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  GR8677 #57
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Verbatim question for GR8677 #57
Quantum Mechanics}Operators

One doesn't really need QM to solve this. Just plug and chug each of the five functions into the following equation,


Applying the operator -i\hbar \frac{\partial}{\partial x} to each function included in the choice, one gets:

(A) i \hbar \sin(k x) ... which isn't an eigenfunction

(B) -i \hbar \cos(k x) ... which isn't an eigenfunction

(C) -\hbar k f ... eigenvalue is off by a sign

(D) \hbar k ... this is the wanted eigenvalue!

(E) -i\hbar k ... off by a sign and imaginary term. Moreover, operators representing observables in QM have real eigenvalues.


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Comments
admin
2013-09-09 22:07:10
testNEC
katajh
2010-10-28 14:26:44
well am i wrong
or i^2 makes -1
so -(-1) gives a positive 1
and says C is the right anwer.
neon37
2010-11-03 09:40:22
you are wrong. i^2 = -1, but \frac{\partial e^{-ikx}}{\partial x} = -ik e^{-ikx}. this multiplied by -i\hbar gives -\hbar k. So C is off by a sign. Perhaps you forgot the minus sign from the operator itself?
NEC

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