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Quantum Harmonic Oscillator

  • Thread starter glederfein
  • Start date
  • #1

Homework Statement


Given a quantum harmonic oscillator, calculate the following values:
[itex]\left \langle n \right | a \left | n \right \rangle, \left \langle n \right | a^\dagger \left | n \right \rangle, \left \langle n \right | X \left | n \right \rangle, \left \langle n \right | P \left | n \right \rangle[/itex]


Homework Equations


Hamiltonian: [itex]H=\frac{P^2}{2m}+\frac{1}{2}m\omega^2X^2[/itex]
Ladder operators:
[itex]a=\sqrt{\frac{m\omega}{2\hbar}}\left ( X + \frac{i}{m\omega}P \right )[/itex]
[itex]a^\dagger=\sqrt{\frac{m\omega}{2\hbar}}\left(X-\frac{i}{m\omega}P \right )[/itex]
[itex]\left [ a,a^\dagger \right ] = 1[/itex]
[itex]\left [ a^\dagger a,a \right ] = -a[/itex]
N operator:
[itex]N=a^\dagger a[/itex]
[itex]N\left | n \right \rangle = n\left | n \right \rangle[/itex]
[itex]a^\dagger \left | n \right \rangle = \sqrt{n+1} \left | n+1 \right \rangle[/itex]
[itex]a \left | n \right \rangle = \sqrt{n} \left | n-1 \right \rangle[/itex]
[itex]\left | n \right \rangle = \frac{(a^\dagger)^n}{\sqrt{n!}} \left | 0 \right \rangle[/itex]


The Attempt at a Solution


[itex]\left \langle n \right | a \left | n \right \rangle = \left \langle n \right | \sqrt{n} \left | n-1 \right \rangle =
\sqrt{n} \left \langle n | n-1 \right \rangle =
\sqrt{n} \left ( \left | n \right \rangle \right ) ^ \dagger \left | n-1 \right \rangle =
\sqrt{n} \left ( \frac{(a^\dagger)^n}{\sqrt{n!}} \left | 0 \right \rangle \right ) ^ \dagger \frac{(a^\dagger)^{n-1}}{\sqrt{(n-1)!}} \left | 0 \right \rangle =
\sqrt{\frac{n}{n!(n-1)!}} \left \langle 0 \right | a^n (a^\dagger)^{n-1} \left | 0 \right \rangle[/itex]

Not sure how to continue from here...
 

Answers and Replies

  • #2
TSny
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The Attempt at a Solution


[itex]\left \langle n \right | a \left | n \right \rangle = \left \langle n \right | \sqrt{n} \left | n-1 \right \rangle =
\sqrt{n} \left \langle n | n-1 \right \rangle =
\sqrt{n} \left ( \left | n \right \rangle \right ) ^ \dagger \left | n-1 \right \rangle =
\sqrt{n} \left ( \frac{(a^\dagger)^n}{\sqrt{n!}} \left | 0 \right \rangle \right ) ^ \dagger \frac{(a^\dagger)^{n-1}}{\sqrt{(n-1)!}} \left | 0 \right \rangle =
\sqrt{\frac{n}{n!(n-1)!}} \left \langle 0 \right | a^n (a^\dagger)^{n-1} \left | 0 \right \rangle[/itex]
.
When you get to [itex]\left \langle n \right | a \left | n \right \rangle = \left \langle n \right | \sqrt{n} \left | n-1 \right \rangle =
\sqrt{n} \left \langle n | n-1 \right \rangle [/itex] you should be able to see the answer.
 
  • #3
When you get to [itex]\left \langle n \right | a \left | n \right \rangle = \left \langle n \right | \sqrt{n} \left | n-1 \right \rangle =
\sqrt{n} \left \langle n | n-1 \right \rangle [/itex] you should be able to see the answer.
Is the answer zero because eigenvectors are always perpendicular to one another?
Doesn't that mean that all the four values in the question are zero?
 
  • #4
TSny
Homework Helper
Gold Member
12,530
2,951
Yes, eigenvectors of a Hermitian operator that correspond to different eigenvalues are orthogonal.

And, yes, all 4 are zero :smile:
 

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