Operator, eigenstate, small calculation

[frerk]

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

1. Homework Statement

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

The Attempt at a Solution

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

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

 

[DrClaude]

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

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 [tex] e^{A} [/tex] is definded bei the Taylor expanion [tex] e^{A} = \sum\nolimits_{n=0}^\infty \frac{A^n}{n!} . [/tex]

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 [tex] e^A |a\rangle = (\sum\nolimits_{n=0}^\infty \frac{A^n}{n!} )|a\rangle ?[/tex]
DrClaude used it and you also used it in the second term.

So I will try now:

[tex] 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 [/tex]

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 :-)