Eigenvalue of the coherent state

In summary, the annihilation operator satisfies the equation \hat{a}|n>=\sqrt{n}|n-1> and \hat{a}|0>=0. The matrix of \hat{a} should be infinite dimensional and zero is not the only eigenvalue. The coherent state, defined by \hat{a}|\alpha>=a|\alpha>, does not always have \alpha=0. The determinant is not needed for infinite dimensional matrices, and for fermions the matrix is finite dimensional and coherent states do not exist. However, they can still be used if the eigenvalues are allowed to be anti-commuting numbers.
  • #1
Tianwu Zang
11
0
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
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:
 
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  • #2
I cannot see the attached matrix, but zero is not the only eigenvalue of the operator a.
 
  • #3
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"
 
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  • #4
Is there anything I forgot to consider?:confused:
Yes, the matrix is infinite dimensional.
 
  • #5
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-[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
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
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.
 
  • #8
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)
 
  • #9
Interesting observation, thanks!
 

1. What is the concept of eigenvalue in the context of the coherent state?

The eigenvalue of the coherent state refers to the value of the operator that corresponds to the state. In other words, it is the value that the state takes when acted upon by a certain operator.

2. How is the eigenvalue of the coherent state calculated?

The eigenvalue of the coherent state can be calculated by using the Schrödinger equation and solving for the eigenvalue that satisfies the equation. Alternatively, it can also be calculated by using the ladder operators in the coherent state representation.

3. What is the significance of the eigenvalue of the coherent state in quantum mechanics?

The eigenvalue of the coherent state is significant because it represents the possible outcomes of a measurement on the state. It also plays a crucial role in determining the time evolution of the state.

4. Can the eigenvalue of the coherent state take on any value?

No, the eigenvalue of the coherent state is restricted to a discrete set of values determined by the state's properties and the operators acting upon it. These values are often referred to as the energy levels of the state.

5. How does the eigenvalue of the coherent state differ from other types of eigenvalues?

The eigenvalue of the coherent state is unique in the sense that it is a continuous variable, unlike the discrete eigenvalues of other states. This is due to the fact that the coherent state is a superposition of all possible energy levels, allowing for a continuous range of eigenvalues.

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