I want to prove that the expectation value of the Hamiltonian $H$ in a state $\Psi$ is an upper bound to the ground state energy.

The proof in my textbook makes sense but does it using integrals. So I tried to do it using bra-ket notation because it just seems a lot more aesthetic that way.

My attempt was this:

Consider a Hamiltonian $H$ with a complete set of orthonormal eigenfunctions $\phi_1, \phi_2, \phi_3, ...,$ with associated eigenvalues $E_1, E_2, E_3, ..., $ such that $E_1 \leq E_2 \leq E_3 \leq ...$.

Then any state of the system $\Psi$ can be expanded as a linearly combinations of the eigenfunctions, so $\Psi = \sum_{n=1}^{\infty} c_n \phi_n$.

Now consider the expectation value of the Hamiltonian in the state $\Phi$:

$$\begin{align} I &= \frac{<\Psi | H|\Psi>}{<\Psi | \Psi>} \\ & = \frac{<\sum_n c_n \phi_n | H | \sum_m c_m \phi_m >}{<\sum_n c_n \phi_n | \sum_m c_m \phi_m>} \\ & = \frac{\sum_{n,m} c_n^* c_m<\phi_n|H|\phi_m>}{\sum_{n, m} c_n^* c_m <\phi_n | \phi_m>}\end{align}$$

and then since the eigenfunctions are orthonormal, $I = \frac{\sum_n |c_n|^2 E_n}{\sum_n |c_n|^2}$. From here it is easy.

But is what I've done so far ok? Am I missing anything?

Thank you.


closed as off-topic by Kyle Kanos, GiorgioP, Jon Custer, Dvij Mankad, ZeroTheHero Apr 15 at 3:56

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  • 2
    $\begingroup$ Seems right to me. $\endgroup$ – Manvendra Somvanshi Apr 11 at 12:45
  • $\begingroup$ @ManvendraSomvanshi yayyyy!! $\endgroup$ – PhysicsMathsLove Apr 11 at 12:45
  • $\begingroup$ The expression you got is the definition of weighted average, which is expectation value in quantum mechanics. $\endgroup$ – Manvendra Somvanshi Apr 11 at 13:05
  • $\begingroup$ Why the downvote? What’s wrong with this post? $\endgroup$ – PhysicsMathsLove Apr 11 at 13:13