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Let's consider an electron as part of a larger system as an atom consisting not only of a nucleus but also of several other electrons. I guess, one can characterize the atom quantum-mechanically in a complete way. Then I am tempted to describe the electron as subsystem of the atom which as part of a complete system would only be possible by a density matrix. But I guess that we can still describe the electron of the atom with a wave function as long as its state is pure. To be a pure state there must exist a complete set of measurements which always give an unique result. However, how do I know that the state of the (this) electron is described by a complete set of measurements which give an unique result ? For instance I can imagine electron states in an atom where I can measure energy, orbital angular momentum and spin (this description seems to be complete) whereas there might be also states I can only measure energy and total angular momentum. So how we know if the description of a state is still complete or not ? In particular I would appreciate to learn about a counter-example, i.e. where indeed the description of the electron is no longer complete as the case where the description of the electron state is still complete seems to be more standard (at least in an atom). My thinking has been simulated by Landau/Lifshitz volume 3, chapter 1, 14 and 67.

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  • $\begingroup$ Actually I am asking for a case where the product decomposition of $N$-electron state into a 1-electron state with a $(N-1)$-electron state is no longer possible, i.e. a density matrix description would be necessary. $\endgroup$ – Frederic Thomas Mar 25 at 11:53
  • $\begingroup$ @Chiral Anomaly: Sorry for my slow response. Your answer was quite elaborate & I wanted to fully understand it before responding. I expected a density matrix as "1-electron state"-description in a complicated atom. Probably to write down such a matrix is complicated. It wasn't my prime interest to see such a matrix. I actually would like to know when can I still expect to describe an electron state as pure state in an atom and when is it no longer possible ? What would be the condition for a switch from a pure electron state to a mixed electron state ? Or doesn't this question have an answer ? $\endgroup$ – Frederic Thomas Mar 29 at 17:51
  • $\begingroup$ The question is answerable, after adjusting the question to account for the fact that electrons are indistinguishable particles (fermions). I posted a new version that is hopefully more clear than the previous one. I also trimmed out the previous example to make the answer a little shorter. $\endgroup$ – Chiral Anomaly Mar 29 at 19:59
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A pure state of an $N$-electron system may be represented by a single wavefunction $$ \psi(x_1,x_2,...,x_N), $$ where $x_n$ denotes the whole collection of variables needed to parameterize both the location and spin of the $n$-th electron. Even though this is a pure state of the $N$-electron system, it might not be decomposable into a one-electron pure state and a $N-1$-electron pure state. The question is how we can recognize when such a decomposition is possible.

Making the question precise is tricky, because electrons are "indistinguishable" particles. Specifically, electrons are fermions. This means that the state of an $N$-electron system never has the form $$ \psi(x_1,x_2,...,x_N) = f(x_1)g(x_2,x_3,...,x_N) \tag{1} $$ where $f$ is a single-electron state and $g$ is the state of the remaining $N-1$ electrons. A multi-electron state must be antisymmetric, meaning that it must change sign when any two of its arguments are exchanged, like this: $$ \psi(x_2,x_1,x_3,...,x_N) =-\psi(x_1,x_2,x_3,...,x_N). \tag{2} $$ For this reason, there are no observables (or measurements) associated with an individual electron. Any measurement that we try to apply to a single electron will automatically be sensitive to all of the electrons that are present. For example, we can count the number of electrons in a given region of space, but we cannot determine whether or not a specific electron is in that region; the latter concept is not even meaningful.

The closest we can come to something like (1) in an $N$-electron system is this: $$ \psi(x_1,x_2,x_3,...,x_N) = f(x_1)g(x_2,x_3,...,x_N) + \text{even} - \text{odd}, \tag{3} $$ where "even" and "odd" refer to all possible even and odd permutations of the indices. For example, the $N=3$ version of equation (3) is \begin{align} \psi(x_1,x_2,x_3) &= f(x_1)g(x_2,x_3) \\ &+ f(x_2)g(x_3,x_1) \\ &+ f(x_3)g(x_1,x_2) \\ &- f(x_2)g(x_1,x_3) \\ &- f(x_1)g(x_3,x_2) \\ &- f(x_3)g(x_2,x_1). \tag{4} \end{align} If a $3$-electron state has the form (4), then we can describe it as the composition of a one-electron state $f$ with a two-electron state $g$, while respecting the fact that electrons are "indistinguishable" particles (more specifically, they are fermions). With that in mind, the question could be interpreted like this:

Given an $N$-electron state, how do we know whether or not it can be decomposed into a pure state $f$ of one electron and a pure state $g$ of $N-1$ electrons, as shown in equation (3)?

If that is the question, then this is the answer: The given $N$-electron state $\psi$ can be written in the form (3) only if $$ \psi(x_1,x_2,...,x_N)\psi(x_{N+1},x_{N+2},...,x_{2N}) +\text{even} - \text{odd} = 0. \tag{5} $$ In particular, if $$ \psi(x_1,x_2,...,x_N)\psi(x_{N+1},x_{N+2},...,x_{2N}) +\text{even} - \text{odd} \neq 0, \tag{6} $$ then we know that the $N$-electron state $\psi$ cannot be written in the form (3), no matter what we try to use for the functions $f$ and $g$.

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    $\begingroup$ Thank you for your answer. Actually I don't understand the expression (5). In particular $\psi(x_{N+1}, \ldots, x_{2N})$. It would suggests to me that the number of electrons has doubled. What is the meaning of it. ? $\endgroup$ – Frederic Thomas Mar 30 at 20:26
  • $\begingroup$ @FredericThomas Intuitively, if any one of the original $N$ electrons has a pure state $f$ of its own, then we would violate the Pauli exclusion principle if we tried to make a system with $2N$ electrons using two copies of the original $N$-electron state, because we'd be trying to put two electrons in the same single-electron state $f$. That's what equation (5) is saying: if we try to make a $2N$-electron state from two copies of the $N$-electron state, antisymmetrized overall as required, then we'll get an invalid state-vec (zero) if any of the original electrons had a pure state of its own. $\endgroup$ – Chiral Anomaly Mar 30 at 21:55
  • $\begingroup$ @FredericThomas ...so you're right: in equation (5), the number of electrons has doubled. The idea is to use that doubling "experiment" as a test to see if it violates the Pauli exclusion principle. If it doesn't, if the $2N$-electron state is still valid (non-zero), then we know that none of the original $N$ electrons has a state $f$ of its own. The idea of a "density matrix" doesn't work with fermions the way it does with distinguishable particles, because we can't choose which electrons to ignore (they're indistinguishable). That's why I didn't express things in terms of a density matrix. $\endgroup$ – Chiral Anomaly Mar 30 at 22:03
  • $\begingroup$ When I asked the question I did not expect such a criterion. Interesting. Thank you for your help. $\endgroup$ – Frederic Thomas Apr 3 at 11:19

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