# Symmetrisation of (fermionic) two-particle system without vs. with spin in wave function

I'm using D. J. Griffiths's textbook Introduction to Quantum Mechanics (3rd ed.) for my introductory university course on the subject. In chapter 5 (starting at section 5.1.1), he discusses the behaviour of identical particles.

For a start, he introduces an elementary spatial wave function for a system of two non-interacting particles where one of them is in state $$\psi_a$$ and the other is in state $$\psi_b$$:

$$\psi(\mathbf{r_1},\mathbf{r_2})=\psi_a(\mathbf{r_1})\psi_b(\mathbf{r_2})$$

He shortly after introduces how identical particles cannot be told apart, so, because "one of them" and "the other" is physically ambiguous, we write the spatial wave function of such a two-particle system as a superposition:

$$\psi_{\pm}(\mathbf{r_1},\mathbf{r_2})=A\,(\psi(\mathbf{r_1},\mathbf{r_2})\pm\psi(\mathbf{r_2},\mathbf{r_1}))$$

He posits that $$\psi_+$$ is the governing wave function for bosons, and $$\psi_-$$ for fermions $$-$$ which make for, respectively, a symmetric spatial wave function $$\psi_+(\mathbf{r_1},\mathbf{r_2})=\psi_+(\mathbf{r_2},\mathbf{r_1})$$, and an antisymmetric spatial wave function $$\psi_-(\mathbf{r_1},\mathbf{r_2})=-\psi_-(\mathbf{r_2},\mathbf{r_1})$$. By this, it makes sense that when $$\psi_a=\psi_b$$, fermionic systems have no sensical spatial wave function (Pauli's exclusion principle).

Now, as Griffiths likes to do to simplify explanations, he leaves spin out of wave functions. One paragraph later, he shows how fermions are expected to be further away from each other than distinguishable particles, and the converse for bosons ("exchange interaction"): this uses only integrals over space, so I assume it's fine to generalise the result to wave functions including spin. If I've interpreted his text later on in the chapter correctly, we can conclude such behaviour merely based on spatial wave functions, so I'll call particles that repel like fermions, and equivalently have can be given a combined spatial wavefunction $$\psi_-$$, "spatially fermionic".

Here's the problem. He adds spin into the discussion of two-electron systems as a spinor factor $$\chi(1,2)$$, and asserts:

It is the whole [$$\psi(\mathbf{r_1},\mathbf{r_2})\chi(1,2)$$], not just the spatial part, that has to be antisymmetric with respect to exchange. (...) Thus the Pauli principle allows two electrons in a given position state, as long as their spins are in the singlet configuration.

This statement confuses me.

• For one: does "not just" imply that fermions still need to be spatially fermionic, as was asserted when spin was not yet included in the discussion, or that only $$\psi(\mathbf{r_1},\mathbf{r_2})\chi(1,2)$$ need be antisymmetric?

• Secondly: is "$$\psi(\mathbf{r_1},\mathbf{r_2})$$" an elementary function $$\psi(\mathbf{r_1},\mathbf{r_2})=\psi_a(\mathbf{r_1})\psi_b(\mathbf{r_2})$$, or is it an artificially (anti)symmetrised wavefunction like $$\psi_+(\mathbf{r_1},\mathbf{r_2})$$ and $$\psi_-(\mathbf{r_1},\mathbf{r_2})$$? If it's the former, that would mean that the spatial factor $$\psi(\mathbf{r_1},\mathbf{r_2})$$ in the combined wave function for our two-fermion system $$\psi(\mathbf{r_1},\mathbf{r_2})\chi(1,2)$$ cannot whatsoever be treated equally to the artificially (anti)symmetrised $$\psi_\pm(\mathbf{r_1},\mathbf{r_2})$$. So, if we can't, and if we assume the answer to question 1 is that the system has to be spatial fermionic as well, then how will we (or nature) ever ensure that $$\psi$$ is properly (anti)symmetrised?

• Thirdly: since $$\psi(\mathbf{r_1},\mathbf{r_2})\chi(1,2)$$ must just be antisymmetric, why can't we take the triplet configuration of the two electrons (which gives a symmetric $$\chi(1,2)$$), and have an antisymmetric spatial wavefunction $$\psi(\mathbf{r_1},\mathbf{r_2})$$? (This thread tries to answer, but I don't think it gives proper closure.)

Note to future readers regarding the third question:

After some discussion in the comments of the accepted answer, and repeatedly having studied the above quote in the context of the chapter again, I came to the correct interpretation of what exactly Griffiths tried to exclude when writing "the Pauli principle allows two electrons in a given position state, as long as their spins are in the singlet configuration".

His claim can be phrased as follows:

If $$\Psi=\psi(\mathbf{r_1},\mathbf{r_2})\chi(1,2)$$, then there exists no mathematical function $$\psi(\mathbf{r_1},\mathbf{r_2})$$ that is antisymmetric w.r.t. interchange of $$\mathbf{r_1}$$ and $$\mathbf{r_2}$$ and uses only one state $$\psi_a$$ instead of a $$\psi_a$$ and a $$\psi_b$$ (if you will, $$\psi_a = \psi_b$$).

In the accepted answer by ZeroTheHero, you'll find the explanation as to why this is true $$-$$ the essence is that antisymmetrisation happens through determinants in permutation group theory, and that those become 0 when any $$\psi_a = \psi_b$$.

The main consequence is, in the end, as stated at first: two identical fermions, e.g. electrons, can't occupy the same $$\psi_a = \psi_b$$ unless and only unless being in an antisymmetric, i.e. singlet, spin configuration, exactly because there exists no separable antisymmetric spatial wave function that would allow for a symmetric, i.e. triplet, spin configuration.

Additionally, after walking through the chapter once more with this claim in mind, it became apparent that my concept of "spatial fermionicity" is indeed a separate property two particles can have. In the accepted answer, it is established that two fermions (e.g. electrons) needn't be spatially fermionic for them to be fermions. However, the system can still have said property, or even its exact opposite: in paragraph 5.2.1 on excited helium states, it is discussed that in parahelium, the electrons are specifically "spatially bosonic" (their expected separation is smaller than for distinguishable particles), making them interact at a closer range on average, measurable in the higher energy for such states.

The total wavefunction needs to be antisymmetric. Thus you can have:

1. symmetric in space, antisymmetric in spin; for example \begin{align} \sim \left(\psi_a(x_1)\psi_b(x_2)+\psi_a(x_2)\psi_b(x_1)\right) \left(\vert +\rangle_1\vert -\rangle_2 -\vert -\rangle_1\vert +\rangle_2\right) \end{align}

2. Antisymmetric in space, but symmetric in spin; for example \begin{align} \sim \left(\psi_a(x_1)\psi_b(x_2)-\psi_a(x_2)\psi_b(x_1)\right) \left(\vert +\rangle_1\vert -\rangle_2 +\vert -\rangle_1\vert +\rangle_2\right) \tag{1} \end{align}

There are merely examples. For instance \begin{align} \psi_a(x_1)\psi_a(x_2) \left(\vert +\rangle_1\vert -\rangle_2 -\vert -\rangle_1\vert +\rangle_2\right) \end{align} or \begin{align} \left(\psi_a(x_1)\psi_b(x_2)-\psi_a(x_2)\psi_b(x_1)\right) \vert +\rangle_1\vert +\rangle_2 \end{align} are also totally antisymmetric. Note that, in this last example, the spin state $$\vert +\rangle_1\vert +\rangle_2$$ is one state of the triplet, and clearly symmetric. The spin part of (1) is another component of the triplet, and the spin state $$\vert -\rangle_1\vert-\rangle_2$$ is the last component. Thus all members of the triplet state are symmetric under permutation, implying in this case that spatial part must be antisymmetric.

In connection to a comment:

To get a fully antisymmetric wave-function for $$n$$ particles, one needs at least $$n$$ distinct functions. The reason for this is root in the theory of the permutation group; at the practical level, these antisymmetric wavefunctions are constructed as determinants since - in the parlance of group theory - this function carries the fully antisymmetric representation of the permutation group. In the $$3$$-particle case, we would have \begin{align} \psi(x_1,x_2,x_3) = \left\vert \begin{array}{ccc} f_a(x_1)&f_a(x_2)&f_a(x_3)\\ f_b(x_1)&f_b(x_2)&f_b(x_3)\\ f_c(x_1)&f_c(x_2)&f_c(x_3) \end{array} \right\vert\, . \end{align} By elementary properties of the determinants, interchanging two columns - this amounts to permuting $$x_i\leftrightarrow x_j$$ introduces a minus sign thus guaranteeing antisymmetry. If two functions are the same - say $$f_b=f_a$$ - then two rows are identical and the determinant is automatically $$0$$.

To get a fully symmetric function, one must use the permanent, which basically is computed as the determinant but with positive signs everywhere. One can construct such permanents using any number of functions.

There are also functions of mixed symmetry (broadly related to immanants), useful when combining spin and spatial degrees of freedom so the result has definite symmetry. One must then construct these using the tools from the symmetric group, such as Young symmetrizers.

How to combine these partially symmetric functions is explained in textbooks with chapters on the symmetric group.

Note that partially symmetric states appear only for $$3$$ or more particles, basically because the permutation group $$S_2$$ only has $$1$$-dimensional irreducible representations, whereas $$S_n$$ for $$n\ge 3$$ has irreps of dimension greater than $$1$$.

Finally, note that the partially symmetric functions constructed in this way are not the same as the Laughlin wavefunctions used in anyonic theories.

• For future reference, I'll summarise: 1. the system does not necessarily have to be "spatially fermionic" (see first example), but it must be totally fermionic; 2. as per the examples, $\psi$ can be arbitrary (both a superposition or a single product are allowed), as long as the total is antisymmetrised; 3. yes, the triplet state is possible. If you allow me to follow up: 1. Do the exchange forces (only proved in the book without spin) still hold when the system is not "spatially fermionic"? 2. Why does Griffiths seem to dismiss the triplet state (surely, he means something)?
– Mew
May 17, 2020 at 0:27
• @Mew I don’t have that book so I cannot say. May 17, 2020 at 1:01
• Okay, to rephrase without referring to the book (it is not enlightening anyway): 1. Given what we've established, are electrons still always expected to be further apart than distinguisable particles? I can only prove this for $\psi_-$ if I pretend like spin doesn't exist in wave functions. 2.Is there something "spooky" going on with the triplet configuration, that would make anyone cautious to mention that electrons could be in such a state?
– Mew
May 17, 2020 at 1:08
• @Mew that would depend on the state and the Hamiltonian. Plus, in what sense are you saying “further apart”? Average sense? If so, average of what? May 17, 2020 at 1:21
• @Mew This is correct. It is not possible to construct an $n$ particle antisymmetric spatial (or spin) wavefunction with fewer than $n$ distinct functions $\psi_a,\psi_b\ldots \psi_n$. Jun 6, 2020 at 14:16

When discussing this area of physics, keep in mind that it is the labels on the identical particles that are swapped during an exchange operation. Keep this distinct from the notion of the location of a particle, for example.

For Fermions, it is the overall state, including both spatial and spin parts, which has to change sign when any given pair of labels are swapped.

The overall state sometimes can be written as a product of (spatial part) and (spin part) but this does not always happen. Let's deal with that case first, though, since it is the simplest. Suppose we have a case involving spatial states $$A$$ and $$B$$ for a pair of electrons. We assign the labels $$1$$ and $$2$$ to the electrons. Then one can have any or all of $$\frac{1}{2}\left( A_1 B_2 + A_2 B_1 \right) ( \uparrow_1 \downarrow_2 - \uparrow_2 \downarrow_1 ),$$ $$\frac{1}{2}\left( A_1 B_2 - A_2 B_1 \right) ( \uparrow_1 \downarrow_2 + \uparrow_2 \downarrow_1 ),$$ $$\frac{1}{\sqrt{2}}\left( A_1 B_2 - A_2 B_1 \right) \uparrow_1 \uparrow_2 ,$$ $$\frac{1}{\sqrt{2}}\left( A_1 B_2 - A_2 B_1 \right) \downarrow_1 \downarrow_2,$$ and also $$\frac{1}{\sqrt{2}} A_1 A_2 ( \uparrow_1 \downarrow_2 - \uparrow_2 \downarrow_1 ),$$ $$\frac{1}{\sqrt{2}} B_1 B_2 ( \uparrow_1 \downarrow_2 - \uparrow_2 \downarrow_1 ).$$

All the above are cases where the spatial and spin parts can be written separately. But there are also further possibilities, such as: $$\frac{1}{\sqrt{2}}( A_1 B_2 \uparrow_1 \downarrow_2 - A_2 B_1 \uparrow_2 \downarrow_1 ) .$$ Introductory treatments often fail to mention this case. I have just given an example; there are many others. To write your own, just write any state without paying attention to exchange symmetry, then put a minus sign, and then write the state again but with the labels swapped. Finally, check whether in fact you then have zero because everything cancelled, and then if it is not zero then check how it has to be normalized.

In the above I adopted a perfectly logical notation, but if you prefer to write something like $$\psi_A({\bf x}_1)$$ and $$\psi_B({\bf x}_1)$$ instead of $$A_1$$ and $$B_1$$ then that is completely ok too. Finally, multiplication (strictly speaking, tensor product) of wavefunctions or state-vector is commutative, so for example one has $$\frac{1}{\sqrt{2}}( A_1 B_2 \uparrow_1 \downarrow_2 - A_2 B_1 \uparrow_2 \downarrow_1 ) \equiv \frac{1}{\sqrt{2}}( A_1 B_2 \uparrow_1 \downarrow_2 - B_1 A_2 \downarrow_1 \uparrow_2 )$$ The first version draws attention to the fact that it is the labels $$1$$ and $$2$$ that are swapped, not the states $$A$$ and $$B$$. But the second version is usually easier for a human to read. Notice that in this kind of state (called entangled, as opposed to the earlier examples which are product states) it is possible to say 'the particle in state $$A$$ has its spin up' without needing to say whether one is referring to particle $$1$$ or $$2$$.

• What is the benefit of writing inseparable wavefunctions, or separable wavefunctions? In most materials that I read, they always take linear combinations of inseparable wavefunctions, to form separable wavefunctions like the ones above. Is there a particular reason for doing that? Sep 3, 2021 at 10:31
• Is the product state wavefunction more fundamental or 'general' than the single-slater determinant wavefunctions which often turn out to be inseparable. You've written that these single slater determinant wavefunctions are called entangled states, as the position and spins have been entangled together. If this is perfectly consistent, then why do most material forcefully convert these into product states, by taking a linear combination ? Sep 3, 2021 at 10:35
• When we talk about Entanglement, we talk about two particles whose total wavefunction can't be separated into the wavefunctions for each particle. But here, it is the spatial wavefunction, that is inseparable from the spin wavefunction. How can this be called entangled state ? Here the spatial part is inseparable from the spin part. Normally we are taught that it is the two particles whose wavefunctions must be inseparable. Can you elaborate on this ? Sep 3, 2021 at 10:49
• @NakshatraGangopadhay The product states give a complete set of states. The symmetrized states (i.e. symmetric and antisymmetric) give a complete set of states. One expands a given state in whichever basis is convenient. For example when you know already you have indistinguishable fermions then you know you will not need the symmetric states so you may as well used a basis of antisymmetric ones. But then for doing integrals it may be useful to write them as sums of product states. ... Sep 3, 2021 at 11:03
• @NakshatraGangopadhay Regarding "entanglement" yes I agree there are various definitions of differing degrees of strictness. For example one could bring in the consideration of whether the degrees of freedom under consideration could in principle be spacelike separated. A particle and its spin cannot be spacelike separated, and in this sense the term "entangled" might be questioned. But "entangled" is also widely used as a synonym for "not expressible as a product" and in this sense you can have entanglement between location and spin of a particle. Sep 3, 2021 at 11:05

One: Only $$\psi(r_1,r_2)\chi(1,2)$$ need be antisymmetric.

Second: Because $$\psi(r_1,r_2)\chi(1,2)$$ need be antisymmetric, if $$\chi$$ is symmetric ($$\chi_+$$), $$\psi(r_1,r_2)$$ is the antisymmetrized wavefunction $$\psi(r_1,r_2)$$, and if $$chi$$ is antisymmetric ($$\chi_-$$), it is the symmetrized wavefunction $$\psi_+(r_1,r_2)$$. A general wavefunction will be a linear combination of both kind of things

Third: Absolutelly yes.

Unrelated, but there is a beautiful theorem on relativistic quantum mechanics that is the spin-statistics theorem https://en.wikipedia.org/wiki/Spin%E2%80%93statistics_theorem