# State Vectors as elements of Hilbert Space

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1. Aug 20, 2015

### devd

The state, $| S\rangle$, say, of a system is represented as a vector in a Hilbert space.

$\psi (x, t)$ is the representation of the state vector in the position eigenbasis; $\psi (p, t)$ in the momentum eigenbasis et cetera. That is, $\psi (x, t) = \langle x|S\rangle$; $\psi (p, t) = \langle p|S\rangle$.

Now, suppose i expand $\psi(x,t)$ as $\psi(x,t) = \sum_{n=0}^\infty a_n \cos(\frac{n\pi x} {L})$.
Similarly for $\psi (p, t)$.
[Assuming Dirichlet conditions hold, $a\leq x\leq b$ etc]

I can now interpret this as: $\psi(x, t)$ is an element of a Hilbert space, with the $\cos(\frac{n\pi x} {L})$'s as basis. [The $\cos(\frac{n\pi p} {L'})$'s as basis for $\psi (p, t)$].

How do i reconcile these two interpretations? $\psi (x, t)$ as the coefficient in the expansion of $| S\rangle$ in the $|x\rangle$ eigenbasis as opposed to an element of a Hilbert space expanded in the $\cos(\frac{n\pi x} {L})$ basis with coefficients $a_n$.

2. Aug 20, 2015

### Truecrimson

It is an expansion of an expansion.
The actual vector in the Hilbert space is $$|S\rangle = \int dx |x\rangle \langle x|S\rangle = \int dx \psi(x) |x\rangle.$$ (I leave out the time for simplicity.)
Now if I have another basis $|n\rangle$, I can insert the identity again in that expression.
$$\int dx \sum_n \langle x|n\rangle \langle n|S\rangle |x\rangle$$
In your case, $\langle x|n\rangle = \cos (\frac{n \pi x}{L})$ and
$$\langle n|S \rangle = \int dx \langle n|x\rangle \langle x|S \rangle = \int dx \cos (\frac{n \pi x}{L}) \psi (x) = a_n.$$

From Ballentine section 4.1
This is like in relativity where people sometimes call a set of coefficients themselves a vector (or a tensor). But pedantically speaking, it is $|S\rangle$ that is an element of the Hilbert space.

3. Aug 20, 2015

### devd

Thanks for the reply. That helped me get a handle on the problem.
One question, what 'set' of functions are we talking about? $\{\psi(x)\}$ is one function, defined by the coefficients at each value of the variable x, right?

4. Aug 20, 2015

### Truecrimson

You're welcome. I think by $\{\psi(x)\}$ (as opposed to just $\psi(x)$) Ballentine means a set of wavefunctions, each coming from a different $|\psi\rangle$. Does that make sense?

EDIT: I looked at Ballentine again and realize he doesn't use the notation for set, but he does say that $\psi(x)$ is a function and "the vector space consists of the functions $\psi(x)$." So I guess I put in the set notation automatically without thinking because that's how I interpret it.

Last edited: Aug 20, 2015
5. Aug 20, 2015

### devd

Yes, i think that is what he means. But, what is that set of functions exactly? What are its elements?

$\psi (x)$ is simple the set of ordered pairs {$(x_1, \psi (x_1)), (x_2, \psi (x_2)), ...$} [x is a continuous variable, so i cant really write it like this. But, just to get the point across.]

So, the thing i'm not being able to comprehend is the nature of the set {$\psi (x)$} as there's just 1 function, $\psi (x)$!

6. Aug 20, 2015

### Truecrimson

To make the notation clear, there should be an additional label like this: $\{\psi_n(x)\}$.

7. Aug 20, 2015

### devd

But, what would that indicate? What is $\{\psi_1(x)\}$ for example? The coefficient of the $|x_1\rangle$ in the expansion of $|S\rangle$? But, that is just a number!

8. Aug 20, 2015

### Truecrimson

By $\{\psi_n(x)\}$, I mean $\{\psi_n(x) = \langle x|\psi_n\rangle|n=1,2,\dots,N\}$ or something like that, with $|\psi_1\rangle, |\psi_2\rangle,\dots$ (Edit: or $|S_1\rangle, |S_2\rangle,\dots$) being vectors in the Hilbert space (and each $\psi_n(x)$ is a function with different values at different $x$, as you understand).

Last edited: Aug 20, 2015
9. Aug 20, 2015

### devd

I think i'll have to mull it over a bit more. Will post when i have a clearer picture of this. Thanks for patiently answering my questions! :)