Eigenvalue Condition

[eNtRopY]

I was recently asked to explain the eigenvalue condition, but I'm sure exactly which condition the inquirer was asking about.

Are any of you nerds familiar with the Eigenvalue Condition?

If so, please enlighten me.

eNtRopY

[jonnylane]

eigenvalues are (in physics) values that define the output of a quantum mechanical operator in an equation.

There is a mathematical meaning, and i think it is pretty much the same thing i.e. mathematical operators in equations.

what was the context of the problem?

[HallsofIvy]

Mathematics: If A is a linear operator on a vector space, the
"eigenvalue" equation is Ax= [lambda]x. x= 0 is a "trivial" solution. If there exist non-trivial (i.e. non zero) solutions [lambda] is an eigenvalue of A.

Physics: As "jonnylane" said, in quantum physics, various possible measurements (position, momentum) are interpreted as linear operators. The only possible specific numerical results of such measurements are eigenvalues of the linear operators. That may be what your inquirer was asking about.

[eNtRopY]

I think that the inquirer was asking about something else. I think that he meant something specifically related to the boundary value of a QM problem. I was just wondering if there was an official Eigenvalue Condition. I see now that there is not.

eNtRopY

[Loren Booda]

Perhaps the operator/eigenvalue postulate of quantum mechanics?

Most likely an eigenvalue boundary condition, though.

[ahrkron]

Originally posted by eNtRopY
I think that the inquirer was asking about something else. I think that he meant something specifically related to the boundary value of a QM problem. I was just wondering if there was an official Eigenvalue Condition. I see now that there is not.

The closest I can think of (as far as an "official" condition) would be the equation
det(A-lambda I) == 0.

Which is used to determine the eigenvalues (lambda) of the linear operator A.

However, just as that, it has nothing to do with QM.