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

  • #1
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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


[/B]
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


[/B]
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.
 

Answers and Replies

  • #2
PeroK
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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.
 
  • #3
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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.
 
  • #4
DrClaude
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##\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.
 
  • #5
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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.
 
  • #6
PeroK
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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}##!
 
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  • #7
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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?
 
  • #8
PeroK
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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!
 
  • #9
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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:
$$ \big \langle (\hat{H} - \langle\hat{H}\rangle) \psi \mid (\hat{H} - \langle\hat{H}\rangle) \psi \big\rangle = 0 $$
$$ \big \langle \hat{H}\psi - \langle\hat{H}\rangle \psi \mid \hat{H}\psi - \langle\hat{H}\rangle \psi \big\rangle = 0 $$
$$ \Rightarrow \hat{H}\psi - \langle\hat{H}\rangle \psi = 0 \Rightarrow \hat{H}\psi = \langle\hat{H}\rangle \psi $$
 
  • #10
PeroK
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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.
 

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