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If you have a wave function $\Psi$ of a system consisting of an electron and the vibrational modes of the crystal, THEN we represent the wavefunction $\Psi%$ to be in the Hilbert Space formed by the tensor product of the Hilbert spaces corresponding to the electron with the Hilbert Space corresponding to the vibrational modes if and only if there isn't an instantaneous interaction between the electrons and the vibrational modes; First of all, this is true right?

The Born-Oppenheimer Approximation technique tells us that we can write the wavefunction $\Psi$ as a product wavefunction- as a product of the electronic ($\phi$) and vibrational modes' ($\zeta$) wavefunctions. We write $\Psi$= $\phi \zeta$ ?

Now my main question:

Is the Born-Oppenheimer Approximation technique equivalent to saying that the Hilbert Space representation of the space that $\Psi$ is situated in is the tensor product space of the electronic and vibrational mode wavefunctions?

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1 Answer

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Whenever two quantum systems are united, the Hilbert space of the wavefunctions of the resulting system is always the tensor product of the Hilbert spaces of the wavefunctions of the source systems.

This does not mean that the total wavefunction is always a product of two wavefunctions from the source spaces. In general case it looks like this: $$ \Psi = \sum_{ij} \phi_i \zeta_j \qquad (1) $$

If the systems do not interact then the total Hamiltonian is the sum of the Hamiltonians of the subsystems: $$ \hat{H}_\Psi = \hat{H}_\phi + \hat{H}_\zeta $$ In this case one can prove that any product of the eigenstates of the subsystems is an eigenstate of the whole system. But even in this case the wavefunction can have the form (1) because this function can describe any state of the system not only a state with determined energy (which is an eigenstate of the Hamiltonian).

In the Born-Oppenheimer approximation the interaction of the subsystems is substantial. The basis consisting of the eigenstates (of the Hamiltonian) of the electron system depends on the state of the crystal lattice. This does not mean though that the Hilbert space of the electron wavefunctions changes. In will neither lose any functions nor get new ones.

So the answers to your questions are:

  1. The total wavefunction is not obliged to be the product $\Psi = \phi\zeta$ even for independent subsystems.
  2. The Hilbert space of the total wavefunctions is always the tensor product of the ones of the subsytems.

These questions are not equivalent.

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If I understand correctly the question, what r.g. is asking is if the Born-Oppenheimer approximation implies that the wave function for a slow $x$ and fast $y$ two particle (or quasi-particle) motion is a tensor product of the form $\Psi(x,y) = \phi(x)\xi(y)$. The answer is that this is not the case. The Born-Oppenheimer approximation would be $\Psi(x,y) = \phi(x)\xi(y;x)$, where $\xi(y;x)$ depends on $y$ but is parametrized on $x$, such that the function changes for each value of $x$. The motion of both particles are therefore not independent and the wave function is not of tensorial form. – perplexity Jan 11 '12 at 11:37
@perplexity: the question is not only about the product of functions but also about the product of spaces. The product of spaces is completely different thing. The functions $\xi(y;x)$ represent the basis of the Hilbert space of the electron system wavefunctions. This space is always the same, say set of functions for which $\left|\xi\right|^2$ is integrable over the crystal volume. The space does not change except some exotic cases when the volume of the crystal changes (this changes the domain of the function). But even then the spaces are isomorphic. – Maksim Zholudev Jan 11 '12 at 12:09
I understand this and your answer is clear and correct. But I thought that the question was centered around the Born-Oppenheimer approximation, and this was missing from your general answer (about properties of the Hilbert space and wave functions of N-particle systems). My comment just tries to put this in the front line. – perplexity Jan 12 '12 at 12:14

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