QM: expectation value and variance of harmonic oscillator

In summary, the expectation value and variance of a one-dimensional harmonic oscillator can be found by first calculating H^2 \psi_n, which is equal to (n + 1/2)^2 \psi_n. Then, using the equation \langle E^2 \rangle = \langle \psi|H^2|\psi \rangle, where \psi is a linear combination of \psi_0 and \psi_1. This method can be applied to any system, and provides an accurate calculation of the variance.
  • #1
renec112
35
4

Homework Statement


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

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

Find the expectation value and variance of harmonic oscillator

Homework Equations


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

The Attempt at a Solution


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

wich of course means i get a bad varians
[tex]\sigma_E = \sqrt{-\frac{1}{4} hbar^2 \omega^2} [/tex]

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.
 
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  • #2
You need to check how you calculated ##\langle E^2 \rangle##.
 
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  • #3
renec112 said:
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 ...
 
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  • #4
PeroK said:
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:
[tex]\langle E^2 \rangle = \hbar^2 \omega^2[/tex]
That's a problem i think because that gives me
[tex]\sigma^2_E = 0[/tex]

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?
Orodruin said:
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.
 
  • #5
renec112 said:
Thank you. I tried writing it in digitally very nicely, and doing i spotted a mistake. Now i get:
[tex]\langle E^2 \rangle = \hbar^2 \omega^2[/tex]
That's a problem i think because that gives me
[tex]\sigma^2_E = 0[/tex]

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##.
 
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  • #6
Thank you very much.
I was doing your steps, and at part b) i saw that i mistakenly have been doing:
[tex]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_-)[/tex]
and that's off course wrong, becuase it implies
[tex](a+b)^2 = a^2 + b^2 + ba[/tex]
it should be
[tex](a+b)^2 = a^2 + b^2 + 2ba[/tex]
giving me
[tex]H^2 =\hbar^2 \omega^2 ((a_+a_-)^2 + \frac{1}{4} + 2\frac{1}{2}a_+a_-)[/tex]

That will change the last line at my attempt to
[tex]\langle E^2 \rangle = \hbar^2 \omega^2 (1/2 + 1/2 + 1/4 = \frac{5}{4} \hbar^2 \omega^2[/tex]
Meaning the varians will be
[tex]\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[/tex]

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?
 
  • #7
renec112 said:
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!
 
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  • #8
PeroK said:
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
 

Related to QM: expectation value and variance of harmonic oscillator

What is the expectation value of a harmonic oscillator in quantum mechanics?

The expectation value of a harmonic oscillator in quantum mechanics is the average value of its position or momentum over time. It is calculated by taking the integral of the wave function multiplied by the operator corresponding to the quantity being measured.

How is the expectation value of a harmonic oscillator related to its energy?

The expectation value of a harmonic oscillator is directly proportional to its energy. This means that as the expectation value increases, the energy also increases. The constant of proportionality is known as Planck's constant.

What is the variance of a harmonic oscillator in quantum mechanics?

The variance of a harmonic oscillator in quantum mechanics is a measure of how spread out the possible values of a quantity, such as position or momentum, are around the expectation value. It is calculated by taking the square of the difference between each possible value and the expectation value, multiplied by the probability of obtaining that value.

How do expectation values and variances change with changes in the quantum state of a harmonic oscillator?

The expectation value and variance of a harmonic oscillator are both dependent on the quantum state of the system. As the quantum state changes, the expectation value and variance will also change. In general, the expectation value will increase or decrease depending on the direction of the change in the quantum state, while the variance will decrease or increase, respectively.

How can the expectation value and variance of a harmonic oscillator be used in practical applications?

The expectation value and variance of a harmonic oscillator are important quantities in quantum mechanics and have many practical applications. For example, they can be used to calculate the energy levels of a physical system, to determine the stability of a molecule, or to predict the behavior of a quantum system under different conditions.

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