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March 31, 2015

March 31, 2015

Posted by **Steven** on Thursday, October 8, 2009 at 9:30pm.

Okay. My understanding is that you use the operator and perform its "thing" on the function. In this case, you will have to find the 1st derivative of sin(x)*e^(ax) ... And if the result is sin(x)*e^(ax) multiplied by some constant, it is an eigenfunction.

Is this correct? I am doing the first derivative but it doesn't show up as an eigenvalue. Neither does the 2nd derivative :(

- Quantum mechanics, eigenfunctions! -
**DrPhysics**, Thursday, October 8, 2009 at 10:05pmYou are absolutely right. To be an eigenfunction, the operator has to reproduce the function with some multiplicative constant. Even without doing a lot of work, you can see for the special case of a=0 it doesn't work because you need to take four derivatives of sine to get back to sine. The problem says "determine if" so you can just say no. :)

- Quantum mechanics, eigenfunctions! -
**Steven**, Thursday, October 8, 2009 at 10:50pmThank you! I thought I was going crazy because I wasn't able to an eigenfunction since 1st deriv. would be

= cos(x)*e^(ax) + a*sin(x)*e^(ax)

which is definitely not an eigenfunction of the operator.

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