# The wave function of a system of two identical particles

For a system of two identical particles, where $$r_1$$ is the position vector of particle 1 and $$r_2$$ is the position vec. of particle 2, the wave function should be one of the plus or minus states: $$$$\psi _ \pm (r_1,r_2) = A [\psi _a(r_1) \psi_b (r_2) \pm \psi_b(r_1) \psi_a(r_2)] ,$$$$ where $$\psi_a$$ and $$\psi_b$$ are the wave functions of particle 1 and 2 respectively [equation 5.10 of Griffith's Intro to Quantum Mechanics 2nd Ed.].

I see that this equation makes the wave function $$\psi_\pm$$ treat the two particles identically, but I don't know of any proof that it is actually the only way of writing this wave equation to treat them identically. For example, why not a wave function like: $$$$\psi _ \pm (r_1,r_2) = A \sqrt{\psi^2 _a(r_1) \psi_b^2 (r_2) \pm \psi^2_b(r_1) \psi^2_a(r_2)}~?$$$$

• It's a valid wavefunction, but you lose the interpretation of having one particle in state a and the other in state b. You just have two particles in some complicated state. – Javier Nov 15 '18 at 21:08

The requirement is $$\psi(x_1,x_2) = \begin{cases} \psi(x_2,x_1) & \text{for bosons} \\ -\psi(x_2,x_1) & \text{for fermions}. \end{cases} \tag{1}$$ This property is required by the spin-statistics theorem in relativistic quantum field theory. Since non-relativistic quantum mechanics is supposed to be an approximation to relativistic quantum field theory, we also enforce it in non-relativistic QM. A special case of equation (1) is $$\psi(x_1,x_2) \approx \begin{cases} f(x_1)g(x_2)+f(x_2)g(x_1) & \text{for bosons} \\ f(x_1)g(x_2)-f(x_2)g(x_1) & \text{for fermions}, \end{cases} \tag{2}$$ but like Lewis Miller's answer said, this is only a special case. The general requirement is equation (1).

The square-root example written in the question does not satisfy the requirement (1).

• Thanks @Dan Yand for mentioning that theorem. If we take $A=i$, wouldn't condition (1) be satisfied by the square-root wave function? – Mathist Nov 16 '18 at 17:46
• @Mathophile-Mathochist For a complex quantity $z$, the function $z\mapsto \sqrt{z}$ is double-valued, with opposite signs. We can choose the sign for one value of $z$, but then the sign for other values of $z$ should be determined by continuity. So, suppose we start with $z=1$ and apply a continuous rotation in the complex plane to get to $z=-1$, say $z=\exp(i\theta)$ with $0\leq \theta\leq \pi$. If we choose $\sqrt{1}=1$, then $\sqrt{\exp(i\theta)}=\exp(i\theta/2)$. Since $\exp(i\pi/2)\neq -1$, this shows that the square-root example does not satisfy the fermion sign-change requirement. – Chiral Anomaly Nov 16 '18 at 18:15
• I see, it makes sense. – Mathist Nov 17 '18 at 22:57

This product form of two-particle wave function is only correct if the particles are not interacting. Nevertheless, it is often used as a first approximation, and if you take the expectation value of the true Hamiltonian and minimize it (by taking a variational derivative) you get the two-body Hartree-Fock equation which is often used to approximate the ground state energy and wave function for many-body systems of Fermions. This approximation is often called the mean field approximation.

• Thanks @Lewis Miller. True, this is only for non-interacting particles. I don't know about the two-body Hartree-Fock equation. I'm familiar with variational calculus, so I would appreciate it if you would explain the relation between this equation and OP. – Mathist Nov 16 '18 at 17:43

You’re forgetting that the wave function must also satisfy the TISE. With this condition the combine wavefunctions must be the sum of a permutation of products.

If we have two particles, one in state $$\psi_a$$ and the other in state $$\psi_b$$, then the state vector would be $$|\psi_a\rangle|\psi_b\rangle$$.

However, if the particles are indistinguishable, then it is equally likely to have the opposite be true (i.e. the "first" particle in state $$\psi_b$$ and the "second" particle in state $$\psi_a$$). Therefore, we would want the entire state to be a linear combination of these two states, each with equal weight. Therefore we end up with

$$|\Psi\rangle=|\psi_a\rangle|\psi_b\rangle\pm|\psi_b\rangle|\psi_a\rangle$$

If we choose to work in the $$|r_1\rangle|r_2\rangle$$ basis, then we end up with the expression you state.

I think the issue with your state is that it is not a "nice" linear combination of the states where one particle is in state $$\psi_a$$ and the other in state $$\psi_b$$. We need this if we want the postulate to hold that when $$|\psi\rangle=\sum c_i|\psi_n\rangle$$, we know that there is a probability of $$|c_i|^2$$ to measure the system to be in state $$|\psi_i\rangle$$