Eigenvalue of the coherent state

  • #1
Hi all,
the annihilation operator satisfies the equation [tex]\hat{a}[/tex]|n>=[tex]\sqrt{n}[/tex]|n-1> and [tex]\hat{a}[/tex]|0>=0
so the matrix of [tex]\hat{a}[/tex] should be
http://www.tuchuan.com/a/2010020418032158925.jpg [Broken]
and zero is the only eigenvalue of this matrix.

The coherent state is defined by [tex]\hat{a}[/tex]|[tex]\alpha[/tex]>=a|[tex]\alpha[/tex]>, yet [tex]\alpha[/tex]are not always equal to zero
Is there anything I forgot to consider?:confused:
 
Last edited by a moderator:

Answers and Replies

  • #2
Demystifier
Science Advisor
Insights Author
Gold Member
12,216
4,557
I cannot see the attached matrix, but zero is not the only eigenvalue of the operator a.
 
  • #3
I cannot see the attached matrix, but zero is not the only eigenvalue of the operator a.

my picture cannot be displayed?
try this weblink: http://www.tuchuan.com/a/2010020418032158925.jpg" [Broken]
 
Last edited by a moderator:
  • #4
172
1
Is there anything I forgot to consider?:confused:
Yes, the matrix is infinite dimensional.
 
  • #5
Yes, the matrix is infinite dimensional.
So solving the eigenvalue of matrix in infinite dimension is not the same with the process to solve the eigenvalue of matrix in finite dimension?
I know the determinant of a-[tex]\alpha[/tex]I is [tex]\alpha[/tex][tex]^{n}[/tex] when the matrix is finite.
How about the determinant when it is infinite? I am not very familiar with it.:blushing:
Thanks!
 
  • #6
DrDu
Science Advisor
6,186
842
You don't need the determinant! Just write down the lower right corner of your matrix eigenvalue equation: setting x1=1 you get
[tex]\sqrt{2}x_2-\alpha x_1=0[/tex], [tex]\sqrt{3}x_3-\alpha x_2=0[/tex], etc. you find that [tex]x_i=\alpha^{(i-1)}/\sqrt{i!}[/tex] for arbitrary alpha!
 
  • #7
Demystifier
Science Advisor
Insights Author
Gold Member
12,216
4,557
Yes, the matrix is infinite dimensional.
Exactly.
By the way, for fermions the corresponding matrix is finite dimensional (only n=0 and n=1 contribute), which is why in field theory coherent states do not exist for fermions but only for bosons.
 
  • #8
172
1
Exactly.
By the way, for fermions the corresponding matrix is finite dimensional (only n=0 and n=1 contribute), which is why in field theory coherent states do not exist for fermions but only for bosons.
That doesn't stop people using fermionic coherent states however. If you allow the eigenvalues to be anti-commuting numbers/grassmann numbers it works just fine.
(see, eg, section 4.1.2 of Altland & Simons)
 
  • #9
Demystifier
Science Advisor
Insights Author
Gold Member
12,216
4,557
Interesting observation, thanks!
 

Related Threads on Eigenvalue of the coherent state

  • Last Post
Replies
17
Views
2K
  • Last Post
Replies
9
Views
2K
Replies
2
Views
694
Replies
5
Views
960
  • Last Post
Replies
15
Views
1K
  • Last Post
Replies
18
Views
3K
Replies
2
Views
2K
Replies
5
Views
822
  • Last Post
Replies
6
Views
3K
Top