Solving the Eigenvalue Problem: Proving \ e ^ A \psi=\ e ^\lambda\psi

[neelakash]

Homework Statement

Consider the following problem:

if \ A\psi=\lambda\psi,prove that

\ e ^ A\psi=\ e ^\lambda\psi

Homework Equations

The Attempt at a Solution

This is my attempt.Please check if I am correct.

If
\ e ^ A\psi=\ e ^\lambda\psi

is correct, we should have:

\ ln\ e^ A\psi=\ ln\ e ^\lambda\psi

or, \ ln \{e^A} +\ ln \psi=\ ln \{e^\lambda} +\ ln \psi

Now cancel \ ln \psi from both sides and post-multiply the resulting equation by \psi

That is---

\ ln \{e^A} =\ ln \{e^\lambda}

or, \ [ln \{e ^ A}]\psi=\ [ln \{e ^\lambda}]\psi

Alternatively,

\ A\psi=\lambda\psi

So, we got the given equation from the equation to be proved.

 

[neelakash]

The LaTeX image is not coming good.But I think everyone will understand my procedure.

 

[Rainbow Child]

...
Now cancel \ ln \psi from both sides and post-multiply the resulting equation by \psi You cannot "cancel". \ln \psi is a matrix.

Do you know the definition of the matrix e^A?

 

[neelakash]

I know the definition of e^A

Let A,B and C be three matrices.

If I have A+C=B+C

what is wrong if I cancel C from both sides?

ln (psi) is a matrix if you consider psi as matrix.I am usimg psi as a complex function like psi=M+iN

 

[neelakash]

I did according to your hint.That works well, but still I am having problem to see why the logaritm of a matrix/operator will not work.

 

[CompuChip]

Also, your way of setting up the proof is flawed, as the following direct analogy shows

Theorem: If 1 = 1, prove that 0 = 1.

Proof:
If 0 = 1 is correct, then we obviously also have 1 = 0. Adding the two equations gives

0 = 1
1 = 0  + 
--------
1 = 1

which is the assumption given.