Consider a quantum-mechanical system : 3 particles in a rigid sphere.

Which of the following is the correct form of wavefunction for this system ?

(a) $\psi(r_1,r_2,r_3,t) = \psi_1(r_1,t) . \psi_2(r_2,t).\psi_3(r_3,t)$

(b) $\psi(r_1,r_2,r_3,t) = \psi_1(r_1,t) + \psi_2(r_2,t) + \psi_3(r_3,t) $

where $r_1,r_2,r_3$ are position vectors of the three particles and $t$ is the time.

If one is correct, then what is wrong with the other?

Also what is justification for the right form - experimental / theoretical ?

(B) Even though I am giving example of 3-particle system above, still I want to ask this : In Schrodinger equation, can $\psi$ refer to more than one particle?

If I am not wrong, Schrodinger himself derived his equation for a single particle only from making changes in "Hamilton-Jacobi equation for a single particle".

Here I am NOT referring to usual explanation given in textbooks - that either repeat the experiment with single particle OR do single experiment with ensemble of particles, prepared in the same state.

  • $\begingroup$ A wave function describing several particles are once is standard fare in quantum many-body physics, which branches into, and becomes the foundation for, quantum chemistry. $\endgroup$ Oct 8, 2021 at 10:44

2 Answers 2


A generic wavefunction for a 3-particle system doesn't have to take any specific form; it is simply a function $\psi(\mathbf r_1,\mathbf r_2,\mathbf r_3)$ such that $\int \mathrm d^3r_1 \mathrm d^3 r_2 \mathrm d^3 r_3 |\psi(\mathbf r_1,\mathbf r_2,\mathbf r_3)|^2 < \infty$. In particular, there's no reason it should need to "factor." A wavefunction which does factor into the form $\psi(\mathbf r_1,\mathbf r_2,\mathbf r_3)=\psi_1(\mathbf r_1)\psi_2(\mathbf r_2)\psi_3(\mathbf r_3)$ is called separable.

With that said, if $\{\phi_i\}$ is a basis for the single-particle Hilbert space $\mathscr H$, then any wavefunction in the 3-particle Hilbert space $\mathscr H\otimes\mathscr H \otimes \mathscr H$ can be written in the form

$$\psi(\mathbf r_1,\mathbf r_2,\mathbf r_3) = \sum_{ijk} c_{ijk} \phi_i(\mathbf r_1)\phi_j(\mathbf r_2)\phi_k(\mathbf r_3)$$ for some constants $c_{ijk}$. That is, the wavefunction need not be separable, but it can be written as a linear combination of terms which are. As a result, both of your proposed wavefunctions are permissible! The first one is immediately separable, and the second can be written

$$\psi(\mathbf r_1,\mathbf r_2,\mathbf r_3) = \psi_1(\mathbf r_1)\mathbf 1(\mathbf r_2)\mathbf 1(\mathbf r_3) + \mathbf 1(\mathbf r_1) \psi_2(\mathbf r_2) \mathbf 1(\mathbf r_3) + \mathbf 1(\mathbf r_1)\mathbf 1(\mathbf r_2) \psi_3(\mathbf r_3) $$ $$= \psi_1(\mathbf r_1)+\psi_2(\mathbf r_2)+\psi_3(\mathbf r_3)$$ where $\mathbf 1(\mathbf r) = 1$ inside the sphere.

Note that this applies to distinguishable particles. If the particles in the system are indistinguishable fermions or bosons, then you need to apply extra symmetry constraints which make the 3-particle wavefunction either antisymmetric or symmetric respectively, and this further constrains the form that wavefunctions may have. For example, a simple product $\psi_1(\mathbf r_1)\psi_2(\mathbf r_2)\psi_3(\mathbf r_3)$ is not permitted for a system of fermions.

  • $\begingroup$ ...You already assumed addition in the equation immediate below the equation containing summation symbol. $\endgroup$
    – atom
    Oct 8, 2021 at 11:57
  • $\begingroup$ @atom I'm not sure what you mean by that. My intention was to demonstrate that $\psi_1(\mathbf r_1)+\psi_2(\mathbf r_2)+\psi_3(\mathbf r_3)$ is the sum of three separable terms and is therefore an allowed 3-particle wavefunction. $\endgroup$
    – J. Murray
    Oct 8, 2021 at 12:00
  • $\begingroup$ "sum of separable terms" ....... I meant to say you started from taking the sum already...introduction of 1(r) =1 is not clear at all $\endgroup$
    – atom
    Oct 8, 2021 at 12:05
  • $\begingroup$ @atom Sorry, I still don't understand the problem. There's nothing to derive here - the definition of the three-particle Hilbert space is essentially the space of wavefunctions which can be written as a linear combination of separable terms. I used the notation $\mathbf 1(\mathbf r)$ to emphasize that each term in the sum $\psi_1(\mathbf r_1) + \ldots$ can be understood as the product of three single-particle wavefunctions. $\endgroup$
    – J. Murray
    Oct 8, 2021 at 12:11
  • $\begingroup$ @atom If I'm understanding you correctly, you are asking for proof that adding separable wavefunctions together is allowed, but that's part of the definition of the tensor product space $\mathscr H \otimes \mathscr H \otimes \mathscr H$. $\endgroup$
    – J. Murray
    Oct 8, 2021 at 12:14

If we consider probability of particle $1$ being at point $x_1$, particle $2$ at point $x_2$ and particle $3$ at point $x_3$, we need to multiply the corresponding one particle probabilities (the word and is telling here, as opposed to or when the probabilities are summed): $$ \rho(x_1,x_2,x_3,t)=\rho_1(x_1,t)\rho_2(x_2,t)\rho_3(x_3,t) = |\psi_1(x_1,t)|^2|\psi_2(x_2,t)|^2|\psi_3(x_3,t)|^2=\\|\psi_1(x_1,t)\psi_2(x_2,t)\psi_3(x_3,t)|^2=|\psi(x_1,x_1,x_3,t)|^2. $$ So the answer is $$\psi(x_1,x_1,x_3,t)=\psi_1(x_1,t)\psi_1(x_2,t)\psi_1(x_3,t).$$

This is hwoever only a part of the story, as we also need to take into account the particle statistics/indistinguishability (as expressed by the famous Pauli exclusion principle for fermions). This is done with the help of Slater determinant (or permanent for bosons): $$ \psi(x_1, x_2, x_3) = \frac{1}{\sqrt{3!}} \left|\begin{matrix} \psi_1(x_1) & \psi_2(x_1) & \psi_3(x_1)\\ \psi_1(x_2) & \psi_2(x_2) & \psi_3(x_2)\\ \psi_1(x_3) & \psi_2(x_3) & \psi_3(x_3)\\ \end{matrix}\right| = ... $$

  • $\begingroup$ Why do you assume that each particle is in the same 1-particle state $\psi$? $\endgroup$
    – J. Murray
    Oct 8, 2021 at 11:04
  • $\begingroup$ @J.Murray a typo. Thanks. $\endgroup$
    – Roger V.
    Oct 8, 2021 at 11:31

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