Operator, eigenstate, small calculation

[frerk]

Hello :-) I have a small question for you :-)

1. Homework Statement

The Operator $$e^{A}$$ is definded bei the Taylor expanion $$e^{A} = \sum\nolimits_{n=0}^\infty \frac{A^n}{n!} .$$
Prove that if $$|a \rangle$$ is an eigenstate of A, that is if $$A|a\rangle = a|a\rangle$$, then $$|a\rangle$$ is an eigenstate of $$e^{A}$$ with the eigenvalue of $$e^{a}.$$

The Attempt at a Solution

I show you a very bad attempt:
$$A|a \rangle = a|a \rangle\quad\quad\quad| : |a\rangle |e^{...}$$
$$e^A =e^a \quad\quad\quad| * |a\rangle$$
$$e^A|a\rangle = e^a|a\rangle$$

I would be glad about an info how to do it right... thank you :)

 

[DrClaude]

I show you a very bad attempt:
$$A|a \rangle = a|a \rangle\quad\quad\quad| : |a\rangle |e^{...}$$
$$e^A =e^a \quad\quad\quad| * |a\rangle$$
$$e^A|a\rangle = e^a|a\rangle$$

I don't even understand what that means.

Start with $e^A|a\rangle$ and use the Taylor expansion of the exponential.

 

[S.G. Janssens]

The Operator $$e^{A}$$ is definded bei the Taylor expanion $$e^{A} = \sum\nolimits_{n=0}^\infty \frac{A^n}{n!} .$$

This is a correct definition when $A$ is a bounded operator. However, operators occurring in QM (is this a QM exercise?) are typically not bounded, but they are usually self-adjoint, so you can use a suitable functional calculus.

I would be glad about an info how to do it right... thank you :)

In view of the above remark, let us assume in addition that $A$ is bounded. Then just apply the series definition, so the first two steps become:
$$
e^A|a\rangle = \left(\sum_{n=0}^{\infty}{\frac{A^n}{n!}}\right)|a\rangle = \sum_{n=0}^{\infty}{\frac{A^n|a\rangle}{n!}}
$$
You continue.

 

[frerk]

Start with $e^A|a\rangle$ and use the Taylor expansion of the exponential.

ok, thank you :-)

This is a correct definition when $A$ is a bounded operator. However, operators occurring in QM (is this a QM exercise?) are typically not bounded, but they are usually self-adjoint, so you can use a suitable functional calculus.

yes it is QM. ok thanks

In view of the above remark, let us assume in addition that $A$ is bounded. Then just apply the series definition, so the first two steps become:
$$
A|a\rangle = \left(\sum_{n=0}^{\infty}{\frac{A^n}{n!}}\right)|a\rangle = \sum_{n=0}^{\infty}{\frac{A^n|a\rangle}{n!}}
$$
You continue.

I think you mean $$e^A |a\rangle = (\sum\nolimits_{n=0}^\infty \frac{A^n}{n!} )|a\rangle ?$$
DrClaude used it and you also used it in the second term.

So I will try now:

$$e^A |a\rangle = (\sum\nolimits_{n=0}^\infty \frac{A^n}{n!}) |a\rangle = \sum\nolimits_{n=0}^\infty \frac{A^n|a\rangle }{n!} = \sum\nolimits_{n=0}^\infty \frac{a^n|a\rangle }{n!} = (\sum\nolimits_{n=0}^\infty \frac{a^n}{n!}) |a\rangle = e^a |a\rangle$$

It think, that looks fine? :-)

 

[S.G. Janssens]

I think you mean

Thank you, fixed. I think you would have known how to do this exercise.

It think, that looks fine? :-)

Yes, much better!

 

[frerk]

thank you both :-)