QM I - Decomposition of countable basic states into coherent states

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SUMMARY

The discussion focuses on the decomposition of countable basic states \(\left|n\right\rangle\) into coherent states \(\left|λ\right\rangle\) in quantum mechanics. The lowering operator \(\hat{a}\) is defined, and the participant attempts to express \(\left|n\right\rangle\) as a sum of coherent states. The solution involves applying the lowering operator multiple times, leading to the realization that the coherent state basis is overcomplete, necessitating the use of a coherent state resolution of unity, which integrates over the complex \(\lambda\) plane.

PREREQUISITES
  • Understanding of quantum mechanics, specifically coherent states
  • Familiarity with operators in quantum mechanics, particularly lowering operators
  • Knowledge of basis decomposition in Hilbert spaces
  • Experience with integrals in complex analysis
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  • Study the properties of coherent states in quantum mechanics
  • Learn about the resolution of unity for coherent states
  • Explore the action of raising and lowering operators on quantum states
  • Investigate the implications of overcomplete bases in quantum systems
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Homework Statement



Consider a quantum system with a countable number of basic states [itex]\left|n\right\rangle[/itex].
Calculate the decomposition into a basis of coherent states [itex]\left|λ \right\rangle[/itex] all obeying [itex]\hat{a}[/itex] [itex]\left|λ \right\rangle[/itex] = λ [itex]\left|λ \right\rangle[/itex]


Homework Equations



[itex]\hat{a}[/itex] is the lowering operator:
[itex]\hat{a} \left|n\right\rangle[/itex] = √n [itex]\left|n-1\right\rangle[/itex]



The Attempt at a Solution



Because [itex]\left|λ\right\rangle[/itex] form a basis, i can equate [itex]\left|n\right\rangle[/itex] = Ʃλ[itex]_{n}\left|λ\right\rangle[/itex].
Applying the lowering operator n-times to both sides of the equation, i get: √n! [itex]\left|0\right\rangle[/itex] = λ[itex]^{n}[/itex] Ʃλ[itex]_{n}[/itex] [itex]\left|λ\right\rangle[/itex]
By equality of two vectors i can say that √n! = λ[itex]^{n}[/itex] and that [itex]\left|0\right\rangle[/itex] = Ʃλ[itex]_{n}[/itex] [itex]\left|λ\right\rangle[/itex].

Now i got kinda stuck. I thought if i get to the [itex]\left|0\right\rangle[/itex] vector, i can just keep applying the raising operator to get any state [itex]\left|n\right\rangle[/itex] written in my new vectors [itex]\left|λ\right\rangle[/itex]. But i realized i do not know how the raising operator acts on them. Neither do i know if i chose the right approach, but it feels like the only thing i could have done, considering the information given.

I would really appreciate some help.
Thanks a lot in advance!
 
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I'm a little confused here as to whether you're trying to express a coherent state as a sum of the |n>'s, or one of the |n>'s as a sum of coherent states :confused:

The former is easy enough, but the latter is trickier - the answer won't be unique, because the coherent state basis is an overcomplete basis. Probably you'll have to use a coherent state resolution of unity, which involves integrating over the complex λ plane.
 

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