Are There Operators in Quantum Physics Without Eigenvalue Properties?

sancharsharma
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I am new to quantum physics.
Is there is any operator in quantum physics that does not show eigenvalue property?
Give example please, thankx...
 
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An operator that has no eigenvalues you mean? Perhaps an operator which has a matrix representation that is invertible? I don't know what kind of things those operators would be...I'm also kind of fuzzy on this linear algebra. Suffice it to say, you don't see those often...
 
sancharsharma said:
I am new to quantum physics.
Is there is any operator in quantum physics that does not show eigenvalue property?
Give example please, thankx...

Yes, the simplest I can think of is the coordinate operator for a free particle.
 
bigubau said:
Yes, the simplest I can think of is the coordinate operator for a free particle.

Aren't the eigenvalues just the coordinate, and the eigenvectors just delta functions for this?

E.g. \hat{x}|x>=x|x>; <x'|x>=\delta(x'-x)

?
 
creation and annihilation operators?
 
Non-Hermitian operators, like the creation and annihilation operators, may not have real eigenvalues...but they may still have complex eigenvalues and eigenvectors, no?

Can you show that no such eigenvalues or eigenvectors exist?
 
The annihilation operator has eigenvectors, namely coherent states with an arbitrary complex number z as eigenvalue; they can be constructed rather easily by using

|z\rangle = \sum_{n=0}^\infty c_n|n\rangle

and solving for the coefficients in

a|z\rangle = z|z\rangle

Trying something similar with the creation operator fails which means that this operator has no eigenvectors.
 
yeh, annihilation and creation operator was a nice idea. thankx.

They are not eigenvalues, nor eigenvectors. See the definition on this page http://en.wikipedia.org/wiki/Eigenva...and_eigenspace
I think the definition given is pretty much mathematical than the quantum mechanics would require.
if we imagine a situation where say only a few positions are allowed (lets say the situation of an electron in a lattice where the probability of being at anywhere else except on anyone of the atoms is negligible), then in that case there is no doubt that the position operator does have eigenvalues as the locations of the atoms.
 
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