Finding eigenstate for the annhilation operator

In summary, the conversation discusses finding the eigenvector of the annihilation operator a and using an arbitrary wavefunction to show how the operator works. The conversation also explores the relationship between n and k, and how to express each term in the sum for |V> in terms of the creation operator a*. Further discussion includes the action of a* on the vacuum state and expressing each term in the sum for |V> in terms of a* acting on the vacuum.
  • #1
Sandrasa
2
0

Homework Statement


Find the eigenvector of the annhilation operator a.

Homework Equations


[itex]a|n\rangle = \sqrt{n}|{n-1}\rangle[/itex]

The Attempt at a Solution


Try to show this for an arbitrary wavefunction:
[itex]|V\rangle = \sum_{n=1}^\infty c_{n}|n\rangle [/itex]

[itex] a|V\rangle = a\sum_{n=1}^\infty c_{n}|n\rangle = \sum_{n=1}^\infty \sqrt{n} c_{n}|n-1\rangle[/itex]

Define a new constant, k = n - 1, and put this in:

- > [itex]\sum_{k=0}^\infty \sqrt{k + 1} c_{k + 1}|k\rangle = \sum_{n=0}^\infty \sqrt{n + 1} c_{n + 1}|n\rangle [/itex]

This was a trick I found in my notes from a lecture, but I do not quite understand how you can say that k = n in the last part. Is it just because k and n are an arbitrary notation for the basis set? Or is it due to the fact that since the sum goes to infity, the difference of 1 does not matter (since I really defined k to be n - 1)?

When I continue with this, I get:

[itex]a( c_{1}|1\rangle + c_{2}|2\rangle + c_{3}|3\rangle + ...) = c( \sqrt{1}c_{1}|0\rangle\ + \sqrt{2}c_{2}|1\rangle + \sqrt{3}c_{3}|2\rangle + \sqrt{4}c_{4}|3\rangle) [/itex]

I am not really sure where to go next. I tried setting the coeffecient before each basisvector equal, like this:

[itex] c\sqrt{1}c_{1} = 0 [/itex]
[itex] c_{1} = \sqrt{2}c_{2}c [/itex]
[itex] c_{2} = \sqrt{3}c_{3}c [/itex]
[itex] c_{3} = \sqrt{4}c_{4}c [/itex]

The problem her is that since the first part does not contain the vector [itex] 0|\rangle[/itex], either c or [itex]c_{1}[/itex] has to be zero in the first equation, and then everything becomes zero, but I know that the annihilation operator has eigenstates.
 
Physics news on Phys.org
  • #2
Sandrasa said:
Try to show this for an arbitrary wavefunction:
[itex]|V\rangle = \sum_{n=1}^\infty c_{n}|n\rangle [/itex]
Is there some reason you didn't start this sum at n=0 ?
 
  • #3
Yes, because when the annihilation operator works on the state zero, we will get zero by definition according to my book so I thought that there was no point in taking it. But I can see that that would get me on the right track again, but I don't know how I can do it?

But by taking the sum from n = 0, I end up with [itex] c_n = c_{n-1}\frac{c'}{\sqrt{n}} [/itex], where c' = [itex] \frac{1}{c} [/itex]. And now I'm not really sure what to do next. Can I just put this in my expression for [itex] |V\rangle [/itex] and call that an eigenstate with c' as the constant? It doesn't look like the expression I found for the eigenstate, which was [itex] |V\rangle [/itex] = [itex] c_{0} [/itex] [itex] \sum\limits_{n=0}^\infty [/itex] [itex] \frac{c^n}{\sqrt{n!}} [/itex][itex]|n\rangle [/itex] in my notation. I tried writting the expression I found out, but then I got something like:
[itex] c_0 + \frac{c'c_0}{\sqrt{1}} + \frac{c_0c'^2}{\sqrt{2}} + \frac{c_0c'^3}{\sqrt{6}} [/itex]. It sort of looks like the equations I got in the start, but I'm stuck in trying to get to one place from the other. Any hints?

And thanks for the help!
 
  • #4
In your "relevant equations", you didn't write out the action of the creation operator ##a^*##. E.g.,
$$a^*|0\rangle ~=~ ~?~$$$$(a^*)^2 |0\rangle ~=~ ~?~$$etc. Try to express each term in your sum for ##|V\rangle## in terms of ##a^*## acting on the vacuum.
 

1. What is an eigenstate for the annihilation operator?

An eigenstate for the annihilation operator is a state in a quantum system that is characterized by a unique energy value and is annihilated when acted upon by the annihilation operator.

2. How is the eigenstate for the annihilation operator found?

The eigenstate for the annihilation operator is found by solving the corresponding eigenvalue problem, which involves finding the values of the state's energy and wavefunction that satisfy the operator equation.

3. What is the significance of finding the eigenstate for the annihilation operator?

Finding the eigenstate for the annihilation operator allows us to determine the energy states and corresponding wavefunctions of a quantum system, which is crucial for understanding the behavior and properties of the system.

4. Can there be multiple eigenstates for the annihilation operator?

Yes, there can be multiple eigenstates for the annihilation operator, each with its own unique energy value and wavefunction. This is known as the degeneracy of the system.

5. How does the eigenstate for the annihilation operator relate to the uncertainty principle?

The eigenstate for the annihilation operator plays a key role in the uncertainty principle, as it allows us to determine the minimum uncertainty in a system's energy and position by measuring the energy of the system in the eigenstate.

Similar threads

  • Advanced Physics Homework Help
Replies
1
Views
820
  • Advanced Physics Homework Help
Replies
1
Views
707
  • Advanced Physics Homework Help
Replies
10
Views
1K
  • Advanced Physics Homework Help
Replies
1
Views
1K
  • Advanced Physics Homework Help
Replies
0
Views
212
  • Advanced Physics Homework Help
Replies
0
Views
103
  • Advanced Physics Homework Help
2
Replies
36
Views
2K
  • Advanced Physics Homework Help
Replies
3
Views
983
Replies
16
Views
523
Replies
4
Views
1K
Back
Top