# Quick question about raising and lowering operators (ladder operators)

1. Mar 14, 2013

### PhysicsGirl90

Reading through my QM text, I came across this short piece on ladder operators that is giving me trouble (see picture). What I am struggling with is how to get to equations 2 and 3 from equation 1.

Can someone point me in the right direction? Where does the i infront of the x go?

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2. Mar 14, 2013

### tiny-tim

Hi PhysicsGirl90!
It disappeared when they multiplied the whole thing by the constant i/√(ωh)

3. Mar 15, 2013

### PhysicsGirl90

Hey tiny-tim,

Thanks for your suggestion. I tried it but i get stuck trying to get the same equation as the text. I have included what i got so far in the picture. Can you give it a look and tell me what I am doing wrong?

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4. Mar 15, 2013

### tiny-tim

Hey PhysicsGirl90!

(just got up :zzz:)

the RHS of what you got is the same as in the text

(apart from a constant, and you can always multiply an eigenvector by any constant!)

5. Mar 15, 2013

### Staff: Mentor

This is a very sloppy "derivation" of the ladder operators. First of all
is not correct, since $\hat{x}$ and $\hat{p}$ are operators. (By the way, it should be $d/dx$, not $d/d\hat{x}$.) You can get inspiration from what algebraic factoring would give, if these were ordinary variables, in order to investigate operators that look like the roots. But $a_+$ and $a_-$ are not "derived" this way.

You can also notice that $$\hat{X} = \sqrt{\frac{m \omega}{\hbar}} \hat{x}$$ and $$\hat{P} = \frac{1}{\sqrt{m \hbar \omega}} \hat{p}$$ are dimensionless.

6. Mar 15, 2013

### PhysicsGirl90

Thanks again tiny-tim and thank you DrClaude for your help

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