Eigenvalue of the coherent state

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Discussion Overview

The discussion revolves around the eigenvalues of the annihilation operator in the context of coherent states, exploring the implications of infinite-dimensional matrices versus finite-dimensional matrices, particularly in relation to bosons and fermions.

Discussion Character

  • Technical explanation
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant asserts that the annihilation operator's matrix has zero as its only eigenvalue, referencing the equation \(\hat{a}|n>=\sqrt{n}|n-1>\).
  • Another participant challenges this claim, stating that zero is not the only eigenvalue of the operator.
  • There is a discussion about the infinite dimensionality of the matrix and how this affects the process of finding eigenvalues compared to finite dimensions.
  • One participant suggests that the determinant is not necessary for finding eigenvalues in this context and provides a method to derive the eigenvalues based on the matrix structure.
  • Participants note that for fermions, the corresponding matrix is finite dimensional, which leads to the conclusion that coherent states do not exist for fermions, although some still explore the concept using anti-commuting numbers.

Areas of Agreement / Disagreement

Participants express differing views on the eigenvalues of the annihilation operator, with no consensus reached regarding the claim that zero is the only eigenvalue. The discussion also highlights the distinction between infinite and finite dimensional matrices, indicating a lack of agreement on the implications of this difference.

Contextual Notes

The discussion includes assumptions about the nature of the matrices involved and the definitions of coherent states, which may not be universally agreed upon. The implications of using anti-commuting numbers in the context of fermionic coherent states are also mentioned but not fully explored.

Tianwu Zang
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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?:confused:
 
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I cannot see the attached matrix, but zero is not the only eigenvalue of the operator a.
 
Demystifier said:
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"
 
Last edited by a moderator:
Is there anything I forgot to consider?:confused:
Yes, the matrix is infinite dimensional.
 
peteratcam said:
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.:blushing:
Thanks!
 
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!
 
peteratcam said:
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.
 
Demystifier said:
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)
 

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