Consider the following problem:

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

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

I did it like this:

\ A \psi=\lambda\psi

Then,\ A \ ln \psi=\lambda\ ln \psi

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


I think I am wrong at the first place...any hint how to aproach?

OK that is awefully wrong...I am trrying...

what about this:

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.

Use the definition of exponent

e^A=I+A+\frac12A^2+\ldots+\frac1{n!}A^n+\ldots

and apply it to \psi. (Logarithm will not help, because again you have to define logarithm for matrices/operators.)

But what if I consider my wave function as psi=M+iN?

But it appears that I have to use logarithm of operators...Is it not defined?I do not know...

You are given that \mathbf A\vec \psi=\lambda\vec \psi. Multiplying a matrix by a scalar is commutative: \mathbf A c = c \mathbf A. While matrix multiplication is not in general commutative, it is associative: (\mathbf A\mathbf B)\mathbf C=\mathbf A(\mathbf B\mathbf C). Use these facts to derive an expression for \mathbf A^n\,\vec \psi. Then use Temur's suggestion.

But what if I consider my wave function as psi=M+iN?

But it appears that I have to use logarithm of operators...Is it not defined?I do not know...



no---temur is right. You use that kind of expansion of operators ALL the time.

Ok, I did it after Temur's suggestion.I asked for clarity...

Thanks to all.