# When a wavefunction collapses on its own what basis does it project onto? (In Quantum Thermo, why do we privilege energy eigenstates?)

Obviously when an experimenter measures it, it projects onto the basis that the experimenter was trying to measure. But this process occurs on it's own, all the time.

Obviously the answer has to do with the exact situation in which it occurs, but what I'm really struggling with is conceptualizing Quantum Thermodynamics because of it. Why do we get our entropy and everything else by counting energy eigenstates? Couldn't the particles in question be in arbitrary combinations of energy eigenstates? If they're entangling and collapsing constantly, why onto the energy eigenbasis?

• there is no such thing as 'wavefunction collapsing on its own' – lurscher Oct 12 '17 at 15:43
• I tried to avoid that terminology but I'm sure you know what I mean. Substitute some jargon about entanglement with environmental degrees of freedom. – DPatt Oct 12 '17 at 15:58
• link.springer.com/article/10.1007/BF02827454 – user167013 Oct 13 '17 at 4:12

We actually force the use of which eigenstates by the macroscopic quantities we want to measure. The starting point is the unknown density operator $D$. We then define its entropy

$$\newcommand{\trace}[1]{\,\mathrm{Tr}\left(#1\right)}S = -k \trace{D\log D}.$$

We wish to minimise it under the constraint that the energy is known,

$$\trace{DH} = U,$$

where $H$ is the hamiltonian. This last equation is the way to encode a measure in the density operator formalism.

Let's continue to the end by writing the density operator that solve the minimisation problem I have just described. This is the Boltzmann-Gibbs

$$D=\frac{1}{Z}\exp\left(-\frac{H}{kT}\right).$$

The key quantity from which pretty much any thermodynamical quantity can be computed is the partition function function $Z$, whose expression comes from the requirement that $D$ shall be normalised, i.e. $\trace{D}=1$,

$$Z=\trace{\exp\left(-\frac{H}{kT}\right)}.$$

Now let's explicit the trace to make it clear,

$$\newcommand{\ket}[1]{|{#1}\rangle}\newcommand{\bra}[1]{\langle{#1}|}Z = \sum_\alpha \bra{\alpha}\exp\left(-\frac{H}{kT}\right)\ket{\alpha},$$

for any orthonormal basis $\{\ket{\alpha}\}$. This is usually tractable only if we choose a basis of eigenstate of $H$,

$$H\ket{\alpha}=E_\alpha\ket{\alpha},$$

because then

$$Z=\sum_\alpha\exp\left(-\frac{E_\alpha}{kT}\right).$$

But note that if we impose another macroscopic constraint than the total energy, then we need more than just the eigenstates of $H$. For example, if we impose a macroscopic magnetic moment, then we will typically need the eigenstates for the spins of our system. The story is the same as above: we will have to take the trace of a Boltzmann-Gibbs distribution featuring an exponential of some spin projector instead of the Hamiltonian, and that will be tractable only if we work with the spin eigenstates.

• So we 'set' the energy at the beginning, forcing them into eigenstates? Wouldn't they start evolving right afterward into combinations? Relatedly, doesn't conservation of energy only work for the expected value? Maybe that's enough for thermo applications. – DPatt Oct 12 '17 at 14:56
• Or we continuously 'measure' the energy to be U, forcing them into eigenstates at all times? I see. [I should add I'm unfamiliar with density operators]. – DPatt Oct 12 '17 at 15:03
• Not really: I have expanded my answer. – user154997 Oct 12 '17 at 15:14

It doesn't actually matter. For example, the microcanonical ensemble would have density matrix $$\rho = \sum_n | E_n \rangle \langle E_n|$$ but this is just the identity, so it is also equal to $$\rho = \sum_n | v_n \rangle \langle v_n |$$ for any basis $|v_n \rangle$. You might ask, is it really physically equal? Well, the full quantum states might be different, but if the density measurements are the same, then the two are completely indistinguishable as long as you're only measuring the system and not its surroundings too. And this is exactly the point of stat mech / ensembles in general -- even in classical stat mech we never care or talk about how the system gets into equilibrium, just the results of measurements on it.

As another example, for spin systems sometimes people say there's a 50% chance of spin up and a 50% chance of spin down, but this is completely equivalent to a 50% chance of spin left and a 50% chance of spin right. For example, the chance of measuring spin down is 50% in the first example, and (2)(25%) = 50% in the second.

• Thanks. I'm not familiar with the density operator formalism yet, however. Is that required for an explanation to this? All I know is that, in thermo, one seeks an ensemble to calculate expected values. We obtained this ensemble by (implicitly) choosing the uniform measure over energy eigenstates. I wasn't sure why. Maybe this is the standard problem of measure, and there's not much Quantum about it? – DPatt Oct 12 '17 at 14:59
• @DPatt Yes, I think the density operator makes it much more clear, because it contains all observable information about the system. But you can also manually check that if you change the measure to be uniform over some other basis, every observable quantity will come out the same. – knzhou Oct 12 '17 at 15:09
• This is a uniquely quantum thing: the ensembles of "uniform position" and "uniform momentum" are totally different classically, but thanks to the magic of quantum superposition they are identical quantumly. – knzhou Oct 12 '17 at 15:10
• That's incredible. Does this mean that quantum mechanics "rescues" us from all the deliberations over measure that we have classically (i.e. why does Boltzmann use dx*dp, etc)? Is this the case even when the energy eigenstates form a continuum? – DPatt Oct 12 '17 at 15:14
• @DPatt Sorry, I've never thought/read about that stuff so I have no idea! I feel like there's nothing wrong with $dxdp$ though, because the only coordinate transformations allowed in Hamiltonian mechanics are canonical ones which preserve the phase space volume $dxdp$. So it really is unique. (You may be able to link the two pictures using the description of a quantum state as a distribution on classical phase space. But I'm not qualified to say any more than that.) – knzhou Oct 12 '17 at 15:35