A question on eigenstates and operators

  • Thread starter vertices
  • Start date
63
0
Why can we say that:

[tex]<x'|e^{i\hat{x}}|x>=e^{ix'}\delta(x'-x) [/tex]

where where [tex]\hat{x}[/tex] is an operator?

I mean if

[tex]\hat{x}|x>=x|x> [/tex]

we may write [tex]<x'|\hat{x}|x>=x<x'|x>=x\delta(x'-x)[/tex]

but in the expression at the top, we have an exponential operator (something I've never come across before) - is |x> an eigenstate of this operator (it seems to be), and why is the eigenvalue of the operator exactly the same form as the operator?

Thanks.
 
1,726
40
You can prove it by using power series expansion

[tex]e^{i\hat{x}} = 1 + i\hat{x} - \hat{x}^2/2! + \ldots [/tex]

Eugene.
 
63
0
You can prove it by using power series expansion

[tex]e^{i\hat{x}} = 1 + i\hat{x} - \hat{x}^2/2! + \ldots [/tex]

Eugene.
Aaahh thanks Eugene. So in general if [tex]{\hat{Z}}[/tex] is any operator and |Z> the eigenfunctions of this operator, we can also say that |Z> are also the eigenfunctions of [tex]e^{i\hat{Z}}[/tex].
 
1,726
40
Aaahh thanks Eugene. So in general if [tex]{\hat{Z}}[/tex] is any operator and |Z> the eigenfunctions of this operator, we can also say that |Z> are also the eigenfunctions of [tex]e^{i\hat{Z}}[/tex].
That's right. |Z> is an eigenfunction of any function [tex]f(\hat{Z})[/tex] of the operator [tex]\hat{Z}[/tex]. The corresponding eigenvalue is [tex]f(Z)[/tex].

Eugene.
 

Related Threads for: A question on eigenstates and operators

  • Posted
Replies
2
Views
1K
Replies
12
Views
1K
Replies
5
Views
672
Replies
5
Views
2K
Replies
14
Views
2K
Replies
17
Views
828

Physics Forums Values

We Value Quality
• Topics based on mainstream science
• Proper English grammar and spelling
We Value Civility
• Positive and compassionate attitudes
• Patience while debating
We Value Productivity
• Disciplined to remain on-topic
• Recognition of own weaknesses
• Solo and co-op problem solving

Hot Threads

Top