# Ladder Operators - I need help.

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#### studiot

Doesn't this arise because the commutator of {a+a-} = 1 ?

#### Pmb

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Doesn't this arise because the commutator of {a+a-} = 1 ?
Not that I could see.

FYI: [A, B] is the commutator of A and B while {A, B} is he anticommutator.

#### studiot

Hopefully I'll be able to do that tomorrow.

As to brackets, let's say it's that shift key at work again.

The number of times I accidentally hit the bloody capslock which is in the wrong place on my laptop and end up with a line of typing I have to redo as it is in the wrong case.

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#### Pmb

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I can place Griffiths text on my website if you want. When you download it let me to so I can delete it. Its the newer second edition too! You should see the massive physics library I have. It takes 2 DVDs to put them all on. I'll mail you a copy if you'd like.

#### benit13

Can you scan page 47 please?

#### Pmb

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Can you scan page 47 please?
Whoops! Done. A friend of mine sent me a description of what's going on. I'll figure it out after I wake up. Last edited:

#### benit13

Thanks for the extra page.

I think it's easier than it looks:

Eq. 2.57:

$$\displaystyle \hbar \omega \left(a_{\pm}a_{\mp} \pm \frac{1}{2}\right) \psi = E \psi$$

Let $$\displaystyle \psi = \psi_n$$ and $$\displaystyle E = E_n = \left(n + \frac{1}{2}\right) \hbar \omega$$ from Eq. 2.61.

Therefore

$$\displaystyle \hbar\omega \left(a_{\pm}a_{\mp} \pm \frac{1}{2}\right) \psi_n = E_n \psi_n = \left(n + \frac{1}{2}\right)\hbar\omega \psi_n$$

Cancel $$\displaystyle \hbar \omega$$ and expand brackets:

$$\displaystyle a_{\pm}a_{\mp}\psi_n \pm \frac{1}{2}\psi_n = n \psi_n + \frac{1}{2}\psi_n$$

$$\displaystyle \mp$$ the term $$\displaystyle \pm \frac{1}{2} \psi_n$$ on both sides then yields

$$\displaystyle a_{\pm}a_{\mp}\psi_n = \left(n + \frac{1}{2} \mp \frac{1}{2}\right) \psi_n$$

which expressed separately yields

$$\displaystyle a_{+}a_{-}\psi_n = n \psi_n$$ and $$\displaystyle a_{-}a_{+}\psi_n = \left(n+1\right) \psi_n$$

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#### Pmb

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Thanks for the extra page.

I think it's easier than it looks:
Thank you very much. Yep. That's what my friend showed me. He added that Griffith wasn't his usual clear self and that's how I got confused. In this particular case it appeared to me from his writing that Eq. (26,5)) followed fom what was before it. And my friend teaches QM at MIT using this text.

I wanted to slap myself once I realized how easy it was.

Thanks for you help. Last edited:

#### studiot

This 'ladder technique' is often presented in QM as something new and cunning but it is really quite old and was originated by messers Newton and Gregory several hundred years ago.

It also goes under the name the shift operator or the displacement operator.
As such it was originally developed as a result of Newton's study of finite differences.

If you think about it is is not so suprising that step up followed by step down leaves you back where you started so if A is the upshift and B is the downshift then

BA = 1.

However AB is more tricky because it depends upon where you start since there is no lower level to descend to if you start with a downshift.