9
$\begingroup$

I have trouble understanding the proof of the Variational Theorem. I'll recall quickly the proof to show my problems (see also this post and the answer by Mateus Sampaio for a detailed proof). Let $H$ be the Hamiltonian operator of a system. Since it is self-adjoint, we can find an orthonormal basis $\{\Psi_n\}$ with corresponding real eigenvalues $E_n$. Suppose that $\min_n E_n$ exists and denote it by $E_0$. We may then order the eigenvalues $E_0\leq E_1\leq\dots$. We can write any normalized state $|\Psi\rangle$ as $$ |\Psi\rangle = \sum_n c_n |\Psi_n\rangle \;\; \text{with} \;\; \sum_n |c_n|^2 = 1. $$ We can now compute for the expected value of the energy $$ \begin{align*} \langle \Psi | H | \Psi \rangle &= \langle \sum_n c_n \Psi_n | H | \sum_m c_m \Psi_m\rangle\\ &= \langle \sum_n c_n \Psi_n | \sum_m c_m H \Psi_m\rangle \\ &= \sum_{m,n} c_n^* c_m \langle \Psi_n | E_m \Psi_m \rangle \\ &= \sum_n |c_n|^2 E_n \geq \sum _n |c_n|^2 E_0 = E_0. \end{align*} $$ What I don't understand is the second step where we pull the Hamiltonian inside the second sum over $m$ to let it act on $\Psi_m$. The Hamiltonian is linear, so interchanging it with a finite sum is not a problem. But here we interchange it with an infinite sum, so in particular we do the following: $$ H\left(\sum_{m=0}^\infty c_m \Psi_m\right) = H\left(\lim_{M\to\infty} \sum_{m=0}^Mc_m\Psi_m\right) = \lim_{M\to\infty} H\left(\sum_{m=0}^M c_m \Psi_m\right). $$ But the Hamiltonian is not continuous. If it was, it would be bounded because it is linear. But it is unbounded because it involves second order derivatives.

My question is: Why can we nevertheless interchange the Hamiltonian with the limit?

$\endgroup$
7
  • $\begingroup$ Can you give some more detail on what you think might go wrong by commuting $H$ past the limit in this case? For example, are you worried about violating the assumptions of a particular theorem, or do you have a case in mind where this wouldn't work? $\endgroup$
    – Andrew
    Mar 7 at 23:58
  • 1
    $\begingroup$ @Andrew For simplicity assume that the Hamiltonian is only given by the kinetic potential energy operator in 1D on the interval $[0,1]$, that is, $H = -\frac{1}{2} \frac{d^2}{dx^2}$. Then $H$ is obviously linear but not bounded as an operator from $L^2([0,1])$ to $L^2([0,1])$ which can be seen by applying it e.g. to the functions $f_n(x) = x^n$. It is straight-forward to see that in this case $$ \frac{\|H f_n\|_{L^2([0,1])}}{\|f_n\|_{L^2([0,1])}} \to \infty \text{ as } n\to \infty. $$ $\endgroup$
    – msaBU
    Mar 8 at 10:29
  • 1
    $\begingroup$ Since a linear operator is continuous if and only if it is bounded (see e.g. here), the Hamiltonian is not continuous, Hence, I don't see a reason why we are allowed to interchange the Hamiltonian with the limit in the proof of the Variational Theorem. Or am I missing something? $\endgroup$
    – msaBU
    Mar 8 at 10:32
  • 1
    $\begingroup$ The mathematically rigorous explanation by @Janik is perfect and is valid in simple cases. However, once you move on to QFT with gauge fields, then several of the basic assumptions of Hilbert spaces (such as separability) break down and those theorems no longer go through (Janik mentions this in his excellent answer!). As physicists we get around this by imposing UV and IR cut-offs so everything is forcibly discrete and finite-dimensional. Then all the theorems are valid. At the end we then remove the cut-offs. $\endgroup$
    – Prahar
    Mar 9 at 10:43
  • 1
    $\begingroup$ Of course, one of the issues is that the final result may then depend on how you imposed the cut-off. We deal with this issue separately through renormalization. $\endgroup$
    – Prahar
    Mar 9 at 10:45

1 Answer 1

4
$\begingroup$

Yes, in general, it is not justified to exchange a limit and an unbounded operator because an unbounded operator is not continuous. However, in your case, you can apply Parseval's identity: If $\{e_n\}_{n\in\mathbb{N}}$ is an orthonormal basis of a separable Hilbert space $\mathcal{H}$, then, for any $x\in \mathcal{H}$, $$x = \sum_{n\in \mathbb{N}}\langle e_n, x\rangle e_n.$$ Thus, if $H$ is a bounded from below self-adjoint operator (defined on $D(H) \subset \mathcal{H}$) with discrete spectrum $\sigma(H) = \{ E_0, E_1, \dots \}$, $E_0 = \min \sigma(H)$, and $\Psi_1, \Psi_2, \dots \in D(H)$ are the corresponding eigenvectors that form an orthonormal basis of $\mathcal{H}$, then, for $\Psi = \sum_{n\in\mathbb{N}} c_n \Psi_n \in D(H)$, $$\begin{align} H\Psi &= \sum_{n\in \mathbb{N}} \langle\Psi_n, H\Psi\rangle \Psi_n \\ &= \sum_{n\in \mathbb{N}} \langle H\Psi_n, \Psi\rangle \Psi_n \\ &= \sum_{n\in \mathbb{N}} E_n \langle\Psi_n, \Psi\rangle \Psi_n \\ &= \sum_{n\in \mathbb{N}} E_n c_n \Psi_n, \end{align}$$ which is the identity you required.

As a remark, let me mention that the given assumptions on $H$ (discrete spectrum and eigenvectors form an orthonormal basis) are pretty strong and in most situations of physical interest not satisfied. A more general proof of the variational theorem employs the spectral theorem. According to the spectral theorem, for every self-adjoint operator $H$, a projection-valued measure $E$ exists such that $$H = \int_{\sigma(H)} \lambda \ \mathrm{d}E(\lambda).$$ Hence, for every $\Psi \in D(H)$: $$\begin{align}\langle \Psi, H \Psi\rangle = \int_{\sigma(H)} \lambda \ \mathrm{d}\langle \Psi, E(\lambda) \Psi\rangle = \int_{\sigma(H)} \lambda \ \mathrm{d}\|E(\lambda) \Psi\|^2 \\ \geq \min \sigma(H) \int_{\sigma(H)} \mathrm{d}\|E(\lambda) \Psi\|^2 = \min \sigma(H) \|\Psi\|^2. \end{align}$$

$\endgroup$
2
  • $\begingroup$ Thanks for this very nice answer and the remark in the end! Can you recommend a textbook or some other source where I can read up on the spectral theorem and the general version of the variational theorem? $\endgroup$
    – msaBU
    Mar 9 at 10:30
  • 2
    $\begingroup$ A decent introductory textbook is Teschl: Mathematical Methods in Quantum Mechanics. Chapter 3 discusses the spectral theorem, and Theorem 2.20 is the variational theorem. $\endgroup$
    – Janik
    Mar 9 at 10:37

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.