Understanding the Schrodinger's equation by variational principle

Question

I reviewed part of my notes in the quantum mechanics class, and still have a few questions about the variational derivation of the Schrodinger's equation:

The variational principle says that the expectation value of $H$ in any state $⟨\psi|H|\psi⟩$ is greater or equal to the ground state energy $E_{min}$. Given the arbitrary state is normalized: $\int\psi(\vec{r})^*\psi(\vec{r}) d\vec{r} = 1$, the claim is that the Schrodinger's equation could be derived from the minimum of the integral: $\int\psi^*(\vec{r})H\psi(\vec{r}) d\vec{r}$. (My paraphrase might not be accurate.)

How can I understand that? How's the derivation relevant to the Lagrangian in classical mechanics? Why do we care about that integral, and how's the integral relevant to the expression of expectation value?

Thanks!!

Answer 1

The spirit of

1. the variational method in QM of the functional $S=\frac{\langle \psi| H-E |\psi\rangle}{\langle \psi|\psi\rangle}$ to find the groundstate wavefunction and groundstate energy, and

2. the principle of stationary action in Lagrangian mechanics

are quite different. The functional $S$ in the QM method (1) is non-local (or in a reformulation it contains an undetermined Lagrange multiplier). Not surprisingly, the corresponding constrained Euler-Lagrange (EL) equation$^1$ essentially boils down to the TISE, which we already knew. In practice, one can not solve the TISE exactly, so that the method (1) instead relies on a class of approximating test functions.

Unrelated to OP's core question, we can't help mentioning that it is possible to construct a local action, whose EL equation is the TDSE, see e.g. this & this Phys.SE post.

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$^1$ A technical note: When we perform the variation, the ket $|\psi\rangle$ and bra $\langle \psi|$ can be treated as independent, cf. e.g. this Phys.SE post.

Answer 2

So as you stated you want to find the Schrodinger equation by minimising the integral

$\int\psi^*(\vec{r})H\psi(\vec{r}) d\vec{r}$

This integral describes the expectation value for the energy, the quantity you want to minimise (since you are considering the ground state, which is $E_{min}$) to get the Schrodinger equation.

In classical mechanics we have the function called the Lagrangian. From Hamilton's Principle of Least Action we minimise the integral of this Lagrangian (which gives us the Euler-Lagrange equations) to find the trajectory of particle(s), or equations of motion, in the system we are investigating (note this process is not limited to classical mechanics - for example it is applicable to fields).

Answer 3

Honestly, the discussion here has been above my head in some respects, so I attempt an answer with great trepidation. But anyway, the expectation of energy is, as you say

$ \int \Psi^*H\Psi d^3x $

Now, we would like to make a variation on $\Psi$ such that

$\int \Psi^* \Psi d^3x = 1$

and remains 1 after our variation. We introduce this restriction with the usual technique of a Lagrange multiplier. So we will actually perform a variation on

$ \int \Psi^* H\Psi d^3x - \lambda (\int \Psi^*\Psi d^3x$ -1) =

$ \int (\Psi^* H\Psi - \lambda \Psi^*\Psi) d^3x +\lambda $

Do a variation on $\Psi^*$ (allowed independently of $\Psi$ as discussed in one of the links in the discussion of your question) and set to zero since we are looking/hoping for an extremal integrand:

$H\Psi - \lambda\Psi = 0$

So the integrand is extremal if $\Psi$ is an eigenfunction of $H$, and $\lambda$ must take on one of the eigenvalues of $H$, which is to say one of the allowed energy values. So we have Schroedinger's time independent equation.