What are the eigenvalues of a non-Hermitian operator?

[soul]

Hi, everyone!

While I was studying for my midterm, I encountered this question.
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Consider the hermitian operator H that has the property that
H⁴ = 1
What are the eigenvalues of the operator H?
What are the eigenvalues if H is not restricted to being Hermitian?
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What I am going to ask is that does it matter this operator have a different power like H⁶ or H⁵? I mean what is the role of the power in this question? Also, I couldn't figure out that how I can find the eigenvalues of such operator when it is not Hermitian?

Thank you.

 

[dextercioby]

Assume a finite dimensional vector space on which H acts. Can you write the characteristic equation for H ?

 

[student111]

I think you can do it simple like this:

Let E be an eigenvector of H with eigenvalue e:

H E = e E

Then H^4 E = e H^3 E = e^2 H^2 E = ...

And use H^4 E = E since H^4 = 1.

This gives you an general equation for the eigenvalues. Now when H is hermitian the eigenvalues must be real, and this restricts the solutions to the equation obtained.

 

[soul]

Assume a finite dimensional vector space on which H acts. Can you write the characteristic equation for H ?

Isn't it H*\Psi = E * \Psi?

 

[soul]

Also, can I say that since H^4 = 1, then one eigenvalue of it is 1 and if H were not a Hermitian operator, one eigenvalue would be i, since H^2 = -1 or 1?

 

[student111]

Let me just post the solution, u are very close anyway:

Let E be an eigenvector of H with eigenvalue e:

H E = e E

Then we have:

H^4 E = e H^3 E = e^2 H^2 E = e^3 H E = e^4 E.

You simply let one H act on E in each step.

Since you have H^4 = 1 we have:

H^4 E = E = e^4 E.

So all you need to solve is:

e^4 = 1.

If H is hermitian it has 4 solutions. But if H is not Hermitian the complex eigenvalues are ruled out and you only have the real solutions.

 

[soul]

Student111, thanks for your answer, but I guess H has 2 values if H is hermitian, if it isn't, it has 4?

 

[student111]

exactly, i mistyped.

Nonhermitian: 1, -1, i, -i.

And only 1,-1 if it is hermitian.

 

[turin]

... I couldn't figure out that how I can find the eigenvalues of such operator when it is not Hermitian?

That's a good thing. Not all operators are guarunteed to be diagnonalizable, but Hermitian is an example that is. Of course, since a power of the operator is unity (i.e. the identity operator), they are hinting that H is at least unitary, which is also diagonalizable.