Is it meaningful in quantum mechanics to speak of charge distribution? Some people say if you sovle the Schroedinger Equation for the hydrogen atom, the eigenfunctions represent a type of charge density...that is, the square of the amplitude is a charge density. Yes, I know some people say it's not charge density, it's just the probability of finding an electron, but let's just call it charge density. That's not the point.

The point is: in more complex systems, does quantum mechanics recognize the existence of a similar quantity to the square of the amplitude in the hydrogen atom...a generalized charge density? or at least, a probability of finding charge. If I look at a glass of water, can I say that according to quantum mechanics, for any given point (x,y,z) there is a certain expectation dQ within any given volume element dV that there is a unit charge within that volume?

Or, as certain correspondents have claimed in a related question which was shut down by the moderators, is it meaningless in quantum mechanics to talk about such a quantity dQ/dV?

See https://physics.stackexchange.com/questions/241247/semi-classical-calculation-gives-wrong-answer-for-emission

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    $\begingroup$ I don't care if it was shut down by Archimedes Plutonium. I know censorship when I see it. $\endgroup$ – Marty Green Mar 4 '16 at 1:04
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    $\begingroup$ Marty, it's clear to me that you don't understand the concept of censorship at all. Yes, I presume that you value my judgement precisely to the extent that I value yours which is to say, none. $\endgroup$ – Alfred Centauri Mar 4 '16 at 1:13
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    $\begingroup$ The "regular contributors" who shut down my other question claimed they couldn't understand what I meant by charge density in quantum mechanics. Then let them answer the question here: is there or isn't there such a thing as charge density? They're quick to shut me down, but not so quick to answer my questions. $\endgroup$ – Marty Green Mar 4 '16 at 1:40
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    $\begingroup$ No one in the comments to the other question claimed that there is no charge density in quantum mechanics. Using $-q\lvert \psi \rvert^2$ is possible for semi-classical computations (particle quantized, field not) for a single charged particle. perturbed by an outer electromagnetic field. What EmilioPisanty correctly said is that the situation you describe there is a thermal state which is by construction stationary and doesn't have a single wavefunction, and thus requires a (quantum) statistical mechanics consideration. $\endgroup$ – ACuriousMind Mar 4 '16 at 2:02
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    $\begingroup$ If you believe there is a charge density in QM, then why the hell did you shut my question down? $\endgroup$ – Marty Green Mar 4 '16 at 3:07

Yes, of course there is. Addressing the specific question you posed in a comment, of

whether underneath it all, in large multi-body systems, there is a real charge density... at least as real as the charge/probability density we get by squaring the wave function for a single electron

there is indeed such a density. Given an $n$-particle system in a pure state $|\Psi⟩$, with the $n$-dimensional position wavefunction $$⟨\mathbf r_1,\ldots,\mathbf r_n|\Psi⟩=\Psi(\mathbf r_1,\ldots,\mathbf r_n),$$ the $n$-dimensional probability density that the electrons will be in a volume $\mathrm d\mathbf r_1\cdots\mathrm d\mathbf r_n$ at $(\mathbf r_1,\ldots,\mathbf r_n)$ is $$\rho(\mathbf r_1,\ldots,\mathbf r_n)=|\Psi(\mathbf r_1,\ldots,\mathbf r_n)|^2=⟨\Psi|\mathbf r_1,\ldots,\mathbf r_n⟩⟨\mathbf r_1,\ldots,\mathbf r_n|\Psi⟩.$$ The probability that the first particle will be at $\mathbf r_1$ is $$ \rho_1(\mathbf r_1) =\int|\Psi(\mathbf r_1,\ldots,\mathbf r_n)|^2\mathrm d\mathbf r_2\cdots\mathbf r_n =⟨\Psi|\big[|\mathbf r_1⟩⟨\mathbf r_1|\otimes I_{(2,n)}\big]|\Psi⟩,$$ where $I_{(2,n)}$ is the identity operator for particles $2$ through $n$ , with analogous formulas for the rest of the particles. The corresponding charge expectation value is $q_1\rho_1(\mathbf r_1)$ for $q_1$ the charge of particle $1$.

Finally, to get the charge density you add all of these together. Thus, you define \begin{align} \rho(\mathbf r) & = \sum_{k=1}^n\int q_k|\Psi(\mathbf r_1,\ldots,\mathbf r_{k-1},\mathbf r,\mathbf r_{k+1},\ldots,\mathbf r_n)|^2\mathrm d\mathbf r_1\cdots \mathrm d\mathbf r_{k-1}\mathrm d\mathbf r_{k+1}\cdots \mathrm d\mathbf r_n \\ & = \sum_{k=1}^n q_k⟨\Psi|\big[I_{(1,k-1)}\otimes|\mathbf r⟩⟨\mathbf r|\otimes I_{(k+1,n)}\big]|\Psi⟩, \end{align} and that tells you the expected value of charge in a volume element $\mathrm d\mathbf r$ at $\mathbf r$.

Moreover, this quantity can be rephrased in the form \begin{align} \rho(\mathbf r) & = \mathrm{Tr}\left( \left[ \sum_{k=1}^n q_k I_{(1,k-1)}\otimes|\mathbf r⟩⟨\mathbf r|\otimes I_{(k+1,n)}\right] |\Psi⟩⟨\Psi| \right), \tag{1a} \end{align} which is then, by linearity, trivially generalized to the case of an arbitrary mixed state $\hat\rho:\mathcal H_n\to\mathcal H_n$ acting on the Hilbert space $\mathcal H_n$ of $n$ particles, \begin{align} \rho(\mathbf r) & = \mathrm{Tr}\left( \left[ \sum_{k=1}^n q_k I_{(1,k-1)}\otimes|\mathbf r⟩⟨\mathbf r|\otimes I_{(k+1,n)}\right] \hat \rho \right).\tag{1b} \end{align} (Note, though, that there is a typographical similarity between $\hat \rho$ and $\rho(\mathbf r)$, but otherwise no relation beyond the stated one.)

Thus far for definitions - but you can always define whatever you want, and it only becomes a relevant quantity if it is useful for something. (It is, on the other hand, as "real" as the squared modulus of a single-particle wavefunction.) In general, this charge density is not particularly useful in full-blown quantum mechanics, because if you have an extra particle of charge $q$ then its electrostatic interaction with the previous $n$ particles will be the operator $$\hat V=\sum_{k=1}^n\frac{qq_k}{\|\hat{\mathbf r}-\hat{\mathbf r}_k\|},$$ which entangles the test particle individually with each of the existing particles. By contrast, the charge density as defined above can only be used to calculate the mean-field effect of the existing particles on the new one, via the potential $$ U(\hat{\mathbf r})=\int \frac{\rho(\mathbf r')\mathrm d\mathbf r'}{\|\hat{\mathbf r}-\mathbf r'\|}, $$ which is an operator solely on the test particle and treats the existing ones as frozen. This can be a good approximation in some situations but it is far from the full interaction.

As a final note, your previous question was closed for other reasons, and pretty much all your claims about what happened on that thread are unfounded. What doesn't exist in that question as currently posed is a time-dependent charge distribution: the charge distribution can be defined and (in principle) calculated.

I suspect I will get nothing but unwarranted acid from you for even attempting this, but here are some comments on your filament.

For a "glowing tungsten filament", the true quantum state is probably (foundational issues aside on whether it is actually possible to create a mixed state without tracing out a physical part of the system) a pure quantum state, but this pure quantum state is highly entangled with all sorts of external systems, including in particular the quantum state of the electromagnetic field it is coupled to.

The approximation you seem to be proposing (to the extent that one can extract a clear and unambiguous proposal from the separate question that you insist on bringing up here) is to ignore this coupling and this entanglement, and to consider simply the quantum state of the filament itself, and its associated charge distribution as per $(1\text{a})$ and $(1\text{b})$ above. That, I may add, is a perfectly reasonable question (when formulated in that way, which you seem to refuse to do).

Now, while the universe so far is (or can be) in a pure state, by choosing to focus on one half of a highly entangled system, you get (you must get) a mixed state (more specifically, a quantum state $\hat\rho:\mathcal H_n\to\mathcal H_n$ of rank bigger than one, which cannot be represented as a projector on a pure state $|\Psi⟩$). As to what that mixed state will be, it's obviously up in the air, particularly if you are looking at a transient. However, quantum mechanics is pretty unambiguous that in this situation the relevant state will be a thermal state.

It should be obvious, but it bears stating: not all mixed states are thermal states, and there are indeed mixed states that do evolve in time. Thermal states, however, are quite narrowly defined: given a temperature $T=1/\beta k_B$ and a system hamiltonian $\hat{H}$, there is a unique thermal state given by $$\hat \rho_{\beta,\hat{H}}=\frac{ e^{-\beta \hat{H}} }{\mathrm{Tr}(e^{-\beta \hat{H}})},$$ which is the obvious generalization of the Boltzmann distribution to a quantum state. In particular, because this state is a function of the hamiltonian, you have $[\hat \rho_{\beta,\hat{H}},\hat{H}]=0$ and by the quantum Liouville equation the thermal state does not evolve in time. That much is essentially incontroversial. Moreover, since the thermal state does not evolve with time, all observable properties are constant over time (also incontroversial). Thus, putting a thermal state as the state of the system would give you zero radiation in this scheme which you term 'semiclassical', an apparent paradox.

Now, you can argue that a thermal state is not an appropriate choice of state for the reduced state of the filament, though an external commentator might wonder why you're expecting thermal radiation from a state that is not in thermal equilibrium. In any case, that argument does not take away the paradox that the thermal state 'ought' to glow but within that approximation it produces no radiation - so there are certainly problems with that approach.

To be sure, the result is certainly surprising, because one intuition does suggest that the 'true' state should be pure, complicated, and time varying - and that much is true. However, it is also the case that the 'true' state is highly entangled, and disregarding this entanglement breaks a lot of the state. (For one thing, it is trivially easy to construct complex time-varying entangled states which produce constant reduced states when either half is traced away.) Personally, I don't think it's reasonable to complain that the time dependence has disappeared after breaking another fundamental part of the state of the system. You may disagree, and I see no value in arguing that point.

Finally, as to whether a thermal state is a reasonable approximation to the reduced state of a glowing filament, you may disagree and that is up to you. I do not particularly care for that debate, which is why I originally answered only this question in its very constrained sense, and not the other one. In any case, do note that the above shows that (i) it is not guaranteed that the reduced quantum state of a glowing filament changes over time, as you state in the other question and in your answer here, and (ii) the thermal state, though not necessarily the only possibility, is nevertheless a perfectly acceptable reduced quantum state of a glowing filament, and the fact that it ought to glow but doesn't in this approximation does show that the approximation has problems.

  • $\begingroup$ I deleted a moderately inappropriate chat discussion (which seemed to be concluded, anyway) (edit) and the significantly more inappropriate discussion which followed. Posting more comments to complain about comment deletion is not OK - you can try taking it to Physics Meta.*(end edit)* Remember to be nice. $\endgroup$ – David Z Mar 4 '16 at 18:13

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