Tuesday, March 13, 2012

Energy measurements in a two-fermion double-well system




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.

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 ?

Your humble correspondent answers:

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}

$$ The system is simply composed of two independent systems – two wells in two different regions – that are not correlated or entangled at all. 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.

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. The single-electron subspace of the Hilbert space is approximately \[

{\mathcal H}_\text{1-electron, 2 wells} = {\mathcal H}_\text{1-electron, 1 well}\otimes {\mathcal H}_\text{2-dimensional}^\text{left or right well}

\] 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. The question 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. 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 classic two-level quantum system; I recommend you Feynman's lectures on physics to train intuition for them.

Technical: If you sometimes see the \(\LaTeX\) symbols at wrong places, with excessive or missing spaces, or with overlapping characters, right-click an equation or e.g. this \(\TeX\) inline maths or symbol, choose Math Settings, Math Renderer, and click at HTML-CSS again (even if it is chosen). MathJax will rebuild the equations and they should be more correct after it's done. You may try to play with the other settings in the right-click MathJax menus, too.


  1. "very simple basic schoolboy quantum mechanics", love the quote

  2. Am I alone in getting 'Math processing Error' for all the equations? That's using Chrome. Firefox never seems to render them.

  3. Dear Lubos,
    Quantum mechanics of two wells
    separated by a large distance between them
    is a very interesting example of an
    entangled state even for
    a one electron system like a simplest
    molecule, hydrogen molecular ion
    H2+ which I studied in:
    PRA 81, 062101 (2010). It is amazing
    that one can separate two protons until 100 Angstrom apart and still have a half electron with positive wave function sitting in the left well whereas another half (with negative sign). This paper proposes measurement of this strange situation occurring in a dissociating ion.

  4. Dear Ron, you're not the only one. The whole Solar system experienced the MathJax.org bug on all servers, even mathjax.org doesn't work.

    I've switched to a Vietnamese mirror, it's version 1.1.7 but hopefully a working one. ;-) You may remind me to switch back when it works again.

  5. Dear Stefan, interesting. I understand in what sense the 1-electron ground state is "entangled".

    But the 2-electron state isn't entangled. There's one electron in the near-ground-state here, one electron there...

  6. Dear Lubos,
    Let's think about H2 neutral molecule
    somehow separated by R>> a_Bohr=0.5 Angstroms.

    The (spin) singlet state solution is:
    psi_left(1)x psi_right(2)+
    psi_left(2)x psi_right(1)
    For triplet state one replaces (+)
    by (-).
    This is an entangled state since it
    is not a product.
    psi_left, pis_right are solutions for a hydrogen atom created at left
    an right centers respectively.

  7. Let me just mention why Stefan is quite right. The one-electron energy eigenstates in the two-well system is the sum (or difference) of one-electron states in the two individual wells which may also be written as states of QFT in the two regions,
    \[ \ket\psi = \frac{1}{\sqrt{2}} \left( \ket 0\otimes \ket 1 \pm \ket 1 \otimes \ket 0\right)\] where 0,1 represent the number of particles in a region and the tensor product always includes the Hilbert space of the left well times the Hilbert space of the right well.

    If you look at the structure of the state above mathematically, you will see that it's isomorphic to the classic maximally entangled EPR states, so a 1-electron eigenstate spread over two wells is entangled in this sense. I suppose that the paper was this one and Stefan has just Westernized his name. ;-)

  8. Oops, I wrote my comment above before I saw Stefan's new comment (now posted above mine) in the moderation queue. I hope that the agreement will help to convince someone and that my \(\LaTeX\) typesetting is prettier! ;-)

  9. Otherwise I obviously agree that the singlet state of a H2 molecule - a two-electron state - is entangled as well. But that's because there are extra local labels (spin).

    Some states one may produce out of the triplet state (which is given by a 3-dimensional Hilbert space) are actually not entangled. Take e.g. the J_z=1 state of it. It has both electrons in the same J_z=1/2 state and there are no various terms how to achieve it. So the full triplet state *is* a tensor product and isn't entangled; it's actually an identical problem to the spinless two electrons in two wells discussed in this blog entry.

    By rotational symmetry, one may write triplet states polarized in all other directions than z, they're not entangled either. On the other hand, the j_z=0 state of the triplet state is entangled, much like the singlet.