# Are combined fermion wavefunctions still antisymmetric after wavefunction collapse?

If we have two electrons in a state $$|\psi\rangle=\frac{1}{\sqrt2}[|\uparrow\downarrow\rangle+|\downarrow\uparrow\rangle]$$ and we measure the spin of the first electron to be up, does the wavefunction collapse into the state $$|\psi\rangle=|\uparrow\downarrow\rangle$$? If so, how is this consistent with the notion that fermion wavefunctions must be anti-symmetric?

I suppose $$|A,B\rangle = |A\rangle |B\rangle$$ is a "2 non-identical particle state" in your notation, so that your first spin state looks symmetric. This is impossible for identical electrons, unless there is a "spatial part" of the wave function that, in this case, must be antisymmetric (the spatial wavefunction must be antisymmetric if the two electrons are in a spin triplet state but symmetric if they are in a spin singlet state).

How identical particles work: you cannot have a state as $$|A\rangle |B\rangle$$, where the first particle is in state $$A$$, but you have to sum more information, so to hide the possibility to know which is the "first particle". The obvious way is to go for
$$|A,B\rangle_{a} \propto |A\rangle |B\rangle - |B\rangle |A\rangle$$ or $$|A,B\rangle_{s} \propto |A\rangle |B\rangle + |B\rangle |A\rangle$$ The states labelled by $$a$$ or $$s$$ are the physical states of identical particles. There is no notion as "the first particle is in $$A$$", so this is not something you can measure. Hence, no collapse to $$|A\rangle |B\rangle$$ or $$|B\rangle |A\rangle$$, as no device can distinguish identical particles in the first place.

• Hmm so you're saying its impossible to measure one of those states that $|\psi>$ is composed of? My exam paper is explicitly saying that you can and asking why it isn't a problem... Oct 8 '20 at 14:24
• Are particles considered identical in your exercise? If particles are NOT considered to be identical (even though real electrons are), then you can have the collapse (because you can distinguish the "first particle"), but the collapsed wave function is not anti-symmetric (which is not a problem IF the particles are non-identical). Conversely, if particles in the exercise are considered identical, then you cannot have the collapse because you cannot even make the measure: the whole statement makes little sense. Oct 8 '20 at 14:54
• But this is obvious: if your electrons are treated as identical particles (as it should be), it makes no sense to say "we measure the spin of the first electron to be up", as in your question. You can only say "we measure one spin to be up". Oct 8 '20 at 14:57

It depends on what the notation $$\left|\uparrow\downarrow\right\rangle$$ means. When talking about Bell states, the left and right positions in the ket normally represent different spatial locations. In this case it doesn't matter whether the particles are of the same type since they can be distinguished by location. Even if both particles are electrons, the states $$(\left|\uparrow\downarrow\right\rangle+\left|\downarrow\uparrow\right\rangle)/\sqrt2$$ and $$\left|\uparrow\downarrow\right\rangle$$ are fine. If you wrote them out in gory detail they would be something like $$(\left|\uparrow_A\downarrow_B\right\rangle-\left|\downarrow_B\uparrow_A\right\rangle+\left|\downarrow_A\uparrow_B\right\rangle-\left|\uparrow_B\downarrow_A\right\rangle)/2$$ $$(\left|\uparrow_A\downarrow_B\right\rangle-\left|\downarrow_B\uparrow_A\right\rangle)/\sqrt2$$

where $$\uparrow_A$$ means an electron with position $$A$$ and spin $$\uparrow$$, and left and right placement in the ket is now an arbitrary label instead of an indicator of spatial position.

I think the answer boils down to this - for electrons that can either exist in single particle spatial wavefunctions $$\phi_A(r)$$ and $$\phi_B(r)$$, and single particle spin states $$|\uparrow\rangle$$ or $$|\downarrow \rangle$$. The 'final measured' state (where the maximum information is measured and known) will be something like 'you measure an electron in box A with spin up and another electron in box B with spin down'.

As @Quillo said, this 'final measured state' has the form (For a set of parameters $$x_1$$ and $$x_2$$ that include both the positions ($$r_1$$ and $$r_2$$) and spins of electrons 1 and 2):

$$\psi(x_1,x_2) = \frac{1}{\sqrt2}[\phi_A(r_1)\phi_B(r_2)|\uparrow_1\rangle|\downarrow_2\rangle - \phi_B(r_1)\phi_A(r_2)|\downarrow_1\rangle|\uparrow_2\rangle]$$,

not a form like this:

$$\psi(x_1,x_2) =\phi_A(r_1)\phi_B(r_2)|\uparrow_1\rangle|\downarrow_2\rangle$$

i.e. I was assuming a 'final measured state' must always have no superposition remaining as 'we have measured all we can' - but this is simply not true for 2 reasons:

1. As I said in my question, this is not an antisymmetric multi-particle wavefunction as electrons (as identical fermions) need to have due to the fact that for identical particles (labelled at any time before measurement) $$P(x_1,x_2)$$ should equal $$P(x_2,x_1)$$. Indeed it is bizarre that in $$\psi(x_1,x_2) =\phi_A(r_1)\phi_B(r_2)|\uparrow_1\rangle|\downarrow_2\rangle$$, the probability of finding 'electron 1' at $$\vec r = (1,0,0)$$ is different to the probaility of finding 'electron 2' at $$\vec r = (1,0,0)$$ seeing as they just differ by an arbitrary label.

2. There is no possible measurable difference between $$\psi(x_1,x_2) =\phi_A(r_1)\phi_B(r_2)|\uparrow_1\downarrow_2\rangle$$ and $$\psi(x_1,x_2) =\phi_B(r_1)\phi_A(r_2)|\downarrow_1\uparrow_2\rangle$$, since both have 'the electron in box A with spin up and the electron in box B with spin down'. The only way you could have a measurable parameter with which to label the particles as 'electron 1' and 'electron 2' is if they were intrinsically different particles, but this is not the case (by definition since we are dealing with an identical particles scenario) so it is unsurprising that the final result will have to be a combination of both of these - as opposed to situations where we could say the proton has spin up in box A and the electron as spin down in box B (in which case we would have the state $$\psi(x_p,x_e)=\phi_A(x_p)\phi_B(x_e)|\uparrow_p\downarrow_e\rangle$$ for these distinguishable particles).

(It turns out these were the 'gory details' @benrg mentioned but I was thrown at the time by his notation.)

My final question then was - if there are no measurable properties by which 'electron 1' and 'electron 2' differ, then why do we use their parameters (e.g. $$r_1$$, $$r_2$$) in the wavefunction - why can't we label 'electron 1 as the electron in box A' and 'electron 2 as the electron in box B'?

It was helpful to realise that the way the combined particle wavefunction works is that we have a tensor product of 2 separate Hilbert spaces - one for 'electron 1' and another for 'electron 2'. This is what defines 'electron 1' and 'electron 2'. Therefore 'electron 1' can just exist in the 'box A' spatial wavefunction, but since this is not an intrinsic property of the electron, it makes no more sense to use try and label 'electron 1' as the 'electron in box A' as 'the electron moving at 5m/s' since this is a variable property - whereas particle type (etc) is an intrinsic property.

Indeed even if we decided to 'relabel' $$r_1$$ and $$r_2$$ as $$r_A$$ and $$r_B$$ after measurement, this would just be semantics and make no real difference since as soon as the electrons are no longer isolated and are allowed to mix with electrons from the rest of the universe (e.g. even if we measure the position of the electrons immediately again with a measurement apparatus composed of some electrons) - since their position is a variable property which can evolve in time, a superposition state will once again be created as to which electron you would measure where (e.g. between one of these electrons and an electron in the measurement apparatus) so ultimately this relabelling would have made no difference except for a semantically different way of writing the wavefunction existing only at the exact time of measurement.

Therefore eventhough there is a difference in the maths between $$\phi_A(r_1)$$ and $$\phi_A(r_2)$$ - the former has particle with Hilbert space 1's spatial state being $$\phi_A$$, the latter has particle with Hilbert space 2's spatial state being $$\phi_A$$, physically no matter when we label the electrons before measurement we can only physically measure them in their superposition state of labels (due to the inevitable mixing before and during measurement). It is also worth noting that much of the 'ugliness' of this form comes from this tensor product of 'electron 1' and 'electron 2' mathematical description in the 'simple' framework of non-relativistic QM. If we use a QFT mathematical description involving 'occupation numbers of electrons in certain states', much of this ugliness disappears.

Ultimately since the non-relativistic QM mathematical framework for describing multiple identical particles (using tensor products for arbitrarily decomposed 'electron 1' and 'electron 2'), we must make up for this arbitrariness by requiring all state to be an (antisymmetric/symmetric) superposition of our arbitrary labels. Nothing changes about this fact even after measurement so we still have a superposition state after 'all the possible measurement can be done'.