QM: expectation value and variance of harmonic oscillator

[renec112]

Homework Statement

A particle is moving in a one-dimensional harmonic oscillator, described by the Hamilton operator:

H = \hbar \omega (a_+ a_- + \frac{1}{2})
at t = 0 we have
\Psi(x,0) = \frac{1}{\sqrt{2}}(\psi_0(x)+i\psi_1(x))

Find the expectation value and variance of harmonic oscillator

Homework Equations

I want to use these equations. For varians:
\sigma_E^2 = \langle E^2\rangle - \langle E \rangle^2
For the energy
E_n = \hbar \omega(n+ \frac{1}{2})
\Rightarrow \langle E \rangle^2 = (\hbar \omega(n+ \frac{1}{2}))^2
and
\langle E^2\rangle = \langle \Psi | H^2 | \Psi \rangle

The Attempt at a Solution

Well i get
\ E = \hbar \omega
\langle E \rangle^2 = \hbar^2 \omega^2
and by using the operators i get
\langle E^2 \rangle = \hbar^2 \omega^2 \frac{3}{4}

which of course means i get a bad varians
\sigma_E = \sqrt{-\frac{1}{4} hbar^2 \omega^2}

Am i using the right method? And can you see where my calculations are wrong? It's quite a lot to write my calculations in with latex, so i would just like to hear if anyone can confirm or disagree with my method. I would love some input.

 

[PeroK]

You need to check how you calculated $\langle E^2 \rangle$.

 

[Orodruin]

And can you see where my calculations are wrong?

It is impossible to see where your computations have gone wrong since you have not provided the actual calculations ...

 

[renec112]

You need to check how you calculated $\langle E^2 \rangle$.

Thank you. I tried writing it in digitally very nicely, and doing i spotted a mistake. Now i get:
\langle E^2 \rangle = \hbar^2 \omega^2
That's a problem i think because that gives me
\sigma^2_E = 0

Here is what i did, I'm sorry it's not latex but i tried writing nice and clear:
https://lh3.googleusercontent.com/TrMK7uq6P6b_G8SHfqwz6RvtnFzTUOYeNUobJJM0zE9EmhhP2TKj0PXPTDV8A20cZlLia7lzulJLwDDAyXlN=w1920-h1012-rw

How am i doing?

It is impossible to see where your computations have gone wrong since you have not provided the actual calculations ...

You are right, it was a bit heavy to write in latex though. I was not sure if my method was correct, that's why i didn't add it in for now.

 

[PeroK]

Thank you. I tried writing it in digitally very nicely, and doing i spotted a mistake. Now i get:
\langle E^2 \rangle = \hbar^2 \omega^2
That's a problem i think because that gives me
\sigma^2_E = 0

Here is what i did, I'm sorry it's not latex but i tried writing nice and clear:
https://lh3.googleusercontent.com/TrMK7uq6P6b_G8SHfqwz6RvtnFzTUOYeNUobJJM0zE9EmhhP2TKj0PXPTDV8A20cZlLia7lzulJLwDDAyXlN=w1920-h1012-rw

How am i doing?

That can't be right either. Here's a suggestion:

a) First calculate $H^2 \psi_n$. Post what you get.

b) Then, calculate $H^2(\alpha \psi_0 + \beta \psi_1)$

c) Then, calculate $\langle E^2 \rangle = \langle \psi |H^2|\psi \rangle$ where $\psi = \alpha \psi_0 + \beta \psi_1$

d) Plug in what you have in this question for $\alpha, \beta$.

 

[renec112]

Thank you very much.
I was doing your steps, and at part b) i saw that i mistakenly have been doing:
H^2 = (\hbar \omega(a_+a_- +\frac{1}{2}))^2 = \hbar^2 \omega^2 ((a_+a_-)^2 + \frac{1}{4} + \frac{1}{2}a_+a_-)
and that's off course wrong, because it implies
(a+b)^2 = a^2 + b^2 + ba
it should be
(a+b)^2 = a^2 + b^2 + 2ba
giving me
H^2 =\hbar^2 \omega^2 ((a_+a_-)^2 + \frac{1}{4} + 2\frac{1}{2}a_+a_-)

That will change the last line at my attempt to
\langle E^2 \rangle = \hbar^2 \omega^2 (1/2 + 1/2 + 1/4 = \frac{5}{4} \hbar^2 \omega^2
Meaning the varians will be
\sigma_E = \sqrt{<E^2> - <E>^2} = \sqrt{\frac{5}{4} \hbar^2 \omega^2 - \hbar^2 \omega^2} = \sqrt{\frac{1}{4} \hbar^2 \omega^2}= \frac{1}{2} \hbar \omega

If that's not correct i'll do your steps. I think it's correct because it looks nice and it's from an exam question. What do you think? does it look realistic?

 

[PeroK]

If that's not correct i'll do your steps. I think it's correct because it looks nice and it's from an exam question. What do you think? does it look realistic?

That's the correct answer, but note that:

$H^2 \psi_n = H(n + \frac12)\psi_n = (n + \frac12)^2\psi_n$

Which avoids getting into the relative complexities of using the expansion for $H$.

Note also that this method is generally valid for all systems:

$H^2 \psi_n = E_n^2 \psi_n$

That said, it was probably good practice to mess about with the $a_+, a_-$ operators!

 

[renec112]

That's the correct answer, but note that:

$H^2 \psi_n = H(n + \frac12)\psi_n = (n + \frac12)^2\psi_n$

Which avoids getting into the relative complexities of using the expansion for $H$.

Note also that this method is generally valid for all systems:

$H^2 \psi_n = E_n^2 \psi_n$

That said, it was probably good practice to mess about with the $a_+, a_-$ operators!

ah much better! Thank you for the help :D