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This question is related but my question here is much more elementary than discussions of the Pauli principle across the universe.

There has been a fair amount of discussion around at the moment on the double well potential model in quantum mechanics. The question I have here is just concerned with some very simple basic schoolboy quantum mechanics that you need to get right in order to progress with the discussions (I think these questions would be too elementary for a physicist to ask - fortunately I'm not a physicist, so my embarrassment is under control).

The scenario is this: I have a double square finite depth potential well with a large distance separating the wells. Call them left hand and right hand wells. The model I want to use is elementary nonrelativistic quantum mechanics. The energy eigenstates of such a system are similar to those of a single well, but what was one energy eigenstate now becomes two with incredibly closely spaced energy levels (ref).

Suppose I have two electrons in this system in the two very closely spaced lowest levels $E_1$ and $E_2$ . The two electrons are indistinguishable, so the state is $${{1}\over{\sqrt{2}}}(|E_1\rangle|E_2\rangle-|E_2\rangle|E_1\rangle)$$ The statement that has often been made in these discussions is something like "an observer in the left hand well now makes an energy measurement of an electron and gets the value $E_1$ ".

In talking about this scenario, in the link I gave above Sean Carroll stated

"Imagine that we have such a situation with two electrons in two atoms, in a mutually entangled state. We measure our electron to be in energy level 1. Is it true that we instantly know that our far-away friend will measure their electron to be in energy level 2? Yes, absolutely true."

This is where I'm confusing myself: By assumption, our system was in an energy eigenstate, and we've made an ideal energy measurement, so by the rules of quantum mechanics after the measurement it ends up still in an energy eigenstate after the measurement. But the observer knows that we have a double well two-fermion system and the correct system state is still the antisymmetrized one $${{1}\over{\sqrt{2}}}(|E_1\rangle|E_2\rangle-|E_2\rangle|E_1\rangle)$$. So for an idealized measurement, the system started in this state and remained in this state. So I ask

question (1): If the measurement proceeds like this, without any extra entanglement-introducing mechanism, won't the observer think "if my far-away friend now makes an energy measurement, he'll have a 50/50 chance of getting $E_1$ or $E_2$" ?

In other words: since this is non-relativistic QM, the particle could have moved to the other well immediately after the first measurement, so the state is still antisymmetrized over the particle identities.

In other other words, the state I described isn't mutually entangled - just antisymmetrized over the electron identities.

question (2): if this correct then how, typically, would I introduce some entanglement between the electrons, as in Sean's statement ?

Finally, when we say "an observer in the left well makes an energy measurement", doesn't this introduce some locality into the picture for the following reason: What it implies is that the observer's measuring equipment is spatially localized in the left well, so that when an energy measurement is made, there is also an implied acquisition of information about the measured electron's location at that time. So I come to:

question (3) Wouldn't this mean that after such a local measurement, the measured electron cannot be in one of the system's energy eigenstates ? So the system state would be something like $${1\over{\sqrt{2}}}(|\Psi\rangle|E_2\rangle-|E_2\rangle|\Psi\rangle)$$.
where $\Psi$ is a spatially localized state, concentrated in the LH well.

This doesn't sound right. Where is my reasoning going wrong ?

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4 Answers 4

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Your confusion is fully appropriate. It is far from schoolboy QM, most wouldn't even notice the discrepancy and live happily with ensuing inconsistencies.

Entanglement is not applicable to this situation, as the Hilbert space in question is not a tensor product.

You can entangle only properties of a composite system that are described by a full tensor product. But states of identical particles only make up the antisymmetric part of a tensor product; so all the common machinery goes awry.

The situation is explained in detail in the section ''Indistinguishable particles and entanglement'' of Chapter B3: Basics on quantum fields of A theoretical physics FAQ, which ends with:

''But if, on a more practical level, one knows already that (by preparation) there are two electrons or two photons, one moving to the left and one moving to the right, one can use this information to distinguish the electrons or photons: The particles are now described by states in the tensor product of two independent, much smaller effective Hilbert spaces with a few local degrees of freedom each (rather than a single antisymmetrized 2-particle Hilbert space with degrees of freedom for every pair of positions), which makes them distinguishable effective objects. Therefore, one can assign separate state information to each of them, and construct arbitrary superpositions of the resulting tensor product states. These effective, distinguishable particles can be (and typically are) entangled.''

[edit after discussion below] I updated my FAQ to give some more details, using spins in place of your energies, which makes the situation more realistic. I quote the relevant passages:

''For example, two electrons are on the fundamental level indistinguishable; their joint wave function is proportional to |12>-|21>; no other form of superposition is allowed. [...] Thus it is impossible to project it by measurement to a separable state, as discussions about entanglement always assume.''

''Realistic discussions about entanglement in quantum information theory usually consider the entanglement of spin/polarization degrees of freedom of photons transversally localized in different rays, electrons localized in quantum dots, etc.. For example, two electrons in two quantum dots $dot_1$ and $dot_2$ are in the antisymmetrized state

$|\psi_1,\psi_2\rangle := |\psi_1,dot_1 \rangle \otimes |\psi_2,dot_2 \rangle- |\psi_2,dot_2 \rangle\otimes|\psi_1,dot_1\rangle$

in 2-particle space (depending on 6 coordinates apart from the spins, the latter represented by $\psi_1$ and $\psi_2$), but as the positions are ignored (being just used to identify which electron is where), they are treated as effective electrons in the 2-particle state $\psi_1 \otimes\psi_2$ residing in the tiny 4-dimensional tensor product $C^2\otimes C^2$. If spin up is measured at $dot_1$, the spin part $\psi_1$ collapses to up, etc.. Note that there is no position measurement involved in observing the spin, as the quantum dots already (and permanently) have measured the presence of a particle in each dot.

Thus in this small effective space, talking about electron entanglement is meaningful, as disregarding position (responsible for the antisymmetrizing) reduces the state space to a tensor product. But in the full 2-particle space, the attempt to reason with entanglement fails on the formal level and is at best paradoxical if just considered informally.

Edit2 (March 13, 2012): The approximate representation mentioned in the answer by Lubos is essentially the same as the exact reduced representation in my answer, if we identify left well = dot_1 and right well = dot_2. It is a reduced description that accounts for the knowledge that one electron is in each well and has to stay there because of the walls of the well. (If the walls were infinitely high, his representation would be exactly the same.) This means having a way to distinguish the electrons, so that one can use a smaller tensor product Hilbert space inside the big antisymmetric Hilbert space to (approximately or exactly, depending on the form of the well) describe the degrees of freedom left.

In my answer above, I took as only dof left the spins, as this is the situation that prevails in quantum information theory. However, the reduced Hilbert space can take other degrees of freedom into account. As long as these additional degrees of freedom are local to each well/dot, one still has the tensor product structure, and hence a setting where entanglement makes sense, and in which one can consider to prepare an entangled 2-electron state.

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Thank you very much for the answer - If I have understood correctly, you are saying that it's inappropriate to talk of entanglement between particle 1 and particle 2 since $ \mathcal{H}\not=\mathcal{H_1}\otimes\mathcal{H_2}$ where $\mathcal{H_1}$ and $\mathcal{H_2}$ are the single-particle Hilbert spaces for particles 1 and 2, but rather it's an antisymmetrized tensor product. This is good, but one small clarification: here we are talking of the situation where the observer in the LH well makes, say, an energy measurement on an electron and an observer in RH well an energy measurement –  twistor59 Mar 5 '12 at 12:54
on an electron there. The desired Hilbert space decomposition in question would be something like $ \mathcal{H}=\mathcal{H_L}\otimes\mathcal{H_R}$ where $\mathcal{H_L}$ is a Hilbert space of states "appropriate" to the observer in the LH well and $\mathcal{H_R}$ is the Hilbert space of states "appropriate" to the observer in the right. I put the word appropriate in quotes because I would like to know if it's possible to give a definition that would make the decomposition of $\mathcal{H}$ possible in this way. If you would be kind enough to clarify this, the bounty is yours! –  twistor59 Mar 5 '12 at 12:54
But now your problem is inconsistent with the form you ascribe to your setting or to your electron state. Entanglement requires that every state in H_L tensor H_R is preparable. In such a state space the electron wvave function has no reason to be antisymmetrized. Antisymmetry is a property guaranteed in the original space of the electron (wherenot all tensor product states exist), but not in an effective substitute. You may not mix two incompatible descriptions without getting paradoxes. –  Arnold Neumaier Mar 5 '12 at 13:33
To create a consistent model for what I think you intended is nontrivial, and I need more time. To model the situation, one needs an appropriate Lindblad equation including simplified measurement interactions; then one can answer what actually happens. It is far from the standard entanglement situation, so simplified reaasoning is fallacious. –  Arnold Neumaier Mar 5 '12 at 13:42
I added more explanations to my answer above. I hope this fully clarifies the situation for you, as i don't think more can be reasonably said about it. So I won't look into the detailed dynamics as I originally thought would be needed. - Many thanks for your question! –  Arnold Neumaier Mar 5 '12 at 16:48

I don't sharply disagree with Dr Neumaier's answer; it is indeed the case that entanglement may only be discussed for Hilbert spaces that are tensor products.

However, if the two parts of the well are sufficiently distant, this is nearly the case of your situation, too. When one looks at it in this approximate way, the answer is that the electrons – assuming that you only occupied one spin state, for example both electrons are spin up – are not entangled.


The Hilbert space with two widely separately wells that can store electrons is approximately the tensor product $$ {\mathcal H} = {\mathcal H}_\text{left well} \otimes {\mathcal H}_\text{right well}$$

The two individual product Hilbert spaces are not quite completely well-defined: one doesn't want to discuss quantum field theory on a "region of space" due to the problems with the boundary conditions (the "big" Hilbert space doesn't constrain the fields near the boundaries around the wells at all while the smaller Hilbert spaces have to impose some boundary conditions, so the factorization above can't be exact).

However, as long as these boundary conditions are not a problem (for example because it's guaranteed that everything is almost totally confined near the well and nothing gets close enough to these boundaries), the Hilbert space does factorize in this way, and so does the state you wrote: $$|\psi\rangle = |\text{1 electron}\rangle_\text{left well} \otimes |\text{1 electron}\rangle_\text{right well} $$ Note that the one-well, one-electron problem only has one ground state: there is no degeneracy here, not even an approximate one.

The system is simply composed of two independent systems – two wells in two different regions – that are not correlated or entangled at all. In quantum field theory, the tensor product state above could be written as $a^\dagger_\text{left well} a^\dagger_\text{right well}|0\rangle$ where the two creation operators don't carry any labels and they are composed of field operators near the two wells, respectively. A non-entangled state is defined as one that can be written as a tensor product and that's exactly what we can do here (in the two-region approximation).

We don't violate the Pauli exclusion principle here in any way because in this approximate two-region description of the system, the binary quantum number "rough position" (which is either "near left well" or "near right well") plays the same role as the spin or other quantum numbers. The two electrons have different eigenvalues of "rough position" which is why they can be in exactly the same state when it comes to energy, spin, and all other quantum numbers.

This extra quantum number is also the reason why you have two nearby energy low-lying states of the two-well problem. There's a two-dimensional Hilbert space for a single electron spanned by energy eigenstates with energies $E_1,E_2$: the corresponding eigenvectors are "even" or "odd" functions of the position (the wave functions either have the same sign in both wells or the opposite sign). In the approximation in which the space between the wells is impenetrable and the boundary conditions for the regions don't pose a problem, we have $E_1=E_2$ and the two-dimensional Hilbert space may also be generated from another basis containing the ground state of the left well and the ground state of the right well. In this approximation, we're just filling two states that only differ by the "rough position" by the maximum number of two electrons.

The inequality $E_1\neq E_2$ in your exact treatment only arises because there's a nonzero probability amplitude for an electron to tunnel from one well to the other one. If it couldn't tunnel, we would have the exact "doubling" of the Hilbert space for a single electron. For the same reason, one can't measure the energy "in one well only" with the accuracy needed to distinguish $E_1$ and $E_2$.

If your measurement apparatus is confined to the vicinity of one well, the error in your energy measurement can't be smaller than $E_1-E_2$ so you won't be able to say "which of the two nearby states" the electron is in. The same holds for the vicinity of the other well which is why the measurement in one well can't influence anything detectable near the other well.

The impossibility to distinguish $E_1$ and $E_2$ by a measurement near a single well is easy to prove; if you measure the electron near the left well, with whatever low-lying energy near $E_1$ or $E_2$, you are proving that this electron is in an eigenstate of the "rough position". But the operator of "rough position" doesn't commute with the total energy; the eigenstate $|\text{left well ground state}\rangle$ is a linear superposition of the $|E_1\rangle$ and $|E_2\rangle$ eigenstates (it's the right linear superposition that vanishes near the other well), something like $$ |\text{left well ground state}\rangle = \frac{1}{\sqrt{2}} \left( |E_1\rangle - |E_2\rangle \right ) $$ If you've measured the "rough position", you are totally uncertain about the eigenvalue of the "exact energy" because these two operators don't commute with one another; a textbook case of the uncertainty principle. If the two wells are equally deep etc., by seeing an electron near the left well, you have 50% odds that its energy was $E_1$ and 50% odds that it was $E_2$ and nothing can be changed about these odds because they follow from the displayed equation above.

In terms of operators, we may say that in the basis "left well ground state" and "right well ground state", the operator of "exact energy" looks like $$ H = \frac{E_1+E_2}{2}\cdot{\bf 1} + \frac{E_1-E_2}{2} \cdot \sigma_1 $$ where the second term is proportional to an off-diagonal matrix similar to the first two Pauli matrices. It isn't diagonal in this basis so if we know that we found an electron near the left well, we know that its "exact energy" (whether it is $E_1$ or $E_2$) is maximally uncertain. And vice versa. If we find an electron in the state $E_1$, and we are sure it is not $E_2$, then this electron must be in a wave function that is nonzero near both well, so we don't learn anything about the "rough position" (left or right) which remains maximally uncertain.

If we make a measurement of an electron near the left well, the right conclusion that the antisymmetry or Pauli's principle allows us to predict is that the other electron is in the right well. It's that simple. But learning that it's in one particular well is incompatible with learning whether or not it is in the $E_1$ or $E_2$ eigenstate because the operators corresponding to these questions don't commute with one another.

If several electrons are in vastly different regions of space, the Pauli exclusion principle becomes inconsequential, of course: the electrons are effectively distinguishable by their location. So the dimension of the Hilbert space for the two separated wells is the simple product of the dimensions of the Hilbert spaces for the individual wells; there's no additional "antisymmetrization" we should do here because we're discussing "off-diagonal blocks" of a matrix and the antisymmetric part of the state is hiding in the convention how we label the two electrons.

But to be able to look at the situation in this factorized way, I had to organize the Hilbert space as a tensor product of pieces that correspond to individual regions. If we organize the Hilbert space according to "individual electrons that may a priori be anywhere", we can't really talk about the entanglement at all because the total Hilbert space of many electrons isn't a tensor product of the individual electrons' spaces: it's the antisymmetrization of it.

The simplest, strict definitions of entanglement don't apply to such antisymmetrized tensor spaces. There's still a natural convention that if we have antisymmetrized (or symmetrized) tensor product Hilbert spaces, we still consider the antisymmetrization (or symmetrization) of a tensor product state to be a non-entangled state. This includes your state. Such a definition will tend to produce similar verdicts as the procedure based on the quantum field (composed of various regions) that I described above.

At any rate, you won't find any helpful way to argue that (and why) these two electrons are entangled: we're not learning any new information (such as the spin) about "the electron in the left well" at all so this "no information" can't be entangled with any information from the right well (which is also empty). The question whether there's entanglement here is either ill-defined or they are not entangled. And even if you found a (contrived) definition that would allow you to say that the simple state is entangled, such an "entanglement" will have no physical consequences. Two highly separated regions (or wells) are independent. In particular, the laws of quantum field theory are exactly local so a measurement or decision done near one well won't immediately influence a spatially separated other well.

To summarize and address your questions:

  1. Finding an electron in the left well ground state means that it has 50% odds to be in the $E_1$ state and 50% to be in the nearby $E_2$ state of the double well problem; we can't simultaneously distinguish left-right as well as $E_1$ vs $E_2$ because the corresponding operators refuse to commute with one another. (I say "refuse", not "fail", because it's a holy right – and the dominant situation – for two operators not to commute. They have no duty to commute in quantum mechanics so a nonzero commutator isn't a failure, isn't bad in any way.) If we find an electron near the left well, what the antisymmetry allows to tell us is that the second electron is near the right well, and vice versa. But measurements linked to one of the two regions can't tell us about the exact energy of one electron (and therefore it tells us nothing about the energy of the other one, either)

  2. In the description of "individual electrons", one can't talk about entanglement because the full Hilbert space is an antisymmetrization (reduced version) of the tensor product, not the full tensor product. In the approximate description of quantum field theory on two regions, the big Hilbert space tensor factorizes and the two-electron state (occupying the two low-lying states) isn't entangled. If the initial state is not entangled and the evolution of the quantum system respects locality (and quantum field theory does), no entanglement may be created by actions done near one well or the other well. Entanglement is always a consequence of the two subsystems' being in contact in the past.

  3. Yes, as I said, you're exactly right: if we know that an electron is near the left well, the odds for its being in the $E_1$ two-well state or the nearby $E_2$ two-well state are exactly 50% for both cases. The left-vs-right and $E_1$-vs-$E_2$ can't be measured simultaneously much like $J_z$ and $J_x$ components of the spin cannot; in fact, these two examples are totally mathematically isomorphic.

A blog version of this answer of mine is here;


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Thanks for the extra clarification. Your deduction <<If your measurement apparatus is confined to the vicinity of one well, the error in your energy measurement can't be smaller than E1−E2 so you won't be able to say "which of the two nearby states" the electron is in.>> is exactly what I was wondering about, but couldn't formulate correctly. I think one of the key elements in the "Coxtroversy" is the belief that the possibility of tunneling between the wells, no matter how infinitesimally improbable, means that local measurements can influence remote ones... –  twistor59 Mar 13 '12 at 10:38
...the above is a quantitative argument that this is not the case –  twistor59 Mar 13 '12 at 10:41
Thanks for your interest, twistor. You're exactly right. The inaccuracy of the total energy may be seen in "subtly different" ways, depending on the procedure. An apparatus that doesn't see the other well has to "cut" its vicinity in some way, and this creates an uncertainty. Alternatively, we may detect the electron in the first well and then measure the energy exactly, taking both wells into account. Then we will find out that the odds are exactly 50% and 50% for $E_1,E_2$ but we need to make a global measurement. –  Luboš Motl Mar 13 '12 at 10:45
...and that needs one heck of a big apparatus ! –  twistor59 Mar 13 '12 at 10:47
The locality in QFT is totally exact, so you won't have even an infinitesimal probability to transfer a "little bit of information" along spacelike intervals of spacetime. Any apparent result that this is possible either pretends that some measurement may be done more accurately than it can, or it pretends that global measurements may be done locally, or does other mistakes. BTW in EPR-entangled pairs, one also transfers exactly 0 information because we can't "order" even our particle how it should be measured. But this double-well case isn't even entangled in any useful sense. It's not EPR. –  Luboš Motl Mar 13 '12 at 10:47

I already answered this question for the "real" case of two electrons with spin, but Lubos assures me that the OP meant for me to consider the case of no spin. I'm not sure what that would look like...if there is no spin, the electrons are bosons, and there is no Pauli exclusion principle. But Lubos further explains in his comment that we are to consider the "embedded" case of all electrons spin-up, where he means this case is "embedded" in the reality of arbitrary spin...we just restrict ourselves to looking at cases where the spin is up.

I've done that now, and it still doesn't look like anything recognizable from the other posted responses. I start by looking at the case of two hydrogen atoms with one electon. That's not too hard; and we do get something like what people are talking about, where there are two energy levels very close together:

One Electron and Two Atoms

The symmetric case is the true ground state, and a single electron in the ground state is equally shared between the two atoms. If you observe the electron in one or the other atoms, then it isn't in the ground state, it's in a superposition. If you find it at the left atom, then its the sum of the symmetric and antisymmetric states. If you find it at the right atom, it's the difference. Since the complex phase of two states precess at different rates, what is now the sum will later be the difference: so an electron found at the left will eventually show up at the right.

If you add a second electron, it's different. Both electrons can occupy the same state if their spins are opposite, but Lubos assures as we are to restrict ourselves to those cases when both electrons are spin up. That brings us back to the answer I posted earlier, for the arbitrary case where there are four energy levels. Only one of those cases allows both spins to be up...it is my State III. Technically, you cannot say that one electron is at A and the other is at B; you have to say that the electrons are in a superposition of states where A is here and B is there, and vice versa.

Either way, there is only one state. There are no superpositions like E1E2 - E2E1, as the OP stated and Lubos and Arnold went to town with. I just can't imagine what people are talking about.

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We are considering the case of two electrons, one in each well. Even with only one space dimension, the wave function has two arguments $x_1$ and$x_2$, not just one. Thus your figure is just the cross section in $x=x_1-x_2$, far from the full picture! –  Arnold Neumaier Mar 15 '12 at 20:13
Arnold, I don't know if you even bother to read what I wrote. I said that the picture was two atoms with one electron. The full picutre with two electrons I posted yesterday, showing four different cases. That was the answer which Lubos ridiculed. –  Marty Green Mar 15 '12 at 20:32
The situation is the same for two scalar electrons, or two electrons both frozen to the up state. One has two spatial degrees of freedom, one for each scalar particle, and one antisymmetrizes these. –  Arnold Neumaier Mar 15 '12 at 20:45
Surely you realize I have no idea what you're trying to say. –  Marty Green Mar 15 '12 at 23:30

It is not clear to me that the answers posted so far address the OP's question. I find the OP does not have a proper representation system for the possible states. There are not two energy eigenstates but four: the true ground state, which is a singlet spin state, and three more states very close to the ground state, which are triplet states. He talks about a superposition of two energy eigenstates as though he has combined them to form an anti-symmetrized state...he seems to be trying to construct the singlet state from the energy states, but it doesn't work that way. You construct the singlet state from the position states...A-up/B-down, etc; and when you combine them properly, you get the singlet state which is in fact the lowest energy eigenstate.

All the fun happens when you get a system of two separated atoms in this singlet state. If you measure the spin of one, then the spin of the other must instantly become opposite. All the wonderful paradoxes that everyone likes to talk about flow from this situation. There are a couple of problems, however. It's not that easy to measure the spin of an electron. I don't care what anyone says: it's not that easy. And I don't think anyone knows of a way, even in principle, to prepare a system of two hydrogen atoms in the spin singlet state. I don't doubt that such pairs exist in nature: I just don't think there's any way to isolate such a pair and determine that it is indeed in the singlet state, at least not without destroying that state in the process.

I put up some sketches on my blog to show what the basis states look like for a system of two hydrogen atoms. I had some trouble getting it right, because it's a little confusing, but I'm pretty sure my final version is correct. This is the picture I came up with: it's supposed to be self-explanatory, but it would probably help to look at my blog. The singlet state is the one boxed in green; the triple states are in purple:

enter image description here

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The context of my question was some arguments being presented in chapter 8 of this book in which a double well system with fermions was considered in which spins were ignored. It was the claims about entanglement purely from the fermionic nature of the particles that I was trying to understand. –  twistor59 Mar 14 '12 at 7:22
It's hard for me to understand how your other correspondents could have given such detailed answers without explicitly taking note of this condition. –  Marty Green Mar 14 '12 at 13:11
Dear Marty, the OP's question makes it absolutely clear that twistor59 is asking about the textbook example of spinless electrons so your discussion of the spins has nothing to do with the original question. In the real world, this "spinless electron" system is embedded in such a way that all the electrons we consider must be e.g. spin-up. In your spin-ful language, one only picks the two-electron states with $s_z=+1$ which really means the unique state from the triplet. Allowing different spins creates more states, and therefore the Pauli's principle becomes even less constraining. –  Luboš Motl Mar 15 '12 at 9:30
I wonder if you could identify any specific way in which the OP makes this point clear? Is it the part where he quotes Sean Carroll talking about two electrons in two atoms? –  Marty Green Mar 15 '12 at 12:55

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