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How to find eigenvalues and eigenfunction

  • Thread starter Another
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  • #1
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OP warned about not using the homework template
defind ## \hat{A}f(x)=f(-x) ## find eigenfunction and eigenvalue

I think

## \frac{d}{dx} ( \hat{A}f(x) ) = \frac{d}{dx} f(-x) ##
## \hat{A} \frac{d}{dx}f(x) + f(x) \frac{d}{dx} \hat{A} = -\frac{d}{dx} f(x)##
## \hat{A} \frac{d}{dx}f(x) + \frac{d}{dx} f(x) = -f(x) \frac{d}{dx} \hat{A}##
## (\hat{A} + 1)\frac{d}{dx} f(x) = -f(x) \frac{d}{dx} \hat{A}##

multiply by ## dx ##

## (\hat{A} + 1)d{f(x)} = -f(x) d{ \hat{A} }##
## ∫ \frac{1}{f(x)}d f(x) = - ∫ \frac{1}{\hat{A} + 1}d \hat{A}##
## \ln{f(x)} = -\ln{(\hat{A}+1)}+\ln{c} ##
## \ln{f(x)} = \ln(\frac{c}{\hat{A}+1}) ##

so...
## f(x) = \frac{c}{\hat{A}+1} ## i think it's wrong
 

Answers and Replies

  • #2
stevendaryl
Staff Emeritus
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Why are you taking derivatives?

The definition of an eigenfunction of an operator is:

##\hat{A} f(x) = \lambda f(x)##

where ##\lambda## is a number.
 
  • #3
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The answer seems to have no relation to the question ie where does the derivative come from. As stated its an interesting question with an interesting answer. Expand f(x) in a power series and write the eigenvalue equation (f(-x) = a*f(x) - a eigenvalue - f(x) eigenfuntion) - then equate terms of the same power.
 
  • #4
ehild
Homework Helper
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defind ## \hat{A}f(x)=f(-x) ## find eigenfunction and eigenvalue
Apply the operator twice - what do you get? What is ## \hat{A}(\hat{A}f(x)) ##? So what are the eigenvalues of ##\hat{A^2}##? What can be the eigenvalues of ##\hat{A}##?
 

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