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

 Quote by Tosh5457 So why is that formalism only used in quantum statistical mechanics, and not on the general theory of quantum mechanics?
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.

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 Quote by DrDu What I wondered, probably because I don't understand qft too well: Wouldn't it be possible to consider the charge density rho as some field, something related to its time derivative as momentum and then doing second quantization with it?
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.

 Quote by A. Neumaier as every equilibrium state is mixed.
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.

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

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

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

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

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 Quote by 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.
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

 This is by far not the only casae you can write this density matrix as a mixture,
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.

 and none of the many mixture representations has any physical relevance as they are observationally undistinguishable.
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.

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 Quote by Jano L. 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.
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.
 Quote by Jano L. In general, different mixtures are in principle distinguishable, if we know something about the preparation of the ensemble.
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.
 Quote by Jano L. 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.
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.

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 Quote by A. Neumaier For two particles, the wave function had already two position arguments, while the charge density has only one. So they cannot be the same.
You are obviously right, here. Nevertheless my question remains: can we quantize charge density, viewed as a field, directly?
 Blog Entries: 9 Recognitions: Homework Help Science Advisor Which charge ? Electric charge ? We surely can.

 Quote by dextercioby Which charge ? Electric charge ? We surely can.
using dirac quantization condition of monopoles or reverse argument?

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