Eigenfunction and Eigenvalue Confusion (1 Viewer)

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1. The problem statement, all variables and given/known data

given the following functions:
y(x)= Acos(kx)
y(x)=A sin(kx)-Acos(kx)
y(x)=Acos(kx)+iAsin(kx)
y(x)=A d(x-x0)

Which are eigenfunctions of the position, momentum, potential energy,kinetic energy, hamiltonian, and total energy operators

2. Relevant equations

y(x) is supposed to be [tex]\psi[/tex](x)

3. The attempt at a solution
I have a list of the operators in my book, but am unsure what the question is asking for...I know that operators are supposed to be multiplied to functions and the result is the functions times a constant. I am a little cloudy on the understanding and need a little bit of help clearing it up.
 

vela

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You're just supposed to see if applying an operator to the function results in the original function times a constant. For example, take the momentum operator

[tex]\hat{p} = \frac{\hbar}{i}\frac{\partial}{\partial x}[/tex]

The function [itex]\psi(x) = \cos kx[/itex] isn't an eigenfunction because applying the operator yields

[tex]\hat{p}\psi(x) = \frac{\hbar}{i}\frac{\partial}{\partial x} \cos kx = i\hbar k\sin kx[/tex]

which isn't equal to cos kx times a constant. The function [itex]\psi(x) = e^{ikx}[/itex], however, is an eigenfunction because you have

[tex]\hat{p}\psi(x) = \hbar k e^{ikx} = \hbar k \psi(x)[/tex]

and the eigenvalue is [itex]\hbar k[/itex].
 
ok, that makes some sense, so since the position operator [x] is just x, then all of the functions are eigenfunctions...similarly, the potential energy operator is just the potential energy function. In this case the function is zero so since its =0 all the functions are eigen functions too right???
 
ok thanks, i think I have bridged the gap in my knowledge, one more quick question though...in Acos(kx)+iAsin(kx) the i doesn't play a role does it? I mean its just like a negative sign??
 

vela

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ok, that makes some sense, so since the position operator [x] is just x, then all of the functions are eigenfunctions
No, that's not correct. For example, you have

[tex]\hat{x}(A \cos kx) = Ax\cos kx[/tex]

That's not in the form of a constant multiplying the original function.
...similarly, the potential energy operator is just the potential energy function. In this case the function is zero so since its =0 all the functions are eigen functions too right???
 
wait, so x is the operator and if it multiplies times Acos(kx) or Acos(kx)+iAsin(kx) then wouldn't it be the same as multiplying x(psi)???
 

vela

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I'm not sure what you're asking. What's psi? Also, since you're using the same symbol for both the operator [itex]\hat{x}[/itex] and variable [itex]x[/itex], you need to be clear which one you're using in your expressions.
 
[tex]\psi[/tex] is psi, and i mean [x] times [tex]\psi[/tex] would be [x]Acos(kx).
 

vela

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What does [x] mean?
 
its the operator...
 

vela

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Yes, applying the operator to a function results in the function multiplied by x (which isn't a constant).
 

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