Raising and lowering operators on a simple harmonic oscillator

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Homework Help Overview

The discussion revolves around the application of raising and lowering operators in the context of the harmonic oscillator Hamiltonian in quantum mechanics. The original poster is trying to understand a specific transformation involving these operators and their relation to the Hamiltonian.

Discussion Character

  • Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to clarify a transformation from one operator expression to another, seeking hints on the method to use. Some participants question the correctness of the initial equation and suggest verifying earlier notes. Others introduce the commutation relation between the operators as a potential method for the transformation.

Discussion Status

The discussion has progressed with participants providing insights into the correct forms of the Hamiltonian and the use of commutation relations. There is a collaborative effort to clarify misunderstandings, and some guidance has been offered regarding the use of the commutator to achieve the desired transformation.

Contextual Notes

The original poster acknowledges a typographical error in their initial statement, which has led to further clarification of the problem context. The discussion reflects a focus on the mathematical properties of the operators involved.

qualal
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Homework Statement


Hi, I'm currently studying for a quantum mechanics exam but I am stuck on a line in my notes:

[tex]Ha\left|\Psi\right\rangle =\hbar\omega\left(a^{t}a a + \frac{a}{2}\right)\left|\Psi\right\rangle[/tex][tex]Ha\left|\Psi\right\rangle =\hbar\omega\left(\left(a a^{t} - 1\right)a + \frac{a}{2}\right)\left|\Psi\right\rangle[/tex]
Where H is the harmonic oscillator hamiltonian

[tex]H = \frac{P^{2}}{2m} + \frac{1}{2}m\omega^{2} x^{2}[/tex]and [tex]a, a^{t}[/tex] are the lowering and raising operators respectively.

[tex]a^{t} = \frac{1}{\sqrt{2\hbar m \omega}}\left(m \omega x - i p\right)[/tex]

[tex]a= \frac{1}{\sqrt{2\hbar m \omega}}\left(m \omega x + i p\right)[/tex]

The part I don't understand is how they got from [tex]a^{t}a a[/tex] to [tex]\left(a a^{t} - 1\right)a[/tex]

I'm not afraid of doing a bit of calculating but I really need a hint as to what method I should use.
 
Last edited:
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Are you sure your first equation is correct?

The Hamiltonian has one of two forms

[tex]H=\hbar\omega(aa^t-1/2)[/tex]

[tex]H=\hbar\omega(a^ta+1/2)[/tex]

Whichever form you use, if you tack on an extra at on the right, you don't end up with what you have. Check your notes farther up.
 
Apologies I mistyped when I wrote

[tex]Ha^{t}\left|\Psi\right\rangle[/tex]

I meant to say [tex]Ha\left|\Psi\right\rangle[/tex]

does this make more sense now?

It now looks just like your second hamiltonian multiplied by "a" to me.
 
Now that you got the first line right, you can get from the first to the second line using the commutator

[tex][a, a^t]=1[/tex]
 
thanks so much! this has been bothering me all day!

I used [tex]\left[ a,a^{t}\right]= a a^{t}-a^{t} a =1[/tex]

and re-arranged to get

[tex]a a^{t}-1=a^{t} a[/tex]

which I then sub straight into the first line :-)
 

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