quantum mechanics - The antisymmetrisation of two identical single particle wave functions is identically zero, why is this important?

Let $f_1,f_2$ be two $\mathbb{R}^3 \to \mathbb{C}$-functions and

$$\mathrm{asym}(f_1,f_2)(x_1,x_2) = f_1(x_1)f_2(x_2) - f_1(x_2)f_2(x_1).$$ If $f_1=f_2$ then $\mathrm{asym}(f_1,f_2)$ is identically zero. Physics professors I have encountered during my studies have used this observation to attempt to convince me of the Pauli exclusion principle, saying something along the lines of —Therefore it is impossible to find two fermions in the same state!

I have never believed this because I cannot see that for every 2-fermion hamiltonian there are two scalar fields $\psi_1,\psi_2$ such that $\mathrm{asym}(\psi_1,\psi_2)$ solves the time independent Schrödinger equation.

If it is indeed false, could you please provide clarification of how this argument can be strengthened such that it becomes fruitful? On the contrary, if this is a mistake on my part, a proof would be much appreciated!

Answer

Physics professors I have encountered during my studies have used this observation to attempt to convince me of the Pauli exclusion principle, saying something along the lines of —Therefore it is impossible to find two fermions in the same state!

That's rather too weak of a statement. The fact that

$$\operatorname{asym}(f,f)\equiv 0,$$ as a consequence of the requirement that fermions have quantum states that are antisymmetric under particle exchange, is the Pauli Exclusion Principle. That's the real, core statement of the principle. The other, hand-wavy formulation ("two particles cannot be in the same place at the same time") is just a placeholder heuristic that's given to students while they learn the necessary mathematics to understand the full formulation of the principle.

I have never believed this because I cannot see that for every 2-fermion hamiltonian there are two scalar fields $\psi_1,\psi_2$ such that $\mathrm{asym}(\psi_1,\psi_2)$ solves the time independent Schrödinger equation.

Too bad. Math doesn't care whether you believe in it or not.

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In any case, though, let's do some math. Let's start with one result that actually matters.

Let $H$ be a two-fermion hamiltonian which is invariant under particle exchange (i.e. which commutes with the particle-exchange operator $\hat{\mathcal P}$), and let $\psi_{1}^{(0)}$ and $\psi_{2}^{(0)}$ be two arbitrary single-fermion initial states.

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  The standard theory guarantees us the existence of a solution $\Psi(t)$ with initial condition $\psi_{1}^{(0)}\otimes \psi_{2}^{(0)}$, given by

  $$\Psi(t) = e^{-i Ht/\hbar} \psi_{1}^{(0)} \otimes \psi_{2}^{(0)},$$ at least formally.
  
  


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  (If $H=H_1\otimes \mathbb I + \mathbb I \otimes H_1$ is non-interacting, then that can be further simplified to $\Psi(t) = e^{-i Ht/\hbar} \psi_{1}^{(0)} \otimes e^{-i Ht/\hbar} \psi_{2}^{(0)}$. If the system does have interactions, there is no guarantee that the evolved state will remain separable; once asymmetrization is introduced, that will imply that the solution cannot be captured by a single Slater determinant. And no, that is not a problem for what you're asking.)

  
  
  


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  Since $H$ commutes with $\hat{\mathcal P}$, then $\hat{\mathcal P}\Psi(t)$ is also a solution of the Schrödinger equation, with initial condition $\psi_{2}^{(0)}\otimes \psi_{1}^{(0)}$.

  
  


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  Similarly, $\Psi_\mathrm{asym}(t) = (1-\hat{\mathcal P})\Psi(t)$ is a solution of the time-dependent Schrödinger equation, with initial condition $\operatorname{asym}(\psi_{1}^{(0)}, \psi_{2}^{(0)})$, and which remains antisymmetric under exchange for all times.

  
  

This is the bit that really matters. Everything else is secondary.

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However, that said, the way you've phrased your question,

I cannot see that for every 2-fermion hamiltonian there are two scalar fields $\psi_1,\psi_2$ such that $\mathrm{asym}(\psi_1,\psi_2)$ solves the time independent Schrödinger equation

is right enough. You are correct: there is no requirement that the eigenstates of arbitrary two-fermion hamiltonians be given by single Slater determinants, and indeed this is rarely the case in the real world.

To be clear, this assumption is indeed a cornerstone of the self-consistent Hartree-Fock method, and this is only an approximation. The exact eigenstates do not have this form, and cannot be found by Hartree-Fock approaches ─ they require post-Hartree-Fock methods that allow the eigenstates to be superpositions of multiple Slater determinants.

(On the real world, of course, we use the HF picture because it's a good approximation. To see how good, have a look at the atomic-level structure on the NIST ASD: pick an atom and an ionization state (say, C I for neutral carbon) and look at the levels. The Leading percentages column shows you the extent to which the HF single-Slater-determinant configuration (say, $2s^2 \,2p^2$ for the C I ground state) captures the fully-interacting post-HF ground state, as well as the contributions for the largest sub-leading determinants where these rise above the noise. Have a look around, and you'll find that, mostly, the single-Slater-determinant HF picture works just fine in practice. Not perfectly (which is why e.g. the Configuration Interaction method dates back to the early thirties), but just fine as a capture of the bulk of the physics.)

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But... you're correct, but it's irrelevant to the quote that bothers you.

That phrasing of the Pauli Exclusion Principle states what happens if you actively want states of two electrons that have both electrons in the same state, and it shows you that those states are impossible.

The principle says nothing about any hamiltonians nor about any eigenstates nor about the time-independent Schrödinger equation. It's a statement about the types of states available on the state space within which the hamiltonians and their eigenstates play, and those aspects are secondary and ultimately irrelevant to the Exclusion Principle proper.

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More generally, perhaps this will help: the implication

$$\psi=\operatorname{asym}(\psi_1,\psi_2) \implies \hat{\mathcal P}\psi = - \psi$$ is true, but its converse isn't. The explicit Slater-determinant form $\operatorname{asym}(\psi_1,\psi_2)$ is one way for a state $\psi$ to satisfy the fermionic-statistics requirement that $\hat{\mathcal P}\psi = - \psi$ but it is not the only way. You explicitly cannot derive the Slater-determinant form from fermionic statistics, because there are plenty of states exchange-antisymmetric that don't satisfy it. (Heck, almost no states have that form.)

The chain of reasoning doesn't go

$\hat{\mathcal P}\psi = - \psi$, therefore $\psi=\operatorname{asym}(\psi_1,\psi_2)$, therefore $\operatorname{asym}(f,f)$ is impossible.

Instead, the chain of reasoning goes

if you wanted to have two fermions in the same state, you'd want something like $f\otimes f$, but the closest you can get to that while respecting $\hat{\mathcal P}\psi = - \psi$ is $\operatorname{asym}(f,f)$, which is impossible.

Yes, there are states that are not of the form $\operatorname{asym}(\psi_1,\psi_2)$ but those just get you even further from the form that you'd want.

(Oh, and also: not all exchange-antisymmetric states are of the form $\operatorname{asym}(\psi_1,\psi_2)$, but they are all a superposition of states of that form. For a simple proof, start with a tensor-product basis, and project down onto the exchange-antisymmetric part by taking $\operatorname{asym}$ of all the basis tensor products, discarding any linearly-dependent terms.)