Find the eigenstates of a basis in terms of those of another basis?

In summary, the conversation revolves around finding eigenstates of the position operator in terms of the number operator's eigenstates. The number operator is given by \hat{N} = \hat{a}^\dagger\hat{a}, with eigenvalues and eigenstates \hat{N}|n\rangle = n|n\rangle. The position operator is given by \hat{x} = \sqrt{\frac{\hbar}{2m\omega}}\Big(\hat{a}^\dagger+\hat{a}\Big) \equiv \gamma\Big(\hat{a}^\dagger+\hat{a}\Big), and its action on one of the number eigenstates is \hat{x}|n\r
  • #1
thecommexokid
70
2

Homework Statement



This isn't exactly a homework question, but I figured this would be the best subforum for this sort of thing. For the sake of a concrete example, let's just say my question is:

Express the position operator's eigenstates in terms of the number operator's eigenstates.

Homework Equations



The number operator is given by
[tex]\hat{N} = \hat{a}^\dagger\hat{a}.[/tex]
It has eigenvalues and eigenstates
[tex]\hat{N}|n\rangle = n|n\rangle.[/tex]
The position operator is given by
[tex]\hat{x} = \sqrt{\frac{\hbar}{2m\omega}}\Big(\hat{a}^\dagger+\hat{a}\Big) \equiv \gamma\Big(\hat{a}^\dagger+\hat{a}\Big).[/tex]
The action of the position operator on one of the number eigenstates is
[tex]\hat{x}|n\rangle = \gamma\Big(\hat{a}^\dagger|n\rangle + \hat{a}|n\rangle\Big)
=\gamma\Big(\sqrt{n+1}|n+1\rangle + \sqrt{n}|n-1\rangle\Big).[/tex]

The Attempt at a Solution



We'd like to find eigenstates of [itex]\hat x[/itex], that is, states [itex]|x\rangle[/itex] satisfying
[tex]\hat x|x\rangle = x|x\rangle.[/tex]
The number basis is complete, so whatever these states we're looking for might be, they are representable as a linear combination of the number states:
[tex]|x\rangle = \sum_n |n\rangle\langle n|x\rangle;[/tex]
I just need to know the coefficients [itex]C_n \equiv \langle n|x\rangle[/itex].

If I look again at the action of [itex]\hat x[/itex],
[tex]\hat x|x\rangle = \sum_n \hat x|n\rangle\langle n|x\rangle \\ \qquad = \sum_n \gamma\Big(\sqrt{n+1}|n+1\rangle + \sqrt{n}|n-1\rangle\Big) \langle n|x\rangle.[/tex]
Suppose I attack this equation from the left with a particular [itex]\langle n'|[/itex]. Then I get
[tex]x\langle n'|x\rangle = \sum_n \gamma\Big(\sqrt{n+1}\langle n'|n+1\rangle + \sqrt{n}\langle n'|n-1\rangle\Big) \langle n|x\rangle \\
\qquad = \gamma\Big(\sqrt{n'}\langle n'-1|x\rangle + \sqrt{n'+1}\langle n'+1|x\rangle\Big),[/tex]
which allows me to develop a recurrence relationship
[tex]C_{n+1} = \frac{x}{\gamma\sqrt{n+1}}C_n-\frac{\sqrt{n}}{\sqrt{n+1}}C_{n-1}.[/tex]

But I don't understand what this says. What is that [itex]x[/itex] doing in there? What does that even mean? And can I use this recurrence relation to get a closed-form answer? (At the very least I'd need to calculate [itex]C_0[/itex] and [itex]C_1[/itex] explicitly, which I don't know how to do.)

Also, I think at some point the position-space wavefunctions ought to come into it:
[tex]\langle x'|n\rangle = \sqrt[4]{\frac{m\omega}{\pi\hbar}}\frac{1}{\sqrt{2^nn!}}H_n\Big(\sqrt{\frac{1}{\sqrt{2}\gamma} x}\Big)e^{-x^2/2\gamma^2}[/tex]
and
[tex]\langle x'|x\rangle = \delta(x-x').[/tex]
 
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  • #2
Hi.
[itex]\hat{x}[/itex]=[itex]\gamma[/itex]([itex]\hat{a}[/itex]†+ [itex]\hat{a}[/itex])
Shows that you have to find common eigenstates of creation and annihilation operators. Are you familiar with quantum coherent states?
If not, you can find out by starting for example with:
|[itex]\alpha[/itex]>=[itex]\sum[/itex]cn|n>,
[itex]\hat{a}[/itex]|[itex]\alpha[/itex]>= [itex]\alpha[/itex]|[itex]\alpha[/itex]>
Get a recurrence relation for the coefficients and see if you obtain a familiar function at the end (hint: in terms of |0>). It might take a while but that's the right path...
 

FAQ: Find the eigenstates of a basis in terms of those of another basis?

What is the concept of eigenstates in linear algebra?

Eigenstates are a set of vectors that do not change direction when acted upon by a linear transformation. In other words, they are the special vectors that are only scaled by a constant factor when transformed.

Why is it important to find the eigenstates of a basis in terms of another basis?

Finding the eigenstates of a basis in terms of another basis allows us to understand how a linear transformation affects the original basis vectors. This information is crucial in many applications of linear algebra, such as solving differential equations and understanding quantum mechanics.

How do we find the eigenstates of a basis in terms of those of another basis?

To find the eigenstates of a basis in terms of another basis, we can use a change of basis matrix. This matrix maps the original basis to the new basis, and by applying it to the eigenstates, we can express them in terms of the new basis.

Can we find the eigenstates of a basis in terms of another basis for any linear transformation?

Yes, the concept of eigenstates and change of basis can be applied to any linear transformation. However, the process may be more complicated for non-diagonalizable matrices, as they may have repeated eigenvalues.

How can finding the eigenstates of a basis in terms of another basis be useful in solving practical problems?

Understanding the relationship between eigenstates of different bases helps us to simplify and solve complex problems in many fields, such as physics, engineering, and computer science. This knowledge can also help us to make predictions and optimize systems in real-world applications.

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