# I have a problem with the variational method approximation in quantum mechanics. Is my issue valid, or am I misunderstanding something?

The variational method for approximating the ground state of a Hamiltonian $H$ by providing a lower bound is simple enough. If we construct any test wave function $|\bar{0}\rangle$ then the claim is $$\frac{\langle \bar{0} | H | \bar{0} \rangle}{\langle \bar{0} | \bar{0} \rangle} \ge \langle 0 |H|0\rangle \equiv E_0$$ with the denominator added because we don't necessarily need to assume that the test function is normalized.

You show this by assuming that $$|\bar{0} \rangle = \sum\limits_k a_k|k\rangle$$

where the sum is over all the eigenstates of $H$ including the true ground state $|0\rangle.$ With that assumption it's pretty trivial to plug this form of the test state $|\bar{0}\rangle$ into the expected value of $H$ and, knowing that $E_k \ge E_0 \forall k$, show that the inequality above is true. So the idea is to take a test function that looks similar to what you think the true ground state wave function looks like, but with adjustable parameters, and try and minimize the matrix element $\langle \bar{0}|H|\bar{0}\rangle$ with respect to those parameters in order to get as high a lower bound as possible.

That's all well and good. But... my problem is with the assumption above, expressed in the second equation. That is, the assumption that any test wave function you come up with can be expressed as a linear combination of actual eigenstates of the Hamiltonian.

The way I see it, when you talk about a vector whose basis of states is something like $|x\rangle$, (eigenstates of $x$) you can construct any conceivable wave function in that matter. However, the basis of eigenstates of $H$ are more restrictive; you can only construct a limited family class of functions from that basis. While both bases are infinite in size, the basis of $x$ has cardinality $\aleph_1$ as opposed to the $\aleph_0$ of the $H$ states, which means that there's no bijective transformation between the functions spanned by the $H$ basis and those spanned by the $x$ basis. The states of $H$ don't cover the entire Hilbert space.

Therefore, I think that the assumption $|\bar{0}\rangle = \sum_k a_k |k\rangle$ is wrong. But it's essential for the derivation of the variational method inequality. You can't just pick a test function of any arbitrary form and assume it belongs to the restricted family of functions that can be expressed by the basis of states $|k\rangle$. But rather, the assumption $|\bar{0}\rangle = \int \psi(x) |x\rangle dx$ would be just fine.

Am I right about this problem, or am I missing something? Is this problem considered by people who use the variational method to try to approximate the ground state energy of a complicated Hamiltonian? If so, how do they get around it?

• Of course you should construct a wavefunction that belongs to the actual space of states, which usually is the condition that it is square-integrable. The states of $H$ do cover the entire Hilbert space, it's just that $x$ has no eigenstates inside the Hilbert space of states (that's why $\langle x \vert x \rangle$ is undefined, by the way!). See Rigged Hilbert space for what's going on there. – ACuriousMind Apr 16 '15 at 15:48
• Every Hilbert space with countable base (called a separable space) is abstractly isomorphic to $L^2(\mathbb{R})$ and all quantum mechanical Hilbert spaces are separable. – ACuriousMind Apr 16 '15 at 15:57
• – Qmechanic Apr 16 '15 at 15:59