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

[Sofie RK]

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.

 

[PeroK]

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.

 

[Sofie RK]

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]

$\hat{H} - \langle\hat{H}\rangle$ 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.

 

[Sofie RK]

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]

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 $\hat{H}$!

 

[Sofie RK]

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]

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!

 

[Sofie RK]

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!

What about this:

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

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

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]

What about this:

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

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

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.