Eigenvalue of the coherent state

[Tianwu Zang]

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

The coherent state is defined by \hat{a}|\alpha>=a|\alpha>, yet \alphaare not always equal to zero
Is there anything I forgot to consider?

 

[Demystifier]

I cannot see the attached matrix, but zero is not the only eigenvalue of the operator a.

 

[Tianwu Zang]

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"

 

[peteratcam]

Is there anything I forgot to consider?

Yes, the matrix is infinite dimensional.

 

[Tianwu Zang]

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-\alphaI is \alpha^{n} when the matrix is finite.
How about the determinant when it is infinite? I am not very familiar with it.
Thanks!

 

[DrDu]

You don't need the determinant! Just write down the lower right corner of your matrix eigenvalue equation: setting x1=1 you get
\sqrt{2}x_2-\alpha x_1=0, \sqrt{3}x_3-\alpha x_2=0, etc. you find that x_i=\alpha^{(i-1)}/\sqrt{i!} for arbitrary alpha!

 

[Demystifier]

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.

 

[peteratcam]

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)

 

[Demystifier]

Interesting observation, thanks!