# A short note on the anti-symmetrisation of electron spin states

$$\newcommand{\lt}{\left} \newcommand{\me}{\middle} \newcommand{\rt}{\right} \newcommand{\la}{\langle} \newcommand{\ra}{\rangle} \newcommand{\ua}{|\hspace{-3 pt}\uparrow\ra} \newcommand{\da}{|\hspace{-3 pt}\downarrow\ra} \newcommand{\uabra}{\la\uparrow\hspace{-3 pt}|} \newcommand{\dabra}{\la\downarrow\hspace{-3 pt}|}$$

When discussing EPRB experiments, often spin states of electrons are used. Suppose we have two spatially separated electrons. The paradigmatic state that is used is the spin-singlet state

$$|\psi_1\ra := \sqrt{\frac12}(\ua_A \da_B – \da_A \ua_B).$$

Then, when one of the spins is measured, one usually talks about, for example, ‘measuring the spin of particle A’. This suggests that A refers to a particle localised at one location (say on the left), and B to a particle localised at another location (say on the right). We can also consider the case where one electron is in a spin-up eigenstate and the other in a spin-down eigenstate:

$$|\psi_2\ra := \ua_A \da_B.$$

There is some confusion in the literature concerning the relation between writing down states in this way and electrons being fermions. Because of the symmetrisation postulate, fermions always have to be in anti-symmetric states. Sometimes it is therefore suggested that a state like $$|\psi_2\ra$$ is strictly invalid, because it is not anti-symmetric. Also, it is sometimes said that $$|\psi_1\ra$$ is a valid state, because it is anti-symmetric. Also, sometimes it is said that in order to write down the correct, anti-symmetric state, the spatial wavefunction has to be included, so that, for example, $$|\psi_1\ra$$ becomes

$$|\psi_3\ra := \sqrt{\frac12}(\ua_A \da_B – \da_A \ua_B) \otimes \sqrt{\frac12}(|L\ra_{A} |R\ra_{B} + |R\ra_{A} |L\ra_{B}),$$

Here, $$|L\ra$$ and $$|R\ra$$ are two (nearly) orthogonal spatial wavefunctions: $$|L\ra$$ with some narrow Gaussian peak on the left, and $$|R\ra$$ with a similar peak on the right. This is, for example, mentioned by Dieks (2016). Dieks also claims that this means that the spins are not ‘localised’ at the locations of the particles, since the spatial- and spin-part of the wavefunction factorise. This would then suggest, Dieks continues, that a single spin measurement is not the measurement of a local property, namely the spin of the localized electron, but of some non-localised spin property of both particles.

In this short note we want to clear up this situation (hopefully, it will get rid of more confusion than it creates). First, we will show that there is nothing wrong with writing down states like $$|\psi_2\ra$$. Mathematically this is perfectly fine, and this is a correct state for fermions. The fact that the anti-symmetry of this state is not visible has to do with the fact that the subsystem decomposition that is used in this notation is a decomposition in ‘electron located left’ and ‘electron located right’. This state can be translated into a state including the spatial part, wherein the asymmetry of the state is clearly visible. Second, we will show that $$|\psi_3\ra$$ is not the correct state when adding the spatial part to $$|\psi_1\ra$$. This is because in $$|\psi_1\ra$$, the subsystems are ‘electron on the left’ and ‘electron on the right’, while in $$|\psi_3\ra$$, the subsystems are ‘electron A’ and ‘electron B’, with both particles having a component in the wavefunction being on the left, and a component being on the right. In fact, it will turn out that adding the spatial part to $$|\psi_1\ra$$ will lead to a state having the same form as $$|\psi_3\ra$$, but the fact that the spin- and spatial parts factorise is coincidental, and we will show that for slightly different states, which also violate the Bell inequalities, no such factorisation occurs.

First, we concentrate on state $$|\psi_2\ra$$. This state means to represent the left electron being in a definite spin-up state, and the right electron being in a definite spin-down state. To emphasise this, we replace the subsystem labels by ‘L’ and ‘R’:

$$|\psi_4\ra := \ua_L \da_R.$$

Translating this state into a state including the spatial wavefunction, we get

$$\sqrt{\frac12}(\ua_1 |L\ra_1 \da_2 |R\ra_2 – \da_1 |R\ra_1 \ua_2 |L\ra_2),$$

where we introduced subscripts 1 and 2 to indicate that these are not localised subsystems. Note that this state is explicitly anti-symmetric. We can do the same thing with three other (orthogonal) states:

\begin{align} \ua_L \ua_R &\mapsto \sqrt{\frac12}(\ua_1 |L\ra_1 \da_2 |R\ra_2 – \da_1 |R\ra_1 \ua_2 |L\ra_2)\\ \da_L \da_R &\mapsto \sqrt{\frac12}(\da_1 |L\ra_1 \da_2 |R\ra_2 – \da_1 |R\ra_1 \da_2 |L\ra_2)\\ \da_L \ua_R &\mapsto \sqrt{\frac12}(\da_1 |L\ra_1 \ua_2 |R\ra_2 – \ua_1 |R\ra_1 \da_2 |L\ra_2) \end{align}

Note that the space spanned by these 4 vectors is a well-defined Hilbert space. Now, we can write down states using the subsystem decomposition ‘particle 1’ and ‘particle 2’, including the spatial parts, as on the right-hand sides of the above formulae. But we might just as well decompose this 4-dimensional Hilbert space into two 2-dimensional subsystems L and R (as on the left-hand sides). So, there is nothing wrong with writing states like $$|\psi_2\ra$$, as long as we keep in mind that the subsystems refer to ‘electron on the left’ and ‘electron on the right’. Of course, this only works if the particles are spatially separated, so that the corresponding wave functions are (nearly) orthogonal.

Now that we know how to translate the four basis states to states including spatial parts, we can also look how state $$|\psi_1\ra$$ looks when translated to a wavefunction including spatial parts. In fact, it becomes

\begin{align} |\psi_5\ra &= \sqrt{\frac14} \lt( \lt( \ua_1 |L\ra_1 \da_2 |R\ra_2 – \da_1 |L\ra_1 \ua_2 |R\ra_2 \rt) – \lt( \da_1 |R\ra_1 \ua_2 |L\ra_2 – \ua_1 |R\ra_1 \da_2 |L\ra_2 \rt) \rt)\\ &= \sqrt{\frac12} \lt( \ua_1 \da_2 – \da_1 \ua_2 \rt) \otimes \sqrt{\frac12} \lt( |L\ra_1 |R\ra_2 + |R\ra_1 |L\ra_2 \rt) \end{align}

Note that the spatial- and spin part factorise, and that the spin part has exactly the same form as the original state $$|\psi_1\ra$$. However, there is a crucial difference: in $$|\psi_1\ra$$ the subsystems refer to ‘left electron’ and ‘right electron’, while in $$|\psi_5\ra$$ the subsystems refer to ‘electron 1’ and ‘electron 2’ (both particles being neither left nor right). Therefore, the subsystem labels need to be different in both states, and therefore $$|\psi_3\ra$$ is not the state one gets by adding the spatial part to $$|\psi_1\ra$$.

Also, the factorisation of the state $$|\psi_5\ra$$ and the spin part having exactly the same form as in $$|\psi_1\ra$$ is rather coincidental. Consider a state with slightly different coefficients (still close enough to a maximally entangled state so that a Bell inequality can be violated with it):

$$|\psi_6\ra := \sqrt{\frac{49}{100}}\ua_1 \da_2 – \sqrt{\frac{51}{100}}\da_1 \ua_2.$$

Adding the spatial part, we get

\begin{align} \sqrt{\frac12} \lt( \lt( \sqrt{\frac{49}{100}} \ua_1 |L\ra_1 \da_2 |R\ra_2 – \sqrt{\frac{51}{100}} \da_1 |L\ra_1 \ua_2 |R\ra_2 \rt) – \lt( \sqrt{\frac{49}{100}} \da_1 |R\ra_1 \ua_2 |L\ra_2 – \sqrt{\frac{51}{100}}\ua_1 |R\ra_1 \da_2 |L\ra_2 \rt) \rt). \end{align}

Note that now, the state does not factorise in a spin- and spatial part (and accordingly, the spin part has not the same form as $$|\psi_6\ra$$).

Note that, when working with the decomposition in ‘electron 1’ and ‘electron 2’, the measurement operators change accordingly. For example, measuring the $$z$$-spin of the left particle, one has to observable

$$\sigma^L_z \otimes \mathbb{I}^R$$

in the left-right decomposition. In the 1-2 decomposition, this becomes

$$\frac{\hbar}{2} (\frac12(\ua_1 |L\ra_1 \da_2 |R\ra_2 – \da_1 |R\ra_1 \ua_2 |L\ra_2) (\uabra_1 \la L|_1 \dabra_2 \la R|_2 – \dabra_1 \la R|_1 \uabra_2 \la L|_2)\\ + \frac12(\ua_1 |L\ra_1 \ua_2 |R\ra_2 – \ua_1 |R\ra_1 \ua_2 |L\ra_2) (\uabra_1 \la L|_1 \uabra_2 \la R|_2 – \uabra_1 \la R|_1 \uabra_2 \la L|_2))\\ – \frac{\hbar}{2} (\frac12(\da_1 |L\ra_1 \da_2 |R\ra_2 – \da_1 |R\ra_1 \da_2 |L\ra_2) (\dabra_1 \la L|_1 \dabra_2 \la R|_2 – \dabra_1 \la R|_1 \dabra_2 \la L|_2)\\ + \frac12(\da_1 |L\ra_1 \ua_2 |R\ra_2 – \ua_1 |R\ra_1 \da_2 |L\ra_2) (\dabra_1 \la L|_1 \uabra_2 \la R|_2 – \uabra_1 \la R|_1 \dabra_2 \la L|_2))$$

One now understands why it is much more convenient to decompose the two-electron system using their spatial location, and not using particle labels!

In fact, if one wants to be always explicit about anti-symmetrisation, one would always have to write down a state for every electron in the universe and anti-symmetrise it! Consider a universe consisting of four electrons. When decomposing into subsystems ‘electron 1, 2, 3, 4’, because of the anti-symmetrisation this state does not factorise into, for example, a pure state for electrons 1+2 and a pure state for electrons 3+4. Therefore, we see that also when writing down states like $$|\psi_4\ra$$ we have already used the fact that the two electrons under consideration belong to some localised subsystem of the universe. One can of course, using a similar procedure as described above, translate the states into states of the four-electron universe, where the antisymmetry is clearly visible. However, this would be unnecessarily complicated, and completely impossible for the actual universe, which, according to the latest estimates, contains about $$10^{80}$$ electrons.

References

Dieks, Dennis (2016) Quantum Information and Locality. [Preprint] URL: http://philsci-archive.pitt.edu/id/eprint/12313 (accessed 2017-05-09)