Is There an Eigenstate for the Creation Operator?

[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.

 

[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>

 

[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?

 

[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.

 

[indigojoker]

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

where do I go from here?

 

[malawi_glenn]

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

[N,a] = Na - aN

 

[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?

 

[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> ?

 

[indigojoker]

\langle a a^{\dagger} \rangle

 

[malawi_glenn]

nope, try:

<n|a

 

[indigojoker]

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

 

[malawi_glenn]

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

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.

 

[Gokul43201]

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

Edit: Not anymore; typo fixed.

 

[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 isn't an eigenstate

Does this mean the eigenstate of the destruction operator is a|n \rangle = 0|n\rangle ? since we want a state that doesn't change right?

 

[malawi_glenn]

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

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

 

[malawi_glenn]

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

should read:

\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

 

[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

 

[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?

 

[malawi_glenn]

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

 

[George Jones]

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

indigo joker: What is the definition of an eigenvector.

 

[indigojoker]

\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.

how do I make this right?

and what exactly is the eigenstate of the destruction operator then?

 

[Gokul43201]

Indigo, make sure you know the definition of an eigenstate/eigenket/eigenvector. Answering George's question will help clarify some of the confusion.

PS: I'm going to step away from here for a bit...too many cooks.

 

[indigojoker]

Definition of eigenstate: Let |\alpha \rangle be a ket and A be an operator, such that A|\alpha \rangle = a |\alpha \rangle, then |\alpha \rangle is called an eigenstate (or eigenket) of the operator A.

Edit: I was writing this before George's post was up (didn't mean to short-circuit you, George).

several comments.

then basically all the notation in books are wrong? since we have the eigenstate of the Hamiltonian to be H|n> =(n+1/2)hw|n>

but since it is an state, shouldn't it be H|n>=n|n>?

And since the destruction operator is a|n>=sqrt(n)|n-1>, then the eigenstate for the destruction operator is |0>?

 

[indigojoker]

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

indigo joker: What is the definition of an eigenvector.

When an operator operates on its eigenvector, we get some constant multiplied by the eigenvector back.

like Ax=cx

where A is the operator, x is the eigenvector, and c is the eigenvalue.

I look at this from a mathematical standpoint, maybe what i don't understand is how this relates physically? not sure.

 

[George Jones]

like Ax=cx.

Right.

So, the problem in your original post, written in QM notation, is to find if \alpha (eigenvalue) and |\psi> (eigenvalue) exist such that

<br /> a^\dagger |\psi&gt; = \alpha |\psi&gt;.<br />

Hint: expand |\psi&gt; as an arbitrary linear combination of energy eigenkets, and use linear independence.

(Actually, there is a shorter, slicker way.)

 

[indigojoker]

well, it asks to find the eigenstate, which I am trying to find out what that means right not, not necessarily finding the eigenvalue or eigenket. It's asking if it has an eigenstate (whatever that means). It was explained to me that an eigenstate is something like A|x>=x|x>

well, I've heard the term energy eigenstate used and I know that an example is H|n> = (n+1/2)hw|n>

Why isn't the |n> = |(n+1/2)hw>?

 

[Gokul43201]

well, it asks to find the eigenstate, which I am trying to find out what that means right not, not necessarily finding the eigenvalue or eigenket.

Eigenstate and eigenket are essentially the same thing (so long as the ket is used to represent the state of some system).

It's asking if it has an eigenstate (whatever that means).

That means "is there a state that the <blah> operator acts on and the result is some scalar multiple of the same state?"

It was explained to me that an eigenstate is something like A|x>=x|x>

That is not an eigenstate; that's an eigenvalue equation. In that equation, |x> is an eigenstate of the A operator.

well, I've heard the term energy eigenstate used and I know that an example is H|n> = (n+1/2)hw|n>

Again, that's an example of an eigenvalue equation for the Hamiltonian operator (specifically for the Hamiltonian of a harmonic oscillator). The state ket |n>, which is basically a state containing n quantized oscillations (this is not by construction, but follows from the solution of the eigenvalue problem), is an eigenket or eigenstate or eigenvector of the Hamiltonian given by p^2/2m + (1/2)mw^2 x^2. In the position(x)-representation, these kets map onto functions (related to the Hermite polynomials) which are called the eigenfunctions of the Hamiltonian operator in position representation (i.e., \cal{H} = -(\hbar^2/2m)\partial_x^2 + (1/2)m\omega ^2 x^2). From your above equation, we see that the eigenvalues of the Hamiltonian (also known as energy eigenvalues) are (n+1/2)hw.

Why isn't the |n> = |(n+1/2)hw>?

The "n" in the ket |n> is only used as a convenient label. For a system with non-degenerate eigenvalues of some operator, it is often convention to label the eigenket associated with each eigenvalue, using some shorthand notation that is indicative of the eigenvalue. For energy eigenkets, one typically writes the eigenket associated with the lowest eigenvalue as |0> and higher numbers are assigned to kets with increasing eigenvalues.

Edit: My LaTeX is showing up weirdly to me. If something looks bizarre, click the TeX and read the script out of the pop-up window.

Edit2: If the LaTeX looks weird to anyone else, please let me know.

 

[indigojoker]

thanks for the explanation.

\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?

so why is this wrong?

 

[Gokul43201]

I don't see what's wrong. Looks fine to me. I guess we'll both have to wait for mg.

 

[malawi_glenn]

Okay, now I see, he did not do the same proof as I am used to do ;)

And sorry indigojoker for not aswering your exact problem, I wanted to give you a more general treatment of how you use these creation operators, and said only that eigenstate is something you exactyl get back when you operate on it. The next thing i wanted to discuss was actually what happens if you take the a operator an let it operate on ground state.

 

[George Jones]

I apologize for the harsh tone of post #25.

Maybe I'm being picky, but I haven't seen anywhere in this thread a demonstration that a^\dagger has no eigenvalues and eigenvectors. Such a demonstration follows directly from

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

but, at this introductory level, I think the details need to be filled in.

Since \left\{ |n&gt; \right\} is a basis for state space, an arbitrary state vector |\psi&gt; can written

|\psi&gt; = \sum_{n=0}^\infty c_n |n&gt;.

Is it possible to find just the right values of the c_n's (e.g., some equal to zero, some equal to -1, some equal to 1/\sqrt{2}, some equal to 42, etc.) such that |\psi&gt; is an eigenvector of a^\dagger?

Vigorous hand waving not allowed.

 

[indigojoker]

Since \left\{ |n&gt; \right\} is a basis for state space, an arbitrary state vector |\psi&gt; can written

|\psi&gt; = \sum_{n=0}^\infty c_n |n&gt;.

Is it possible to find just the right values of the c_n's (e.g., some equal to zero, some equal to -1, some equal to 1/\sqrt{2}, some equal to 42, etc.) such that |\psi&gt; is an eigenvector of a^\dagger?

Vigorous hand waving not allowed.

I don't think so, but I'm not sure how to prove why.