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I'm struggling with the concept of identical particles in QM. Say I have two electrons, with one trapped on my left $\left|\uparrow\right\rangle_1$ and one trapped on my right $\left|\downarrow\right\rangle_2$. The 2-particle wavefunction is then clearly $\left|\uparrow\right\rangle_1 \otimes \left|\downarrow\right\rangle_2$. If an exchange happens, clearly the wavefunction would be $\left|\downarrow\right\rangle_1 \otimes \left|\uparrow\right\rangle_2$, which is totally different and I can detect this exchange by measuring in the appropriate basis. The 2-particle system does not seem to violate any law of QM to me, and I can do all kind of operations on them including entanglement.

Yet, why is it that when I open QM textbook like Griffiths and Sakurai etc. in the Identical Particles chapters, they always say that this is impossible, that in order to construct a many-particle wavefunction, one must know that

God doesn't know which is which, because there is no such thing as "this" electron, or "that" electron

and

we (can't) follow the trajectory because that would entail a position measurement at each instant of time, which necessarily disturbs the system

I mean, am I tracking the electrons if I trap them on 2 sides of my table? Even if I am tracking the electron, what is the bad thing about it that makes my many-particle state $\left|\uparrow\right\rangle_1 \otimes \left|\downarrow\right\rangle_2$ invalid to QM? What am I misunderstanding?

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    $\begingroup$ I think what you're looking for is the concept of "Effectively distinguishable particles". $\endgroup$ Commented Dec 27, 2022 at 19:07
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    $\begingroup$ See for example section 3.2. of this PDF or this section 2. $\endgroup$ Commented Dec 27, 2022 at 19:19
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    $\begingroup$ Are the 1 and 2 subscripts supposed to denote position, or are they meaningless labels? This is a crucial point. If they are positions, then several current answers are wrong. $\endgroup$
    – benrg
    Commented Dec 28, 2022 at 19:43
  • $\begingroup$ I think I got the point. The 1 and 2 are Hilbert spaces, and the exchange operators switch these Hilbert spaces (or the order of the tensor product), meaning effectively it'd switch all properties (left/right, up/down) were the particles have an identity associated to these label/Hilbert spaces. $\endgroup$
    – Kim Dong
    Commented Dec 29, 2022 at 21:27
  • $\begingroup$ I am unsure but > No-cloning_theorem at wikipedia (maybe can help someone here) ""In physics, the no-cloning theorem states that it is impossible to create an independent and identical copy of an arbitrary unknown quantum state,"" (..) $\endgroup$ Commented Jan 1, 2023 at 23:34

7 Answers 7

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The state of the electrons in the situation you describe is actually not $|\uparrow, {\rm left}\rangle_1 \otimes |\downarrow, \rm{right}\rangle_2$. It is instead \begin{equation} |\Psi\rangle = \frac{1}{\sqrt{2}}\Big(|\uparrow, {\rm left}\rangle_1 \otimes |\downarrow, {\rm right} \rangle_2 - |\downarrow, {\rm right} \rangle_1 \otimes |\uparrow, {\rm left} \rangle_2 \Big) \end{equation} where the label "left" or "right" refers to the spatial location of the particle, and the indices $1$ and $2$ refer to the "identity" of the particle. In particular, you cannot say whether electron 1 or electron 2 is the one that is trapped at your left, with spin up.

This may feel surprising and counterintuitive. However, if you think through the experimental setup carefully, you will find there is no way for you to say whether electron 1 or 2 is "really" the one in the left trap. There is no way for you to "mark" either of the two electrons.

If it helps, you can view indistinguishable particles as a mathematical possibility that quantum mechanics allows for. Maybe you don't intuitively believe electrons must be indistinguishable, but you have to accept that quantum mechanics logically allows indistinguishable particles. Indistinguishable particles live on a subset of Hilbert space where the state is fully symmetric or fully antisymmetric under exchange of particles.

It then becomes an experimental question about whether real-world electrons are described by the quantum mechanics distinguishable or indistinguishable particles. Electrons exhibit Fermi-Dirac statistics, a fact that forms the basis of modern condensed-matter theory, and therefore in some sense is experimentally tested every time you use a conductor or insulator, or every time a material scientist designs electronic properties of a material. So as an experimental question, it is settled that electrons behave as indistinguishable fermions.

The fact that electrons are indistinguishable fermions is also understood theoretically using relativistic quantum field theory.

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    $\begingroup$ I am very much struggling with this... Isn't the $S_z$ operator commute with $H$, therefore if I measure it once it should technically stay the same unless if I evolve or measure it in some noncommuting basis? How is it that after I measure and make sure that it is $|\uparrow\rangle_1 \otimes |\downarrow\rangle_2$, it actually isn't that anymore but the one you given above? $\endgroup$
    – Kim Dong
    Commented Dec 27, 2022 at 19:13
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    $\begingroup$ @KimDong You can measure the spin of the left particle, but not of particle #1. In the state $\Psi\rangle$ in my answer, there are a superposition of two terms. In the first term, $|\uparrow\rangle_1 \otimes | \downarrow\rangle_2$, particle #1 is measured on the left with spin up. In the second term, $|\downarrow\rangle_1\otimes | \uparrow\rangle_2$, particle #2 is measured on the left with spin up. That is an eigenstate of the $S_z$ operator since there is definite spin for the particle on the left. You just can't say whether particle 1 or 2 is the one with spin up. $\endgroup$
    – Andrew
    Commented Dec 27, 2022 at 19:23
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    $\begingroup$ @Andrew Put differently, it makes no sense, a priori, to speak of "particle 1" and "particle 2". However, if you can effectively distinguish them, then this can make sense - and whether you call the particle "left particle" or "particle 1" does not make a difference then: You can, effectively, label it. See also the references I commented under the question (and the references therein). $\endgroup$ Commented Dec 27, 2022 at 19:26
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    $\begingroup$ @KimDong Ah that's interesting. I am not an expert with quantum computing, but I think it is possible that they suppress this subtlety to simplify descriptions of quantum algorithms. It is true that in some situations, you can treat electrons as if they were distinguishable and get away with it. (For example, we don't antisymmetrize the wavefunction over every electron in the Universe even though technically we should). However, that would only be an approximation that would work in some cases; in general you need to account for the Fermi-Dirac statistics to describe electron behavior. $\endgroup$
    – Andrew
    Commented Dec 27, 2022 at 19:41
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    $\begingroup$ @KimDong Consider two qubits, each made of two boxes plus one electron in either box. This is more or less the question situation: you can say "the electron in qubit 0 is in the 0 box" but not "electron 0 is in qubit 0." You encode the computational states into the physical states: e.g. $|10\rangle\mapsto\frac1{\sqrt2}(|0_11_0\rangle-|1_00_1\rangle)$ "qubit 0 is 1, qubit 1 is 0" is a superposition of "electron 0 in qubit 0 box 1, electron 1 in qubit 1 box 0" and "electron 0 in qubit 1 box 0, electron 1 in qubit 0 box 1". It's not even an "approximation", just a different choice of basis. $\endgroup$
    – HTNW
    Commented Dec 28, 2022 at 8:44
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Tracking the two electrons you're talking about would requiere knowledge of both position and momentum, which contradicts Heisenberg uncertainty principle. Since you cannot track them you don't know which electron is (spin) up $|\uparrow\rangle$ and which one is down $|\downarrow\rangle$. As a consequence, your state for the system must be a combination of both situation, i.e.:

$$ |{\rm state\ of\ 2\ electrons}\rangle = C_1|\uparrow\rangle_1\otimes |\downarrow\rangle_2 + C_2 |\downarrow\rangle_1\otimes |\uparrow\rangle_2 $$

Because of normalization and extracting a global phase (that according to QM is irrelevant) we get

$$ |{\rm state\ of\ 2\ electrons}\rangle = \frac{1}{\sqrt{2}}(|\uparrow\rangle_1\otimes |\downarrow\rangle_2 + e^{i\theta} |\downarrow\rangle_1\otimes |\uparrow\rangle_2) $$

And now, as an empirical fact, nature tells us that $e^{i\theta} = -1$ for fermions (odd half-integer spin: electrons and similar particles) and $+1$ for bosons (integer spin particles).

From the theory point of view, this empiral fact is proven in quantum field theory and it's related to commutation and anticommutation rules of the fields describing such particles.

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  • $\begingroup$ Hmm, I think I'm improving, but not quite there yet. Then what is the "qubit 0" and "qubit 1" that we encounter in quantum computing, which is often made up of electron or photon? I mean, clearly, I can go to the IBMQ or Xanadu, and prepare q0 at 1, and q1 at 0, and they say it'd be $|01\rangle$. But as you said, isn't it actually $|01\rangle-|10\rangle$ if it's made of ions or electron, and something else if it's photon? $\endgroup$
    – Kim Dong
    Commented Dec 27, 2022 at 19:34
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    $\begingroup$ My knowledge in quantum computing is not very advanced, but it is not needed to answer your question. Learn step by step, quantum computing is a very advanced and complicated subject $\endgroup$
    – Vicky
    Commented Dec 27, 2022 at 22:06
  • $\begingroup$ Can you explain how your first sentence is relevant? That's something that's also true for distinguishable particles, so it doesn't seem like it should be able to say anything about the difference in behavior between distinguishable and indistinguishable particles. In addition, the transition from equation (1), in which the $C$'s are arbitrary, to equation (2), where they have equal magnitude is not really justified by either the requirement of normalization or the irrelevance of a global phase. $\endgroup$
    – march
    Commented Dec 27, 2022 at 22:45
  • $\begingroup$ 1) Regarding the 1st sentence: it's motivation for eqs. (1) and (2) that apply for both distinguisable and undistinguisable particles. Now if they're distinguisable both states on RHS are equal and the phase is +1. Antisymmetry or symmetry is a fact coming from experiments and from QFT. $\endgroup$
    – Vicky
    Commented Dec 27, 2022 at 22:57
  • $\begingroup$ 2) Regarding going from (1) to (2). I think normalization and global phase are enough for this purpose. Normalization gives you $|C_1|^2 + |C_2|^2 = 1$ so that $C_i = (\sqrt{2})^{-1}\times e^{i\theta_i}$ is a solution. Are there maybe other solutions? Mathetically it is possible, but physically there's no reason a priori to assume that one state or another in RHS of eq(1) is more likely, so the modules of the $C$s must be equal and hence the solution above for $C_i$. Maybe, it is true that this last reasoning should be included, I let it here. Thanks for the feedback $\endgroup$
    – Vicky
    Commented Dec 27, 2022 at 23:02
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Your mistake is that you assume that you may put a number according to the positions. But actually the identical electron state would look like, \begin{equation} \frac{1}{2}\Big(|\uparrow, \text{left}\rangle_1\otimes|\downarrow, \text{right}\rangle_2-|\downarrow, \text{right}\rangle_1\otimes|\uparrow, \text{left}\rangle_2\Big) \end{equation} I.e. exchange operation exchanges all properties between two particles. Or equivalently it simply exchanges the number you associate with particles between them.

As to where this principle comes from, what prohibits the symmetric states for fermions and anti symmetric for bosons, you won't get the answer in QM. It may be understood only from the spin-statistic theorem in QFT.

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    $\begingroup$ I am very much struggling with this... Isn't the $S_z$ operator commute with $H$, therefore if I measure it once it should technically stay the same unless if I evolve or measure it in some noncommuting basis? How is it that after I measure and make sure that it is $|\uparrow\rangle_1 \otimes |\downarrow\rangle_2$, it actually isn't that anymore but the one you given above? $\endgroup$
    – Kim Dong
    Commented Dec 27, 2022 at 19:17
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    $\begingroup$ @KimDong The thing is all the observables must commute with the exchange operator. So you may have e.g. $S_{z,\text{left}}=S_{z, \text{left}1}+S_{z, \text{left}2}$. The state I wrote is its eigenstate. So you may in fact talk about "spin of the particle on the left". But if you measure it later you can't say that it was the same particle you measured earlier. In fact this question makes no sense as you don't have any observables that would correspond to the number of the particle as such observable would not commute with exchange operation $\endgroup$
    – OON
    Commented Dec 27, 2022 at 19:34
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I mean, am I tracking the electrons if I trap them on 2 sides of my table?

Not really. Your trap isn't perfect. You can't literally have an infinite square well to ensure the particle stays trapped. Hence, you can't really know if the particles tunneled and exchanged positions, for example. The only way for you to know whether the particle actually stayed in the same trap the whole time is by measuring its position continuously. If they do switch places, you won't know, because there is (by definition) no way to distinguish.

The main point is that if you stop measuring the position of a particle, you have no way of knowing whether that is still the same particle. Traps aren't perfect, so there is a slight probability your particle tunneled out of it and was replaced by a different one. There are no observational differences, but it is not the same particle anymore.

Hence, in QM, it doesn't make sense to speak of a particular particle when you have another identical particle. You are unable to distinguish them, so you can only describe the pair, or perhaps the left particle. But there exist no tags on the particles telling you which is $1$ and which is $2$. Unless you are willing to continuously measure their positions and hence affect the experiment.

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    $\begingroup$ Hmm, I think I'm improving, but not quite there yet. Then what is the "qubit 0" and "qubit 1" that we encounter in quantum computing, which is often made up of electron or photon? I mean, clearly, I can go to the IBMQ or Xanadu, and prepare q0 at 1, and q1 at 0, and they say it'd be $|01\rangle$. But as you said, isn't it actually $|01\rangle-|10\rangle$ if it's made of ions or electron, and something else if it's photon? $\endgroup$
    – Kim Dong
    Commented Dec 27, 2022 at 19:34
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    $\begingroup$ @KimDong I don't know much about quantum computing, but I guess they probably refer to the sites rather than the specific particles. For example, I can't talk about electron $1$ (I can't tag it with a $1$, after all), but there surely is an electron on the left and it has a state. The specific electron might change, but I'm concerned only with the state of the left electron. This is perfectly okay. $\endgroup$ Commented Dec 27, 2022 at 19:42
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    $\begingroup$ Notice this is okay because I'm not distinguishing the particles, but the sites. I might not be able to say whether electron $1$ is up or down because I don't know who is electron $1$, but I can ask whether the electron on my left is up or down $\endgroup$ Commented Dec 27, 2022 at 19:42
  • $\begingroup$ @NíckolasAlves Just a question: When you say this : Hence, in QM, it doesn't make sense to speak of a particular particle when you have another identical particle , How do one know if a particular particle (really) is identical to another? $\endgroup$ Commented Dec 27, 2022 at 20:20
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    $\begingroup$ @WilliamMartens they're identical if they're of the same kind: one electron is identical to any other electron, one photon is indistinguisable from antoher photon, and so on. From the POV of quantum field theory, every particle of a particular kind is created by the same field, hence identical $\endgroup$
    – Vicky
    Commented Dec 27, 2022 at 22:09
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I'm struggling with the concept of identical particles in QM. Say I have two electrons... What am I misunderstanding?

Not every solution to the Schrodinger equation is a "physical state." (Not every solution describes the natural world as it actually is.)

Consider, for example, a solution to the two-particle Schrodinger equation: $$ \Psi(\vec r_1, s_1;\vec r_2, s_2)\;. $$

If the above function is a solution, then so too is $$ \Psi(\vec r_2, s_2;\vec r_1, s_1) $$ a solution of the Schrodinger equation having the same energy, since the Hamiltonian is symmetric with respect to particle index exchange. If the states are different then this degeneracy is called "exchange degeneracy."

It is a fact of nature that the wave functions that actually describe two physical particles always obey either: $$ \Psi(\vec r_2, s_2;\vec r_1, s_1) =\Psi(\vec r_1, s_1;\vec r_2, s_2)\tag{bosons} $$ or $$ \Psi(\vec r_2, s_2;\vec r_1, s_1) =-\Psi(\vec r_1, s_1;\vec r_2, s_2)\tag{fermions}\;. $$

Therefore a state with only spin dependence such as $$ |\Psi\rangle \sim |\uparrow; \downarrow\rangle $$ is not a physical fermion state.

But a state like $$ |\Psi\rangle \sim |\uparrow; \downarrow\rangle - |\downarrow; \uparrow\rangle $$ is a physical fermion state, since exchanging the indices returns the same state (multiplied by negative one).


Here is a quote from Hans Bethe to make you feel better:

"Now it is a fact of nature, established by many observations, that all actual wave functions in physics obey $P_{ij}\Psi = \pm\Psi$, either with the plus or the minus sign. In other words, from the great multitude of mathematical solutions of $H\Psi = E\Psi$, nature has selected only the nondegenerate ones... Which symmetry applies depends on the type of identical particles involved; experiment shows that $\Psi$ is antisymmetric for electrons, protons, neutrons, mu-mesons..."

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  • $\begingroup$ What does it mean for a state to be a "physical state"? I think the proposed $|\uparrow\rangle_1 \otimes |\downarrow\rangle_2$ is very much measurable? $\endgroup$
    – Kim Dong
    Commented Dec 27, 2022 at 19:07
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    $\begingroup$ It means a state that actually describes the world. This is basic multi-particle quantum mechanics. Every state is either completely symmetric (bosons) or completely anti-symmetric. $\endgroup$
    – hft
    Commented Dec 27, 2022 at 19:10
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    $\begingroup$ But I can measure 1 particle to be up, and the other to be down, and it's also an experimental fact right? Then isn't in that moment, I would have $|\uparrow\rangle_1 \otimes |\downarrow\rangle_2$? Don't they do this in quantum computing often, with a particle up and one is down? $\endgroup$
    – Kim Dong
    Commented Dec 27, 2022 at 19:20
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Exchange symmetry doesn't apply to measurable properties like position and spin. It only applies to unphysical labels.

Your wave function is

$$\left|\uparrow\right\rangle_1 \otimes \left|\downarrow\right\rangle_2$$

Each particle in this wave function has three properties: a spin ($\uparrow$ or $\downarrow$), a subscript ($1$ or $2$), and a position in the tensor product (left or right). But your particles have only two physical properties, spin and position. Obviously spin is spin, but it's not clear what you mean by the other two properties in the wave function. The answer to your question depends on what they mean.

Other answers have assumed that they are both unphysical labels, redundantly encoding the same which-particle information. If they are, then the correct antisymmetrized wave function would be $\left|\uparrow\right\rangle_1 \otimes \left|\downarrow\right\rangle_2 - \left|\downarrow\right\rangle_1 \otimes \left|\uparrow\right\rangle_2$ (up to a scalar factor), where I obtained the second term from the first by reversing the product and swapping $1$ and $2$. But this doesn't really make sense, because physical position, which is crucial to your thought-experiment, is not represented in your wave function at all, so you can't model the swapping of the particles.

If only the tensor-product order is meaningless, and the subscripts represent physical position, then the antisymmetrized wave function would be $\left|\uparrow\right\rangle_1 \otimes \left|\downarrow\right\rangle_2 - \left|\downarrow\right\rangle_2 \otimes \left|\uparrow\right\rangle_1$, where I obtained the second term from the first by reversing the product only. If you physically swap the particles, then the wave function becomes $\left|\uparrow\right\rangle_2 \otimes \left|\downarrow\right\rangle_1 - \left|\downarrow\right\rangle_1 \otimes \left|\uparrow\right\rangle_2$, which I obtained from the previous wave function by swapping $1$ and $2$.

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You seem to be talking about two particles in the same quantum well or in the same atom. If you had two particles trapped in two adjacent quantum wells, they wouldn't necessarily be bound by the Pauli exclusion principle to have opposite up and down spins. I don't know if quantum entanglement changes that situation.

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