# Prove that ##\psi## is a solution to Schrödinger equation

## Homework Statement

For a wavefunction ##\psi##, the variance of the Hamiltonian operator ##\hat{H}## is defined as:

$$\sigma^2 = \big \langle \psi \mid (\hat{H} - \langle\hat{H}\rangle)^2 \psi \big\rangle$$

I want to prove that if ##\sigma^2 = 0##, then ##\psi## is a solution to the Schrödinger equation (##\hat{H}\psi = E\psi##).

##\psi## is a normalized wavefunction.

## Homework Equations

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Let ##\phi## be a wavefunction. The norm ##\parallel \phi \parallel = \sqrt{\langle \phi \mid \phi \rangle} ## is equal to zero if and only if the wavefunction ##\phi## is equal to zero (##\phi = 0 \Leftrightarrow \parallel \phi \parallel = 0) ##

Given hint: ##\hat{H} - \langle\hat{H}\rangle## is a hermitian operator.

## The Attempt at a Solution

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I think maybe the equation can be written as:

$$\sigma^2 = (\hat{H}\psi - E\psi)\psi$$

I know that ##\sigma^2 = 0##.

##\psi## is a normalized wavefunction ##\Rightarrow \sqrt{\langle \psi\mid\psi \rangle} = 1##, and ##\psi \neq 0##. Then, ##\hat{H}\psi = E\psi = 0##, and ##\psi## is a solution to the Schrödinger equation.

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PeroK
Homework Helper
Gold Member
What you've written doesn't make much sense to me, I'm afraid. Note that ##\langle \hat{H} \rangle## is a real number. Although in this notation it also means the operator which is this multiple of the identity operator.

You need to look at what ##(\hat{H} - \langle \hat{H} \rangle)^2## means. And use the hint.

What you've written doesn't make much sense to me, I'm afraid. Note that ##\langle \hat{H} \rangle## is a real number. Although in this notation it also means the operator which is this multiple of the identity operator.

You need to look at what ##(\hat{H} - \langle \hat{H} \rangle)^2## means. And use the hint.
##\hat{H} - \langle\hat{H}\rangle ## is a hermitian operator, which means that its eigenvalues must be real. But I don't understand what happens when a hermitian operator is squared.

DrClaude
Mentor
##\hat{H} - \langle\hat{H}\rangle ## is a hermitian operator, which means that its eigenvalues must be real. But I don't understand what happens when a hermitian operator is squared.
An operator squared is simply the operator applied twice. So you can split it, ##\hat{A}^2 = \hat{A} \hat{A}## and then use its hermiticity.

An operator squared is simply the operator applied twice. So you can split it, ##\hat{A}^2 = \hat{A} \hat{A}## and then use its hermiticity.
Hermitian operators have real eigenvalues and orthogonal eigenfunctions.

$$(\hat{H} - \langle \hat{H} \rangle)\psi = \lambda\psi,$$ where ##\lambda## is a real number. Is this something I can use?

Then,

$$\sigma^2 = \big\langle \psi \mid (\hat{H} - \langle\hat{H}\rangle)(\hat{H} - \langle\hat{H}\rangle)\psi \big\rangle$$

$$\sigma^2 = \lambda \big\langle \psi \mid (\hat{H} - \langle\hat{H}\rangle) \psi \big\rangle$$

$$\sigma^2 = \lambda^2 \big\langle \psi \mid \psi \big\rangle$$

But this does not makes sense either because then ##\lambda## has to be zero.

PeroK
Homework Helper
Gold Member
Hermitian operators have real eigenvalues and orthogonal eigenfunctions.

$$(\hat{H} - \langle \hat{H} \rangle)\psi = \lambda\psi,$$ where ##\lambda## is a real number. Is this something I can use?

Then,

$$\sigma^2 = \big\langle \psi \mid (\hat{H} - \langle\hat{H}\rangle)(\hat{H} - \langle\hat{H}\rangle)\psi \big\rangle$$

$$\sigma^2 = \lambda \big\langle \psi \mid (\hat{H} - \langle\hat{H}\rangle) \psi \big\rangle$$

$$\sigma^2 = \lambda^2 \big\langle \psi \mid \psi \big\rangle$$

But this does not makes sense either because then ##\lambda## has to be zero.
You need to get some concepts clear:

A Hermitian operator, ##\hat{H}## is defined as an operator for which:

##\forall \psi, \phi: \ \langle \phi | \hat{H} \psi \rangle = \langle \hat{H} \phi | \psi \rangle##

You can show that the eigenvalues of a Hermitian operator are real and that the eigenvectors corresponding to different eigenvalues are orthogonal.

But, Hermitian operators do not have a monopoly on real eigenvalues, or othogonal eigenvectors!

Also, the question did not say that ##\psi## was an eigenvector of ##\hat{H}##, which is what you effectively assumed. In other words, if you have a Hermitian operator, ##\hat{H}##, you cannot just pick the first vector you find lying around, ##\psi## say, and declare that ##\psi## is an eigenvector of ##\hat{H}##!

DrClaude
You need to get some concepts clear:

A Hermitian operator, ##\hat{H}## is defined as an operator for which:

##\forall \psi, \phi: \ \langle \phi | \hat{H} \psi \rangle = \langle \hat{H} \phi | \psi \rangle##

Suggestion:

$$\Big\langle \psi \mid (\hat{H} - \langle\hat{H}\rangle)(\hat{H} - \langle\hat{H}\rangle)\psi \Big\rangle = \Big\langle (\hat{H} - \langle\hat{H}\rangle)\psi \mid (\hat{H} - \langle\hat{H}\rangle)\psi \Big\rangle$$

$$= \Big\langle \hat{H}\psi - \langle\hat{H}\rangle\psi \mid \hat{H}\psi - \langle\hat{H}\rangle\psi \Big\rangle$$

Now, I remember that an expectation value is not an observable, but a mean value of the observables/eigenvalues given the probability to get the value. I want to prove that if ##\sigma^2 = 0##, then ##\psi## is a solution to the Schrödinger equation, which means that ##\hat{H}\psi = E\psi##. I am thinking that if I can argue that ##\hat{H}\psi = \langle\hat{H}\rangle\psi## or/if ##\langle\hat{H}\rangle\psi = E\psi##, then I have proved that ##\psi## is a solution to the Schrödinger equation.

Are these equations and arguments right? Am I missing anything so far?

PeroK
Homework Helper
Gold Member
Suggestion:

$$\Big\langle \psi \mid (\hat{H} - \langle\hat{H}\rangle)(\hat{H} - \langle\hat{H}\rangle)\psi \Big\rangle = \Big\langle (\hat{H} - \langle\hat{H}\rangle)\psi \mid (\hat{H} - \langle\hat{H}\rangle)\psi \Big\rangle$$

$$= \Big\langle \hat{H}\psi - \langle\hat{H}\rangle\psi \mid \hat{H}\psi - \langle\hat{H}\rangle\psi \Big\rangle$$

Now, I remember that an expectation value is not an observable, but a mean value of the observables/eigenvalues given the probability to get the value. I want to prove that if ##\sigma^2 = 0##, then ##\psi## is a solution to the Schrödinger equation, which means that ##\hat{H}\psi = E\psi##. I am thinking that if I can argue that ##\hat{H}\psi = \langle\hat{H}\rangle\psi## or/if ##\langle\hat{H}\rangle\psi = E\psi##, then I have proved that ##\psi## is a solution to the Schrödinger equation.

Are these equations and arguments right? Am I missing anything so far?
First, there's something about the way you do the maths that does sort of mean you miss things. In this case, you are given:

$$\sigma^2 = \big \langle \psi \mid (\hat{H} - \langle\hat{H}\rangle)^2 \psi \big\rangle$$

And, you need to show that if ##\sigma^2 = 0##, then ##\psi## is a solution to the Schrödinger equation (##\hat{H}\psi = E\psi##).

But, you never once wrote, as your starting point:

$$\big \langle \psi \mid (\hat{H} - \langle\hat{H}\rangle)^2 \psi \big\rangle = 0$$

You seem to leave ##\sigma^2 = 0## in the background somewhere. It never quite makes it into the equations somehow!

First, there's something about the way you do the maths that does sort of mean you miss things. In this case, you are given:

$$\sigma^2 = \big \langle \psi \mid (\hat{H} - \langle\hat{H}\rangle)^2 \psi \big\rangle$$

And, you need to show that if ##\sigma^2 = 0##, then ##\psi## is a solution to the Schrödinger equation (##\hat{H}\psi = E\psi##).

But, you never once wrote, as your starting point:

$$\big \langle \psi \mid (\hat{H} - \langle\hat{H}\rangle)^2 \psi \big\rangle = 0$$

You seem to leave ##\sigma^2 = 0## in the background somewhere. It never quite makes it into the equations somehow!

$$\sigma^2 = \big \langle \psi \mid (\hat{H} - \langle\hat{H}\rangle)^2 \psi \big\rangle$$

$$\sigma^2 = 0 \Rightarrow \big \langle \psi \mid (\hat{H} - \langle\hat{H}\rangle)^2 \psi \big\rangle = 0$$

Hermiticity:
$$\big \langle (\hat{H} - \langle\hat{H}\rangle) \psi \mid (\hat{H} - \langle\hat{H}\rangle) \psi \big\rangle = 0$$
$$\big \langle \hat{H}\psi - \langle\hat{H}\rangle \psi \mid \hat{H}\psi - \langle\hat{H}\rangle \psi \big\rangle = 0$$
$$\Rightarrow \hat{H}\psi - \langle\hat{H}\rangle \psi = 0 \Rightarrow \hat{H}\psi = \langle\hat{H}\rangle \psi$$

PeroK
Homework Helper
Gold Member

$$\sigma^2 = \big \langle \psi \mid (\hat{H} - \langle\hat{H}\rangle)^2 \psi \big\rangle$$

$$\sigma^2 = 0 \Rightarrow \big \langle \psi \mid (\hat{H} - \langle\hat{H}\rangle)^2 \psi \big\rangle = 0$$

Hermiticity:
$$\big \langle (\hat{H} - \langle\hat{H}\rangle) \psi \mid (\hat{H} - \langle\hat{H}\rangle) \psi \big\rangle = 0$$
$$\big \langle \hat{H}\psi - \langle\hat{H}\rangle \psi \mid \hat{H}\psi - \langle\hat{H}\rangle \psi \big\rangle = 0$$
$$\Rightarrow \hat{H}\psi - \langle\hat{H}\rangle \psi = 0 \Rightarrow \hat{H}\psi = \langle\hat{H}\rangle \psi$$
Yes, that's it.