# Eigenstate and creation operator

1. Oct 22, 2007

### indigojoker

I'm a little bit confused in general about what an eigenstate is. So say we have something like: H|n>=hw(N+1/2)|n>

|n> is the eigenket, hw(N+1/2) is the eigenvalue, but what exactly is an eigenstate?

The entire question asks if there are eigenstate to the creation operator and to prove it.

I thought that a*|n>=sqrt(n+1)|n+1>

Isnt that the eigenstate? I guess I'm pretty lost in general. Any help would be appreciated.

2. Oct 22, 2007

### malawi_glenn

eigenstate is when you get exactly the same state back again.

and what is

a*|n>=sqrt(n+1)|n+1>

??

you mean

$$a^{\dagger}|n>=\sqrt{n+1}|n+1>$$

Then I would say:

HINT: Consider the comutator between N and $$a^{\dagger}$$ or $$a$$, and use the fact that $$N = a^{\dagger} a$$

and $$N|n> = (\hbar \omega /2 + n)|n>$$

Last edited: Oct 22, 2007
3. Oct 22, 2007

### indigojoker

oh, so then the creation operator does not have eigenstates because the state will aways be raised to n+1

where as the destruction operator gives the same state of 0 when it acts on the lowest state. right?

4. Oct 22, 2007

### malawi_glenn

yes

but you want to prove this, right?

Use that the number operator N is $$a^{\dagger} a$$ and consider the commutator between N and a and a-dagger.

5. Oct 22, 2007

### indigojoker

I get [N,a]=-a and [N,a*]=a*

where do I go from here?

6. Oct 22, 2007

### malawi_glenn

what happens if you do N(a|n>), and are smart to use the commutator?

[N,a] = Na - aN

7. Oct 22, 2007

### indigojoker

Na*|n>=(n+1)a*|n>

so this says that the eigenvalue increases by one... so thus, could i now conclude that the eigenvalues are continuously increasing when applying the creation operator, and hence the operator does not have an eigenstate?

8. Oct 22, 2007

### malawi_glenn

GOOD!

it means that you get the same thing when operating with N.

$$N (a^{\dagger} |n>) = N |n+1>$$

right? Now its easy to identify what $$a^{\dagger} |n>$$ becomes; i.e $$a^{\dagger} |n> = \lambda |n+1>$$ where $$\lambda$$ is a complex constant.

now to determine the constant, what is the corresponding bra to $$a^{\dagger} |n>$$ ?

9. Oct 22, 2007

### indigojoker

$$\langle a a^{\dagger} \rangle$$

10. Oct 22, 2007

### malawi_glenn

nope, try:

$$<n|a$$

11. Oct 22, 2007

### indigojoker

doesnt
$$\langle a a^{\dagger} \rangle = \langle n|a a^{\dagger}|n \rangle$$

12. Oct 22, 2007

### malawi_glenn

no, if $$a^{\dagger}|n \rangle = \lambda |n+1>$$ what is the corresponding bra, and what is the corresponding value to $$\lambda$$ is ? $$\lambda$$ is a complex number.

Last edited: Oct 22, 2007
13. Oct 22, 2007

### Gokul43201

Staff Emeritus
mg, that last post is confusing: the LHS is a ket and the RHS is a c-number.

Edit: Not anymore; typo fixed.

Last edited: Oct 22, 2007
14. Oct 22, 2007

### indigojoker

i think i got it...
$$a^{\dagger} |n \rangle = c |n+1 \rangle$$
$$\langle n|a a^{\dagger}|n \rangle = \langle n+1| c^* c | n+1>$$
$$\langle n|a a^{\dagger}|n \rangle = \langle n+1| c^* c | n+1>$$
$$\langle n|a^{\dagger} a +1 |n \rangle = |c|^2$$
$$\langle n|a^{\dagger} a |n \rangle +\langle n|1 |n \rangle = |c|^2$$
$$\langle n|N|n \rangle +\langle n|1 |n \rangle = |c|^2$$
$$n+1 = |c|^2$$
$$\sqrt{n+1} = c$$

$$a^{\dagger} |n \rangle = \sqrt{n+1} |n+1 \rangle$$

An so now we can conclude that even if n was in ground state, the creation operator will increase the energy and keep increasing it if operated, and so there isnt an eigenstate

Does this mean the eigenstate of the destruction operator is $a|n \rangle = 0|n\rangle$ ? since we want a state that doesnt change right?

Last edited: Oct 22, 2007
15. Oct 22, 2007

### malawi_glenn

yes of course, i was sloppy but i think he got my point :)

16. Oct 22, 2007

### malawi_glenn

$$\langle n+1|a a^{\dagger}|n+1 \rangle = \langle n+1| c* c | n+1>$$

$$\langle n|a a^{\dagger}|n\rangle = \langle n+1| c^* c | n+1>$$

then you have

$$\langle n|a a^{\dagger}|n\rangle = c^* c \langle n+1| n+1> = |c|^2$$

17. Oct 22, 2007

### indigojoker

ok, this can be similarly done for the destruction operator, but what exactly is the definition of an eigenstate? For the destruction operator, is the eigenstate:

$$a|n \rangle = \sqrt{n} |n-1 \rangle$$

or
$$a|n \rangle = 0 |n-1 \rangle$$

18. Oct 22, 2007

### malawi_glenn

$$\langle n|a^{\dagger} a +1 |n \rangle = |c|^2$$

and

$$\langle n|a^{\dagger} a |n \rangle +\langle n|1 |n \rangle = |c|^2$$

Is not right. Where did you get that from?

Last edited: Oct 22, 2007
19. Oct 22, 2007

### malawi_glenn

the definition of an eigenstate is if you operate on it exactly the same state is left.

20. Oct 22, 2007

### George Jones

Staff Emeritus
Unless I've missed something, this thread has gone completely off the rails.

indigo joker: What is the definition of an eigenvector.