What is the Missing Step to Prove the Ladder Operator Equation?

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Homework Statement

The problem is to show that,

\hat{a_{+}}|\alpha&gt;=A_{\alpha}|\alpha+1&gt;

using

\hat{a_{+}}\hat{a_{-}}|\alpha&gt;=\alpha|\alpha&gt;It's not hard to manipulate \hat{a_{+}}\hat{a_{-}}|\alpha&gt;=\alpha|\alpha&gt; into the form,

\hat{a_{+}}\hat{a_{-}}[{\hat{a_{+}}|\alpha&gt;}]=(1+\alpha)[\hat{a_{+}}|\alpha&gt;]

But I am unable to make the connection from this to,

\hat{a_{+}}|\alpha&gt;=A_{\alpha}|\alpha+1&gt;

I know it's just using the properties of the eigenfunctions/values of a Hermatian operator at this point, but I seem to be missing exactly what that is.What am I missing?

 

[gabbagabbahey]

It's not hard to manipulate \hat{a_{+}}\hat{a_{-}}|\alpha&gt;=\alpha|\alpha&gt; into the form,

\hat{a_{+}}\hat{a_{-}}[{\hat{a_{+}}|\alpha&gt;}]=(1+\alpha)[\hat{a_{+}}|\alpha&gt;]

Well, this is an eigenvalue equation for the operator \hat{a}_+\hat{a}_-, with eigenvalue \alpha+1 and eigenstate \hat{a}_+|\alpha\rangle...but compare this to your original eigenvalue equation for this operator...surely if \alpha+1 is the eigenvalue, the eigenstate must be |\alpha+1\rangle (or at least a scalar multiple of it)...doesn't that tell you everything you need to know about \hat{a}_+|\alpha\rangle?

 

[jdwood983]

From what you've done, using both ladder operators on the ket, what does that say about N=\hat{a}_+\hat{a}_- and \hat{a}_\pm??

 

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Thanks very much for the replys.

gabba, That did occur to me, but I wasn't willing to make the concession that,

\hat{a_{+}}\hat{a_{-}}|\alpha&gt;=\alpha|\alpha&gt;

Was a general property and not \alpha specific.Is this a property of NORMALIZED eigenvectors(for which I should have specified |alpha> is defined as)? If so I suppose that would explain the A_{\alpha} as a normalization constant.


jd, I know \hat{a_{+}}\hat{a_{-}} is Hermatian although neither are individually... I'm not sure if that's what you mean.

 

[jdwood983]

jd, I know \hat{a_{+}}\hat{a_{-}} is Hermatian although neither are individually... I'm not sure if that's what you mean.

I was trying to guide you with less words than what gabba said: If

<br /> N\hat{a}_\pm|n\rangle=(n\pm1)\hat{a}_\pm|n\rangle<br />

and N|n\rangle=n|n\rangle, then \hat{a}_\pm|n\rangle are multiplicative eigenstates of |n\pm1\rangle.

 

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Thank you that helps, I'll have to stare at that for awhile to let it sink in.