3
$\begingroup$

It is usually said that the wave function for $N$ particles cannot be imagined in 3 dimensions, but only in $3N$ dimensions. What is the exact argument? Is there a loophole?

Schrödinger, Feynman and many others picture single-particle wave functions in 3 dimensions. So there should be a way to picture many-particle wave functions in 3 dimensions as well. For example, two non-interacting, distant particles can be imagined, each with its own wave function, to "live" in 3d space.

About a possible loophole: I recall reading that many-particle wave functions can be pictured in 3d if one allows them to be multi-valued (in a nice way). But I cannot find any reference on this. (For example, one could assign, at every point in space, a separate value for the wave function of each particle. The total wave function would then be "multi-valued" at each point in space.)

$\endgroup$
  • $\begingroup$ I'd also like to find a textbook reference. $\endgroup$ – Christian Jun 17 at 6:19
  • $\begingroup$ The explanation, as to why everything cannot be handled in a simple 3D configuration space for the N-particles, could lie in the construction of the N-particle quantum state space i.e. Fock space from single particle wave-function in Hilbert space. $\endgroup$ – Ishika_96_sparkle Jun 17 at 7:33
3
$\begingroup$

No, there isn't really a loophole. Consider two particles which live in 1D, and let them be bosonic. The wavefunction of the composite system can most generally be written as some symmetric function $\psi(x_1,x_2)=\psi(x_2,x_1)$.

If we make the assumption that $$\psi(x_1,x_2) = \frac{\psi_A(x_1)\psi_B(x_2) + \psi_B(x_1)\psi_A(x_2)}{\sqrt{2}}$$ and further assume that $\psi_A$ and $\psi_B$ are such that they only have support in regions $R_A$ and $R_B$, $R_A\cap R_B = \emptyset$, then the probability density for the position of the first particle is

$$|\psi_{eff}(x_1)|^2 = \int|\psi(x_1,x_2)|^2 dx_2 = \frac{|\psi_A(x_1)|^2 + |\psi_B(x_1)|^2}{2}$$

and the same holds for $\psi_2$. Therefore, one could imagine a "reduced" wavefunction $$\phi(x) = \psi_A(x)+\psi_B(x)$$ which has been projected down into 1D space, and whose norm squared gives the probability density for finding a particle (the identity of which is meaningless) at position $x$.


There were two massively-simplifying assumptions here that do not hold in general. First, we assumed that the composite wavefunction could be cleanly separated into single-particle wavefunctions $\psi_A$ and $\psi_B$, which is generally not the case. Consider for example

$$\psi(x_1,x_2) = e^{-(x_1+x_2)^2}$$

For such a wavefunction, the probability of finding the a particle at some $x_1$ is an inextricable function of $x_2$; there is no sense of separate probability distributions for $x_1$ and $x_2$ individually.

Second, we assumed that the single-particle wavefunctions $\psi_A$ and $\psi_B$ were completely separated, in the sense that their respective supports had no overlap. Were this not the case, then the full probability density would not have cleanly separated into positive-definite parts, and the interference effects would become too important to ignore.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ The original question was "Can many-particle wave functions be imagined in 3 dimensions - instead of 3𝑁 dimensions?" For me you answer is confusing: is it a yes or a no? $\endgroup$ – Christian Jun 19 at 4:57
  • $\begingroup$ @Christian The answer is no, typically not. My answer explains the circumstances under which it is (in some sense) possible, but those circumstances don't hold in general. $\endgroup$ – J. Murray Jun 19 at 5:01
2
$\begingroup$

This answer is simply to complement the one provided by J. Murray and discuss density functional theory, a much-used theory to study quantum systems with a 3-dimensional function: the electron density.

For all N-electron systems, the kinetic energy and electron-electron interaction terms are the same. Therefore, all you need to specify such a system is the electron number N and the external potential $V_{\mathrm{ext}}$. Hohenberg and Kohn proved that the external potential is uniquely determined by the electron density:

$$ n(\mathbf{r})=N\int d\mathbf{r}_2\cdots d\mathbf{r}_N|\Psi(\mathbf{r},\mathbf{r}_2\,\ldots,\mathbf{r}_N)|^2, $$

which is a 3-dimensional function, as opposed to the wave function $\Psi$ which is a 3N-dimensional function. The proof, by reductio ad absurdum is pretty simple: you start assuming that there are two external potentials that differ by more than a trivial constant and that have the same electron density $n(\mathbf{r})$. Using the variational principle for both in turn, you arrive at a contradiction.

This simple idea forms the foundation of density functional theory, the workhorse of modern electronic structure calculations of molecular systems and materials. The ground state of any quantum system is completely determined by the electron density $n(\mathbf{r})$. Practical use of this idea is much more involved, and it is a very active area of research in chemistry, condensed matter physics, and materials science.

| cite | improve this answer | |
$\endgroup$
  • 1
    $\begingroup$ Very interesting. What kind of equation does $n$ satisfy, if any? $\endgroup$ – thermomagnetic condensed boson Jun 17 at 7:52
  • 1
    $\begingroup$ @thermomagneticcondensedboson We can write the energy of a many-particle system as a functional of the density, $E[n]$, and the value of $n$ that minimises this functional corresponds to the ground state density and gives the ground state energy. The challenge: we don't know the general form of this functional, only approximations. This is complementary to the wave function problem; for wave functions we know the equation to solve, but it is in terms of a very complicated 3N-dimensional object, whereas here we know the solution is a simple 3D object, but we don't know the equation is obeys. $\endgroup$ – ProfM Jun 17 at 8:02
  • $\begingroup$ Alright, thanks. I still have to grasp the relationship between $n$ and $\psi$. At first glance they seem to indicate the same thing, i.e. they give a probability (density) to find a particle in a particular region, however this isn't quite right and the relationship is given by the equation you've written and that Hohenberg and Kohn have derived. $\endgroup$ – thermomagnetic condensed boson Jun 17 at 8:59
  • $\begingroup$ The question was "Can many-particle wave functions be imagined in 3 dimensions - instead of 3𝑁 dimensions?" - Is your answer a yes or a no? $\endgroup$ – Christian Jun 18 at 16:14
  • $\begingroup$ @Christian, I agree with J. Murray's answer; all I wanted to do was to add that, although the wave function depends on 3N variables, there is another quantity, the density, which depends on 3 and that can also be used to do some quantum mechanics. $\endgroup$ – ProfM Jun 18 at 18:05

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.