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A problem with Ladder operators

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Homework Statement



The problem is to show that,

[tex]\hat{a_{+}}|\alpha>=A_{\alpha}|\alpha+1>[/tex]

using

[tex]\hat{a_{+}}\hat{a_{-}}|\alpha>=\alpha|\alpha>[/tex]


It's not hard to manipulate [tex]\hat{a_{+}}\hat{a_{-}}|\alpha>=\alpha|\alpha>[/tex] into the form,

[tex]\hat{a_{+}}\hat{a_{-}}[{\hat{a_{+}}|\alpha>}]=(1+\alpha)[\hat{a_{+}}|\alpha>][/tex]

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

[tex]\hat{a_{+}}|\alpha>=A_{\alpha}|\alpha+1>[/tex]

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?
 

Answers and Replies

  • #2
gabbagabbahey
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It's not hard to manipulate [tex]\hat{a_{+}}\hat{a_{-}}|\alpha>=\alpha|\alpha>[/tex] into the form,

[tex]\hat{a_{+}}\hat{a_{-}}[{\hat{a_{+}}|\alpha>}]=(1+\alpha)[\hat{a_{+}}|\alpha>][/tex]
Well, this is an eigenvalue equation for the operator [itex]\hat{a}_+\hat{a}_-[/itex], with eigenvalue [itex]\alpha+1[/itex] and eigenstate [itex]\hat{a}_+|\alpha\rangle[/itex]...but compare this to your original eigenvalue equation for this operator....surely if [itex]\alpha+1[/itex] is the eigenvalue, the eigenstate must be [itex]|\alpha+1\rangle[/itex] (or at least a scalar multiple of it).....doesn't that tell you everything you need to know about [itex]\hat{a}_+|\alpha\rangle[/itex]?:wink:
 
  • #3
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From what you've done, using both ladder operators on the ket, what does that say about [itex]N=\hat{a}_+\hat{a}_-[/itex] and [itex]\hat{a}_\pm[/itex]??
 
  • #4
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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,

[tex]\hat{a_{+}}\hat{a_{-}}|\alpha>=\alpha|\alpha> [/tex]

Was a general property and not [tex]\alpha[/tex] 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 [tex]A_{\alpha}[/tex] as a normalization constant.


jd, I know [tex]\hat{a_{+}}\hat{a_{-}}[/tex] is Hermatian although neither are individually..... I'm not sure if that's what you mean.
 
  • #5
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jd, I know [tex]\hat{a_{+}}\hat{a_{-}}[/tex] 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

[tex]
N\hat{a}_\pm|n\rangle=(n\pm1)\hat{a}_\pm|n\rangle
[/tex]

and [itex]N|n\rangle=n|n\rangle[/itex], then [itex]\hat{a}_\pm|n\rangle[/itex] are multiplicative eigenstates of [itex]|n\pm1\rangle[/itex].
 
  • #6
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Thank you that helps, I'll have to stare at that for awhile to let it sink in.
 

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