# Symmetry of two-particle system of electrons

Consider a two particle system consisting of two electrons. The complete state of the electron includes its position wave function and also a spinor describing the orientation of its spin: $$\psi(r) \otimes \chi(s).$$ Why does it follows that for the the two particle system that if we have an anti-symmetric spin state of the two electrons such as the singlet state $$\frac{1}{\sqrt{2}}(| {\uparrow} \rangle \otimes | {\downarrow} \rangle - | {\downarrow} \rangle \otimes | {\uparrow} \rangle)$$ then this has to be joined with a symmetric spatial function (and similarly if we have a symmetric state of two electrons such as $$| {\uparrow} \rangle \otimes |{\uparrow} \rangle$$ then this has to be joined by an anti-symmetric spatial wave function?

Also if two electrons occupy the singlet spin state then the spatial wave function describing the two particle state would be symmetric, but I thought that for identical particles which are fermions (such as electrons), the spatial wave function is always antisymmetric?

Thanks for any assistance.

• An odd state multipled by an even state gives you an odd state. It's been a while but I think it is analogous to that with symmetery, to preserve anti symmetry. Anyway, I am sure someone will sort us both out.
– user140606
Jan 16, 2017 at 18:48

For fermions, the total wave function, including both the spatial wave function and the spin state, must be antisymmetric under exchange. Since the product of two antisymmetric functions is symmetric (as is the product of two symmetric functions), it is necessary that either the spin is antisymmetric or the spatial wave function is antisymmetric but not both.

• Thanks for your answer. Do you mean that say the spatial wave function $\psi(x_{1}, x_{2})$ of a two electron system is symmetric under the exchange operator then we require that the spin would be in the singlet state $| 0 0 \rangle =\frac{1}{\sqrt{2}}(| \uparrow \rangle \otimes | \downarrow \rangle - | \downarrow \rangle \otimes | \uparrow \rangle)$ hence we would have a total wave function $$\Psi = |\psi \rangle \otimes |0 0 \rangle?$$
– Alex
Jan 16, 2017 at 20:30
• That's correct: if the spatial wave function for two electrons is symmetric, the spin state must be antisymmetric - ie., the singlet state. If instead the spatial wave function is antisymmetric, the spin state would have to be a triplet state. Jan 16, 2017 at 20:33
• Okay thanks. One other query... Would I be right in stating the following: Given that the state of the two particle system is $$\Psi = | \psi \rangle \otimes |00 \rangle$$ could I then project the tensor product onto the position basis by using the linear operator $\langle x_1, x_2 | \otimes I$, thus resulting in the tensor product $$\psi(x_1, x_2) \otimes |00\rangle$$, since the left hand side of the tensor product is a complex number and the right side is a matrix, we can consider the state $$\psi(x_1,x_2)|00 \rangle$$ as simply matrix times a complex scalar?
– Alex
Jan 17, 2017 at 10:32

I will extend existing answer to cover all cases.

The Exchange Symmetry requirement is a requirement on the state and the use of labels such as 1, 2 to label particles of identical type. For fermions the requirement is that if the labels are exchanged the state must change sign, and have no other change. That is: $$| \psi \rangle_{1,2} = - | \psi \rangle_{2,1}$$ Note that the Exchange Symmetry is not a statement about exchanging the particles, despite the fact that it is often stated that way. It is really a statement about exchanging the LABELS. Clarity on this point will help avoid mistakes later on.

Now fermions have position and spin degrees of freedom. The state I have called $$|\psi\rangle$$ above includes both the position and spin. Such a state can always be written $$| \psi \rangle_{12} = \sum_{i = 1}^4 a_i | \phi_i \rangle_{12} \otimes | \chi_i \rangle_{12} \tag{1}$$ where each $$| \phi_i \rangle$$ is purely spatial and $$| \chi_i \rangle$$ is purely spin and the sum runs over a set of basis states such as the spin up/down states or the singlet and three triplet spin states (for a pair of spin-half particles). The $$a_i$$ are coefficients which tell you which state you have. The symbol $$\otimes$$ is the tensor product. It is often omitted.

In an introductory treatment it is almost always proposed that in the first instance we consider a product state, or one which can be factorized into a spatial part in a product with a spin part. Such a state takes the form $$| \psi \rangle_{12} = | \phi \rangle_{12} \otimes | \chi \rangle_{12}.$$ Ok, now let's swap the LABELS. We get $$| \psi \rangle_{21} = | \phi \rangle_{21} \otimes | \chi \rangle_{21}.$$ The Exchange Symmetry requirement for fermions is $$|\psi \rangle_{21} = -|\psi \rangle_{12}$$ so we deduce: $$| \phi \rangle_{21} \otimes | \chi \rangle_{21} = - | \phi \rangle_{12} \otimes | \chi \rangle_{12}.$$ Now we can, if we wish, consider states where the spatial and spin parts each have definite symmetry separately (this is the case most commonly considered). In this case the requirement is:

EITHER the spatial part changes sign and the spin part does not
OR the spatial part does not change sign and the spin part does

There are three symmetric spin states for a pair of spin-half particles: $$|\uparrow_1 \uparrow_2\rangle,\qquad \frac{1}{\sqrt{2}}\left(|\uparrow_1 \downarrow_2\rangle + |\downarrow_1 \uparrow_2\rangle \right), \quad |\downarrow_1 \downarrow_2\rangle$$ Together they are called the triplet states (it is often said, somewhat loosely, that the spins are then 'aligned'). There is one antisymmetric spin state for a pair of spin-half particles; it is the singlet state (it is often said, somewhat loosely, that the spins are then 'anti-aligned'): $$\frac{1}{\sqrt{2}}\left(|\uparrow_1 \downarrow_2\rangle - |\downarrow_1 \uparrow_2\rangle \right).$$ Hence we find that a pair of identical spin-half particles can be in a singlet spin state only if their spatial part is symmetric w.r.t. exchange, and they can be in a triplet spin state only if their spatial part is antisymmetric w.r.t. exchange. Conversely, if the spatial part is symmetric (e.g. because both particles are in the same spatial state) then the spin part has to be the singlet. This case commonly arises in the ground state of atoms with two valence electrons, and in close collisions between particles of the same type. It is the reason why the ground state of helium is a singlet state.

So far so good, and you will find the above (or some looser version of it) in all the introductory treatments.

But for complete generality we should also acknowledge that the complete state does not necessarily have to be a product of spatial and spin parts. There can be an entanglement between spatial and spin degrees of freedom. This aspect confused me when I first learned about exchange symmetry, which is why I want to include it here. For example suppose we have two electrons, and suppose the state has a spin-up electron moving to the left and a spin-down electron moving to the right. As an opening gambit we propose for the complete state: $$| \psi \rangle_{12} \stackrel{?}{=} | \text{L} \rangle_1 | \uparrow \rangle_1 | \text{R} \rangle_2 | \downarrow \rangle_2 .$$ Then we note that this state is impossible because it lacks the correct exchange symmetry (i.e. antisymmetry), so we fix it up by making a superposition of this with the same state with labels switched: $$| \psi \rangle_{12} = \frac{1}{\sqrt{2}} \left( | \text{L} \rangle_1 | \uparrow \rangle_1 | \text{R} \rangle_2 | \downarrow \rangle_2 - | \text{L} \rangle_2 | \uparrow \rangle_2 | \text{R} \rangle_1 | \downarrow \rangle_1 \right) . \tag{2}$$ where I used L and R for left and right. This is where the concept of swapping labels rather than particles comes into its own: it makes it very clear how to construct the state with the correct overall anti-symmetry. Having done that then we can, it we like, adopt a more succinct notation such as $$| \psi \rangle_{12} = \frac{1}{\sqrt{2}} \left( | \text{L} \uparrow\,, \, \text{R} \downarrow \rangle - | \text{R} \downarrow \,,\, \text{L} \uparrow \rangle \right) . \tag{3}$$ Notice that in this state we have the situation, "whichever particle is moving left, it has spin up, but we don't know which it is" or, better, "whichever particle is moving left, it has spin up, but the quantum state contains strictly no information about which particle (1 or 2) that may be."

It is now a useful exercise to express the state (3) in terms of singlet and triplet states. In order to do this (exercise for the reader) it is clearest to return to (2) and work it out from there. The answer is $$| \psi\rangle_{12} = \frac{1}{2}\left( |LR\rangle + |RL\rangle \right) | \chi_0 \rangle + \frac{1}{2}\left( |LR\rangle - |RL\rangle \right) |\chi_{1,0} \rangle$$ where $$|\chi_0\rangle$$ is the singlet state and $$|\chi_{1,0}\rangle$$ is the triplet state with zero $$z$$-component. This exercise tells you much of what you need to know about exchange symmetry!

For the most general case we have to return to eqn (1). We can take as a convenient basis of spin states the singlet and the three triplet states. We then have a superposition like the one we just wrote, but now all three triplet states appear, and for each of them the corresponding spatial state is antisymmetric (but it does not have to be the same spatial state for each).

A final remark on "exchange interaction". The terminology "exchange interaction" has entered into the physics lexicon but some of us want to banish it again because it is misleading for beginners. Exchange symmetry requirements DO NOT INTRODUCE ANY NEW INTERACTIONS INTO THE HAMILTONIAN. Rather what they do is RESTRICT THE SET OF AVAILABLE STATES. This often has the effect of restricting the spatial state for a given spin state. A common case is a singlet where the spatial state has to be symmetric, as compared to a triplet where the spatial state must be antisymmetric. Since these spatial states will have different energies (owing to Coulomb repulsion between electrons, for example) it looks as though there is an energy associated with the spin. So we get the idea of an "exchange interaction". But it is really (in this example) an ordinary electromagnetic interaction. It should not surprise us that two different states of motion have different energies. That is observed in simple classical examples too, such as coupled pendula.