can we talk about the velocity and acceleration observables for a moment? Actually, we can't, because they fundamentally have to do with time evolution, and we can't have observables that depend on the time-evolution of the state -- observables must act on the Hilbert space. But we can define observables that predict how the state will evolve (like the Hamiltonian with the Schrodinger equation).

Doing this systematically is where the Heisenberg picture comes in.

What does this mean? Everything we've discussed so far is the Schrodinger picture, where the state evolves on a fixed background basis created by the observables' eigenvectors -- so observables represent active transformations. Instead, we can have a completely different picture of reality, the Heisenberg picture, where we view time-evolution as simply viewing the state in a different basis -- then the observables represent passive transformations.

OK, so how do we do this? Remember how every question in quantum mechanics can fundamentally be asked in terms of expectation values (specifically those of Hermitian projections). The expected value of an observable at time $t$ of course evolves as:

$$\langle A\rangle(t) = \langle\psi|U^*(t)AU(t)|\psi\rangle$$
In the Schrodinger picture, we attach the $U(t)$ to $|\psi(0)\rangle$ to make $\langle A\rangle(t)=\langle\psi(t)|A|\psi(t)\rangle$. In the Heisenberg picture instead, we attach the $U(t)$ to the $A(0)$, writing $\langle A\rangle(t)=\langle\psi|A(t)|\psi\rangle$.
From differentiating conjugation in $A(t)=U^*(t)A(0)U(t)$, we get:

$$\frac{dA}{dt}=\frac{i}\hbar [H, A]$$
This is the Heisenberg equation. Immediately, it yields:

$$\begin{array}{l}\frac{{dX}}{{dt}} = \frac{i}{\hbar }\left[ {H,X} \right]\\\frac{{{d^2}X}}{{d{t^2}}} =  - \frac{1}{{{\hbar ^2}}}\left[ {H,\left[ {H,X} \right]} \right]\end{array}$$
Thinking of the evolution of $X$ as a translation of the co-ordinate system, etc., what this does is give us two conditions on what the Hamiltonian should look like for a "Euclidean" system:

$$\begin{array}{l}\left[ {H,X} \right] =  - \frac{{i\hbar }}{m}P\\\left[ {H,\left[ {H,X} \right]} \right] = \frac{{{\hbar ^2}}}{m}U'(x)\end{array}$$
This gives us yet another strong reason (besides the fact that the Hamiltonian generates time-translations, that the "eigenvectors" of $\partial/\partial t$ are the energy states by the de Broglie theorem (but not really), etc.) to suspect that the Hamiltonian represents the energy of the system. Indeed if we use:

$$H=\frac1{2m}P^2+U(x)$$
We can confirm those conditions above. Well, this is certainly not the only Hamiltonian compatible with classical mechanics, so at this point, I'll just say that this is confirmed by experiment, and is an axiom of the quantum theory of Euclidean systems.

Exercise: By taking expectation values in the Heisenberg equation, show that $m\frac{d}{dt}\langle x\rangle =\langle p\rangle$ and $\frac{d}{dt}\langle p\rangle = -\langle U'(x)\rangle$ under the Euclidean Hamiltonian. This is called the Ehrenfest theorem.

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I'll discuss one final application of the Heisenberg formalism: it makes Noether's theorem completely trivial.

Indeed, $dA/dt=0$ iff $[H,A]=0$ iff $\forall\tau, H=e^{-i/\hbar A\tau}He^{i/\hbar A \tau}$. Then $A$ is a conserved quantity and conjugation with it as an infinitesimal generator represents a symmetry of the Hamiltonian.