This interactive question will help us develop a deeper understand of mixed states (and eigenstates).
Suppose you are given 1000 quantum systems all prepared in the same exact mixed state. For concreteness, suppose that: it is an electron in a 1-D harmonic oscillator, and that the coefficients c_0 through c_5 may be non-zero while all other c's are zero. Suppose you want to learn about the energy of the electron and the nature of the state it is in.
What people do in practice to study the energy of electrons in bound state systems is to shine monochromatic UV or x-ray frequency light on them, so that an electron absorbs one photon in an energy conserving process; this give the electron enough energy to become free of the bound state system (i.e., if hf exceeds the "work function"). Then one detects the energy of the free electron (absorbing it into a detector) and inferring the energy of the bound electron state via conservation of energy.
Suppose that you do this with, say, 100 of your 900 mixed-state quantum systems. What will your results be? How would you display them? What would they tell you about the nature of the other 900 quantum systems you have?
PS. It might be useful to make up specific numerical values for the non-zero coefficients, and then discuss things for that specific case. This may help clarify what you are saying to others.
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For the number of ground states that you'd find, wouldn't it be |<(C_0)Y_0|Y>|^2, where Y is the mixed state and Y_0 is the ground state of energy? I guess after that you'd need to know if the different HO eigenstates were orthogonal. Are they? Does <(Y_0)|(Y_1)>= zero, etc.? I don't have the HW#5 solutions yet. If they are orthogonal, it would seem that
ReplyDelete|<(C_0)Y_0|Y>|^2= |(C_0)|^2, and then that the number of ground states found would be
(100)(|(C_0)|^2). But this is of course not exact, becuase it's a probability, so this ratio would become more and more accurate as the sample size increased. But whats the margin of error and percentage chance that you'd step outside of it? I know we studied this in 133, but I've forgotten how to do it.
To Hefeweizen:
ReplyDeleteThe eigenstates of the Hamiltonian operator are always orthogonal regardless of the potential. So I think you are right in saying the number of times that we might expect the particle to be "observed" in a particular state would be given by the square of state's corresponding normalization coefficient multiplied by the number of times we have done the experiment.
What's interesting here is that this implies that even though the systems have been prepared in a mixed state they each will be "observed" to have been in a particular eigenstate distributed presumably in agreement with their respective probabilities. I'm not sure what we could say about the nature of the other 900 quantum systems but it seems like, assuming the systems are undisturbed, they would be distributed on a plot (number of times "observed" vs eigenstate) in much the same way. The most contribution to the probability in these fourier series expansions comes from the lowest energy wave form. Perhaps this has something to do with the electron wanting to be in a lower energy state.
If we're running non-zero coefficients up to n=5, then aren't we talking about linear combinations of state functions? If so, then wouldn't it be impossible to 'prepare' 1000 electrons to be in any exact mixed state?
ReplyDeleteRegardless, if this madness were true, then in horrible violation of quantum mechanics, we would actually KNOW something! Two electrons in exactly the same mixed state would yield exactly the same bound state energy. How could they differ in this problem?
I don't think two electrons in a mixed state necessarily have to have the same energy--it's just that the probability distribution of energies is the same for both. For example, let's say two states were both 1/Sqrt[2](Psi0 + Psi1), ie they're both composed of equal amounts ground state and first excited state. The energy of one of the electrons, when measured as described above, won't be some combination of E0 and E1, but will be one or the other--the chances for each are 50%. The same holds true for the other electron. However, the two don't have to have the same energy--just because they're in the same mixed state doesn't mean they'll both have the same energy, just the same probability of energies. In fact, the chances of both energies being the same are only 50% (both E1 or both E0), while the chances of them being different are also 50% (one E0, the other E1, and vice versa).
ReplyDeleteWhat exactly does the Pauli exclusion principal say about this problem? My understanding is that no two fermions (electrons in this problem), can simultaneously be in the same quantum state (as defined by the four quantum numbers).
ReplyDeleteSo wouldn't that mean that there is some finite set of possible configurations for our problem? What effect would only having the first 5 states accessible have (is that what C_1 through C_5 means?), would it be to limit the total number of configurations? What if this total number of configurations is less than 1000, is this problem possible? Surely there are two fermions that at some point in time in all of the universe have simultaneously been in the same quantum state -- how does spatial separation factor in to the Pauli exclusion principle?
Not sure if this is relevant, am I missing some fundamental idea here?
To Jason:
ReplyDeleteI believe the pauli exclusion principle is not violated in this problem because the localization of the wave function is one of the defining factors of a particle's quantum state. I'm not sure about the details but I think two electrons orbiting the same atom are localized around that atom thus have the same localization. Something like that :)
Correction: around the same nucleus*
ReplyDeleteI think a good way to look at it actually is a particle in a 1000 finite square well potentials. The electron's different states "look" the same if the finite square wells are separated far enough. So although the states "look" the same, they are still in different states.
ReplyDeleteI took the problem to mean that we have 1000 identical systems--1000 boxes,each of which contains a single electron. That is to say, 1000 boxes and 1000 electrons. There should be no need to worry about the exclusion principle, because each electron is isolated in its own system.
ReplyDeleteI think that the comment hefeweizen made about the number of ground states found being (100)(|(C_0)|^2) is very reasonable. Is there anything that this theory violates?
ReplyDeleteto display these results, I would say we need some sort of graph involving the number of wells/electrons counted. is c_0 a constant?
And it seems to me like if we kept knocking electrons out of potential wells, we'd eventually have unoccupied potentials. If the states were interacting, wouldn't the electrons try to spread out into these vacant wells to lower their energy?
ReplyDeleteWhich would mean the energies of the freed electrons would decrease (Because we keep zapping the systems with constant energy hbar*f, and with electrons lowering their energy, you'd need more energy to unbind them, so more of the hbar*f would go towards freeing them, and less of the energy would be present outside the well).
Or am I crazy?
Crazy is a strong word. You are very creative. Did you see Kelsy's comment(kcollier). What do you think about that.
ReplyDeleteTo Gatlin and kcollier:
ReplyDeleteJason however brought up a good point, is this experiment even possible? I think we don't have to worry so much about the pauli exclusion principle as long as the fermions are isolated well enough (either separated very far or sit in strong potential wells). So to a good approximation it is like having 1000 isolated infinite square wells, each with an electron in a superposition of states.
Also in response to Gatlin, in reality it seems like if all we had were 1000 finite square well potentials, the electrons would want to lower their energy into another potential as soon as one of the fermions left it. I think this would take a long time if they were isolated well enough.
kelsy's comment makes sense, assuming the 1000 systems are non-interacting.
ReplyDeleteby the way, how do you "prepare a mixed state"? any ideas?
Jason,
ReplyDeleteI assume the answer to your question lies in the notion of expectation value for a mixed state. From my understanding if you have N identical mixed states the mean value of the measured values will be the same. It goes back to what Mike was saying earlier about the probabilities.
If I had to "prepare" this experiment, I would seriously consider taking 1000 boxes with a Schrodinger's cat inside of each one. I think the problem is somewhat similar to what we have here (somebody, correct me if I'm wrong).
Max
Harking back to the question about Pauli exclusion principle, I hate to disappoint you, but sadly 139a is supposed to be a "one-electron" class. That means we just look at one-electron eigenstates, and there no Pauli exclusion principle in play.
ReplyDeleteThe 2- and multi-electron problems are supposed to be reserved for 139b. I guess we could maybe talk about multi-electron atoms after we examine the eigenstates of the hydrogen atom or something? What do you think?
PS. I am afraid cats may have more than one electron.
ReplyDeleteso basically, what we are seeing here, is that given a mixed state and probabilities of finding an electron with a specific energy for several states of the mixed, we will always see the electron in one state with a discrete energy.
ReplyDeleteif we observe knocking out many electrons we should see over time the number of times an electron has a certain energy is directly related to the certain energy's state's potential.
if we do this for a hundred boxes all of the same mixed state, then we should see similar distributions for each system. we could then assume that any other box of the same potential and mixed state would give similar results.
does this sound right? it's easy to say in pictures and numbers, but harder to write out.
I wrote a small program to calculate the mixed state probabilities/energies for any combination of coefficients. You can download it here:
ReplyDeletehttp://people.ucsc.edu/~akunapul/Mixedstate.exe
I think everyone one understands the bare-bones of the answer here, but a nuanced discussion of the meaning and related issues would be ideal.
ReplyDelete(in this class)
ReplyDeletelol
ReplyDelete