Basic question about quantum: Number operator usage

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Homework Help Overview

The discussion revolves around the use of the number operator in quantum mechanics, specifically in the context of calculating the energy of a quantum harmonic oscillator. The original poster is exploring the relationship between the wave function and the number operator to determine the energy state of the system.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to apply the number operator to the wave function to find the energy state, questioning whether this method is more efficient than direct calculation. Some participants inquire about the use of LaTeX for clarity in mathematical expressions. Others suggest that the wave function must be an energy eigenstate for the approach to be valid.

Discussion Status

The discussion is ongoing, with participants providing feedback on formatting and clarifying the conditions under which the number operator can be applied. There is no explicit consensus yet, as different interpretations of the wave function's status are being explored.

Contextual Notes

Participants are discussing the definitions and implications of the number operator and its application to a wave function that may not be an energy eigenstate. There are also mentions of homework constraints and the need for clear mathematical communication.

knightpraetor
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So i was working on a homework problem to calculate energy, and i can calculate it directly. But i was thinking it should be easier to use the number operator on the wave function and find out what it's energy should be. The function involved is the initial wave state of a quantum harmonic oscillator. Anyways, it's getting late but i thought i'd ask to see if I'm thinking the right way, and try to solve it more tomorrow. so

ANyways, Asub+ * Asub- *Psi(x) = n * Psi(x)
where Psi is the wave function.

by the definitions of Asub+ and Asub-, i will end up getting
(2m *(Hamiltonian operator ) -m*omega* hbar) / (2m*omega*hbar)) * Psi = n *Psi

i then insert the hamiltonian operator, which is
- hbar^2/ 2m * d^2/dx^2 + V(x), where V(x) for a harmonic oscillator is
kx^2/2

if i do this and multiply by the wave function should it come out to N * the wave function so that i can learn what energy state the function is in?

the wave function at issue is

Psi (x,0) = A (1-2 *sqrt (mw/hbar)*x)^2 * e^ (-mwx^2/2hbar)

from griffith 2.41

i'm pretty sure i can solve it just by calculating the integral manually, but that seems inefficient..i thought since that in my notes it gave me a "number operator" and I have the energy for any given N, then that should be faster
 
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Could you write everything using LaTex ? I have troubles understanding what you wrote above.

Daniel.
 
i would love to, except i don't know what latex is or how to use it. Can you link me a tutorial on it?
 
found it..working on it
 
So i was working on a homework problem to calculate energy, and i can calculate it directly. But i was thinking it should be easier to use the number operator on the wave function and find out what it's energy should be. The function involved is the initial wave state of a quantum harmonic oscillator. Anyways, it's getting late but i thought i'd ask to see if I'm thinking the right way, and try to solve it more tomorrow. so

ANyways, [tex]a_+ a+- /Psi (x) = n * /Psi (x)[/tex]
where Psi is the wave function.

by the definitions of [tex]a_+ and a_-[/tex], i will end up getting
[tex]/frac{2m/hat(H) -m/omega */bar(h}{2m/omega/bar(h)} /Psi = n /Psi [tex] <br /> i then insert the hamiltonian operator, which is <br /> [tex]-/frac{/bar(h)^2} {2m} * /frac {d^2}{dx^2} + V(x), where V(x) for a harmonic oscillator is <br /> /frac{kx^2}{2}[/tex]<br /> <br /> if i do this and multiply by the wave function should it come out to N * the wave function so that i can learn what energy state the function is in?<br /> <br /> the wave function at issue is <br /> <br /> [tex]/Psi(x,0) = A (1-2x /sqrt(/frac{m/omega}{/bar(h)}))^2 * e^ -/frac{m/omega x^2}{2/bar(h)}[/tex]<br /> from griffith 2.41<br /> <br /> i'm pretty sure i can solve it just by calculating the integral manually, but that seems inefficient..i thought since that in my notes it gave me a "number operator" and I have the energy for any given N, then that should be faster[/tex][/tex]
 
ANyways,

[itex]a_+ a_- \psi(x) = n \psi(x)[/itex]
where [itex]\Psi[/itex] is the wave function.

by the definitions of [itex]a_+[/itex] and [itex]a_-[/itex], i will end up getting
[itex](2m \hat H -m \omega \hbar) / (2m \omega \hbar)) \Psi= n \Psi[/itex]

i then insert the hamiltonian operator, which is
[itex]- \frac{\hbar^2} {2m} \frac{d^2}{dx^2} + V(x)[/itex], where [itex]V(x)=\frac{kx^2}{2}[/itex] for a harmonic oscillator is

if i do this and multiply by the wave function should it come out to N * the wave function so that i can learn what energy state the function is in?

the wave function at issue is

[itex]\Psi (x,0) = A (1-2 \sqrt {\frac{m \omega}{\hbar}} x)^2 e^ {-\frac{m \omega x^2}{2 \hbar}}[/itex]

from griffith 2.41

i'm pretty sure i can solve it just by calculating the integral manually, but that seems inefficient..i thought since that in my notes it gave me a "number operator" and I have the energy for any given N, then that should be faster

I tried to do it for you - if something's wrong, perhaps you can quote from my posting and try to correct it yourself? Latex is very straightfroward, if you know some things. :smile:
 
Last edited:
[itex]\Psi (x,0) = A (1-2 \sqrt {\frac{m \omega}{\hbar}} x)^2 e^ {-\frac{m \omega x^2}{2 \hbar}}[/itex]
 
Last edited:
You have to do "\", not "/". :wink:
 
lol you i didn't notice it either..i just looked for other things..slashes are all the same to me:)
and thanks for the help..i'll figure it out bit by bit
 
  • #10
I think this only works if [itex]\Psi[/itex] is an energy eigenstate, which it doesn't have to be. Consider for example the wavefunction [itex]\Psi(x) = \phi_1 + \phi_2[/itex], where [itex]\phi_n[/itex] is the nth energy eigenstate, but [itex]\Psi[/itex] is no (energy) eigenstate at all.
 
Last edited:

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