1. Aug 4, 2009

Hi,

I was wondering how the state vector for a particle in a 1-D box can be expanded as a linear combination of the discrete energy eigenkets as well as a linear combination of the continuous position eigenkets. It seems to me that this is a contradiction because one basis is countable whereas the other is uncountable. Is there any way of reconciling this?

2. Aug 4, 2009

### Staff: Mentor

That is an interesting point. I dont know the answer, but I wonder if it has something to do with the fact that the energy eigenvalues are unbounded while the position eigenvalues are bounded. It is not immediately apparent to me how a bounded but uncountable set could be mapped to an unbounded but countable set.

3. Aug 5, 2009

### schieghoven

The state space for this system is the space of square-integrable functions on [0,1]. This has a countable base. You might like to look at some books on measure theory, Lebesgue integration and functional analysis, e.g. Rudin or Lang. It's a big subject and a cornerstone of modern pure math.

There are ways of defining position eigenkets, but they are not functions and you can't construct a Hilbert space of states out of them.

Cheers

Dave

4. Aug 5, 2009

### George Jones

Staff Emeritus
Take a look at this thread:

5. Aug 5, 2009

Thanks very much, that was really help helpful. I think what you're saying is that since the generalized position eigenstates lie outside the Hilbert space they cannot be considered a basis by definition . This means that the size of $$\{|X\rangle\}$$ says nothing about the dimension of the subspace of $$H$$ containing physically viable functions. Is this correct?

If this is correct, it still seems weird to me that any element belonging to the subspace can be expanded as a linear combination of an uncountable orthogonal set as well as a countable orthogonal basis set. Am I thinking about this in the wrong way? Should I just accept that these two expansions are completely separate entities and that only one of them says something about the size of space?

Last edited: Aug 5, 2009
6. Aug 5, 2009

### George Jones

Staff Emeritus
Sorry, but I'm not an expert on this.
I think that the nuclear spectral theorem applied to the position operator gives this result.

7. Aug 7, 2009

### Staff: Mentor

Shouldn't the eigenkets of any Hermitian operator form an orthonormal basis? I thought that was a basic property of all Hermitian operators.

8. Aug 7, 2009

### Petar Mali

No! This is correct only for bounded operators. And $$\hat{x}$$ is unbounded.

9. Aug 7, 2009

### Staff: Mentor

Shouldn't x be bounded for the particle in a box?

10. Aug 7, 2009

### Petar Mali

$$\hat{x}$$ is multiplicative operator. Take for exaple function $$\frac{x}{1+x^2}$$. This function is in $$L_2(\mathcal{R})$$.

And if you multiplicate this function with $$x$$ you will get function which isn't in

$$L_2(\mathcal{R})$$ - $$\frac{x^2}{1+x^2}$$. So $$\hat{x}$$ is unbounded.

11. Aug 7, 2009

Is the domain of every bounded operator on a Hilbert space the entire space? I think I remember reading that so I'm wondering if this is what you're talking about.

Also, a few small asides:

1) Many people in my theoretical chemistry department say a bounded operator is one which has bounded eigenvalues. It's easy to prove that a bounded operator has bounded eigenvalues but do you know if the converse true as well?

2) Do you know how to reconcile the fact that any wavefunction can be expanded in a countable orthonormal basis as well as in the uncountable generalized eigenfunctions of the position operator? Should I just accept that the latter says nothing about the size of the wavefunction space and leave it at that?

Last edited: Aug 7, 2009
12. Aug 7, 2009

### Staff: Mentor

That makes sense.

That is what I was thinking. I guess Petar's example shows that the converse is not always true. For the particle in a box the x operator has bounded eigenvalues but is not a bounded operator.

13. Aug 7, 2009

### Petar Mali

Operator $$\hat{A}$$ is bounded if exist $$C$$ such that

$$||\hat{A}\varphi||<C||\varphi||$$

$$\forall\varphi$$

2. Because there exist isomorphism between $$L^2$$ and $$l^2$$.

14. Aug 7, 2009

### strangerep

There's an uncountable infinity of functions f(x) that can be defined on [0,1].
But only a (countable) subset of them also satisfy the Schrodinger equation,
which is what's important.

HTH.

15. Aug 8, 2009

### George Jones

Staff Emeritus
Did you mean to write "But the subset of them that also satisfy the Schrodinger equation is separable, which is what's important."?

16. Aug 8, 2009

### George Jones

Staff Emeritus
It depends what is meant by "eigenkets." This is true for Hermitian operators on finite-dimensional Hilbert spaces, but the jump in subtlety that occurs in the move from finite-dimensional Hilbert spaces to infinite-dimensional Hilbert spaces is substantial. In terms of honest mathematics, even a bounded Hermitian operator on an infinte-dimensional space need not have any eigenvectors.
Yes, the position operator for a particle in a box is bounded (the momentum operator is unbounded), but this operator doesn't have any eigenvectors.
You have shown that the position operator is unbounded for state spaces function defined on all of $\mathbb{R}$, but, for a particle in a box, this operator is bounded.

The mathematics of general relativity has a reputation of being difficult, but, in my opinion, the mathematics, as honest mathematics, of non-relativistic quantum mechanics is much more subtle and difficult.

17. Aug 8, 2009

### Staff: Mentor

Meaning that the functions which solve the eigenvalue equation are not themselves properly considered vectors in the space of states? And that is because the Dirac delta is not really a function?

18. Aug 8, 2009

### George Jones

Staff Emeritus
Right. Delta functions are sometimes called eigendistributions or weak eigenvectors of the multiplication operator.

19. Aug 9, 2009

### strangerep

Oops! Yes, I was inexcusably sloppy, wasn't I.

(For the benefit of other readers, a metric space is said to be "separable"
if it has a countable dense subset. A Hilbert space is separable iff it has
a countable orthonormal basis.)