A wavefunction is inherently a multi-particle function. If you have a container that is perfectly isolated from the external universe (not possible, but just imagine it) and filled with $n$ particles, then the information necessary to perfectly describe everything there is to know about that system would be contained in $\Psi(\vec{r_1}, \vec{r_2}, ..., \vec{r_n}, t)$.

The probabilities of all obtainable measurements could be derived from that function.

However, I frequently read about wavefunctions of individual particles within a system such that the assumption

$$\Psi(\vec{r_1}, \vec{r_2}, ..., \vec{r_n}, t) = \Psi_1(\vec{r_1}, t) \Psi_2(\vec{r_2}, t) ... \Psi_n(\vec{r_n}, t)$$

is made. One of my professors has said that individual particle wavefunctions are a more fundamental description of reality; however, I did not think this was correct because it would completely exclude the effects of quantum entanglement (then again, this was a graduate chemical engineering course).

Where I'm going with this is: in what situations is it okay to separate the complete wavefunction like that? How do you know that an isolated electron in an experiment isn't influenced by the rest of the particles in the universe?

It's frequently said that when a measurement of a system is made, the wavefunction collapses into one of its eigenstates. But how is this possible if there is no such thing as a single-particle (or few-particle) wavefunction? Wouldn't it instead be the case that the overall

$$\Psi\left(\text{particle in experiment} + \text{measurement apparatus} + \text{all other particles}, t\right)$$

has evolved in time?

I realize my question is somewhat vague, but any kind of clarification on this issue would be appreciated.

• Aug 7 '13 at 8:06
• Out of curiosity, if the particles are in two separated and sealed containers, could they still form an entangle state? Is there any experimental result? Aug 7 '13 at 10:31

1) "In what situations is it okay to separate the complete wavefunction like that?"

As a simple product, only for distinguishable non-interacting particles. If you add antisymmetry, as Ruslan notes, you get a Slater determinant wavefunction which is exact for indistinguishable non-interacting fermions. Since electrons interact through the Coulomb operator, this is indeed never exact for many-electron wavefunctions, as Dan mentions. The simple product wavefunction was first used by Douglas Hartree; Vladimir Fock added antisymmetry resulting in the Hartree-Fock method (and I guess the Professor in your class is taking you in that direction).

Now you say that you heard this in a graduate chemical engineering course, and Hartree-Fock is still very useful in chemistry. But, as you note, this wavefunction completely ignores entanglement. I'll try to explain why is the product wavefunction still useful despite this shortcoming with an example of chemistry.

Consider a hydrogen molecule at equilibrium bond length. This is a two-electron wavefunction and we can write a wavefunction which considers entanglement as \begin{equation} | \Psi \rangle = c_1 | \sigma_\alpha\sigma_\beta \rangle + c_2 | \sigma_\alpha^*\sigma_\beta^* \rangle \end{equation} where $| \sigma_\alpha\sigma_\beta \rangle$ is the Slater determinant formed from the bonding orbitals $\sigma_\alpha$ and $\sigma_\beta$, whereas $| \sigma_\alpha^*\sigma_\beta^* \rangle$ is a Slater determinant made with the antibonding orbitals $\sigma_\alpha^*$ and $\sigma_\beta^*$ ($\alpha$ and $\beta$ represent the spin of the orbital of course, and I'm assuming that you are familiar with molecular orbitals). By the variational principle, $c_1$ and $c_2$ must minimize the energy of $| \Psi \rangle$. However, at equilibrium bond length the energy of $| \sigma_\alpha\sigma_\beta \rangle$ is much lower than that of $| \sigma_\alpha^*\sigma_\beta^* \rangle$ resulting in $c_1 >> c_2$ and hence $| \Psi \rangle \sim | \sigma_\alpha\sigma_\beta \rangle$. Thus, this is a case in which Hartree-Fock (i.e. a single Slater determinant) can yield a qualitatively correct result. If you add perturbation theory to $| \sigma_\alpha\sigma_\beta \rangle$, you get very accurate (quantitative) results.

However, Hartree-Fock may fail terribly in some cases. Suppose now that we separate the atoms that compose the Hydrogen molecule at such a large distance that their atomic orbitals do not interact anymore. At this point, $| \sigma_\alpha\sigma_\beta \rangle$ and $| \sigma_\alpha^*\sigma_\beta^* \rangle$ become degenerate and $c_1 = c_2$. Thus, in this case we have a multideterminant wavefunction which can't be described by Hartree-Fock. Indeed, in this case the Hartree-Fock energy is unphysical and much larger than that of two Hydrogen atoms.

The lesson of this example is that (antisymmetrized) product wavefunctions are useful in cases where the entanglement is small, but fail terribly if the entanglement is strong. In general in chemistry, this failure occurs when one has degenerate or near degenerate orbitals, which occurs mostly in molecules at dissociation, transition metals, and singlet diradicals. However, simple closed shell species from the first two rows of the periodic table can often be well described by antisymmetrized product wavefunctions (single Slater determinants).

2) "How do you know that an isolated electron in an experiment isn't influenced by the rest of the particles in the universe?"

The electron interacts with the rest of the particles in the universe, but the interaction is negligible. Recall that this interaction must be through the Coulomb operator $1/|r_1 - r_2|$ (in atomic units) and usually $|r_1 - r_2|$ will be large enough so that we can ignore interactions of (say) a molecule in gas phase with the rest of the universe (we can't ignore interactions of molecule with a solvent when dealing with solutions though, but there are approximate methods for dealing with those).

In most cases representing $\Psi=\prod_i\psi_i$ is an approximation, which appears exact only for non-interacting particles.

This is however a working approximation (Hartree-Fock method if we add symmetrization) in many applications of quantum physics, e.g. condensed matter physics. Still, exact wavefunction is almost never separable in individual particle states.

• Should the particles also be distinguishable? Aug 26 at 19:04
• @Anton if you want separability in this exact form of a simple product, then yes. Otherwise, for indistinguishable particles you need to (anti-)symmetrize the state, which will yield a more complicated form (see e.g. Slater determinant) Aug 26 at 19:19

You're right about this. Fermionic wavefunctions, for example, can't be exactly represented by product states. Most states can't be expressed as a single product of one-electron wavefunctions, or even a single Slater determinant.

It is, however, useful to think of general entagled states in terms of sums of product states.