Why not formulate QM in terms of |ψ| squared?

A. Neumaier

Because outside statistical mechanics, one can pretend that a mixed state is a mixture of pure states (though not uniquely, and the decomposition has therefore no physical meaning), and the pure state computations look slightly more familiar. Whereas in statistical mechanics you really _need- the mixed states, as every equilibrium state is mixed.

A. Neumaier

Actually, in QFT, charge density is a field, the zero component of the corresponding current.

But charge density is not the same as |psi|^2, except in the single particle case.

DrewD

Sorry, this is off topic, but where can I find more info on this. It came up in a paper I was reading, but I don't understand why. I don't think the standard QM books I have cover this. Any advice where to look would be appreciated. Sorry to disrupt the thread.

Jano L.

But doesn't quantum statistical physics pretend the same? Take, for example, the density matrix for harmonic oscillator [itex]\rho_{mn} = \frac{e^{- n\hbar \omega/ (k_B T)}}{Z} \delta_{mn}[/itex] (in the basis of its Hamiltonian). Is not this a good example of a mixture of pure states - eigenstates of the Hamiltonian?

DrDu

I was assuming a non-relativistic setting. There, I think, charge density and psi square are identical.

A. Neumaier

My statement holds both in the relativistic and the nonrelativistic case, where things look identical in the second-quantized formulation.

For a single particle, charge density and absolute psi square are identical.

For two particles, the wave function had already two position arguments, while the charge density has only one. So they cannot be the same.

A. Neumaier

This is by far not the only casae you can write this density matrix as a mixture, and none of the many mixture representations has any physical relevance as they are observationally undistinguishable.

A. Neumaier

Look an any introduction to statistical mechanics, which doens't treat only the classical case.
e.g., Part II of http://lanl.arxiv.org/abs/0810.1019

Jano L.

That is true. There are potentially many different ways to express that density matrix. And I agree that preferring one mixture above the others seems artificial and ungrounded.

However, the fact is that no other representation is used. Statistical physics relies on the properties of Hamiltonian, not some other operator, and really treats energy representation with distinction.

I am not so sure. In general, different mixtures are in principle distinguishable, if we know something about the preparation of the ensemble. Of course, simple counting of frequencies of results of measurement will not reveal that; but there is more to the ensemble than just results of measurements on it. This is more easily shown for spins than for energy, but why assume that canonical density matrix is different in this respect?

I think the relevance of H eigenstates is in that we need to assume ensemble of them in order to use Boltzmann's probability distribution to find the canonical density matrix in the first place. For other mixtures, how would you arrive at density matrix that is diagonal in H representation? I do not see how that is possible, but perhaps you know.

A. Neumaier

Statistical mechanics doesn't care at all about the basis in which things are represented, except for computational reasons. One sees it clearly once one leaves the equilibrium situation.
One can distinguish ensembles only if they are prepared artificially, by mixing observations made at different times. But the ensembles of statistical mechanics are of a different nature; they are instantaneous, a single time already defines the ensemble. Then nothing can distinguish a decomposition into subensembles.
All that is needed is that e^{beta H} commutes with H, which is a triviality, independent of the basis used to express it. No diagonalization is needed or assumed.

DrDu

You are obviously right, here. Nevertheless my question remains: can we quantize charge density, viewed as a field, directly?

dextercioby

Which charge ? Electric charge ? We surely can.

andrien

using dirac quantization condition of monopoles or reverse argument?

A. Neumaier

Together with charge density, you need st least the associated charge current operaots, which complement the density to a conservation law. Quantizing this is called current algebra. There have been varous attempts and there is significant literature, but apart from the 1+1D case (where current algebra essentially turns into Kac-Moody algebra) no real breakthrough.