# From wave to ket

1. Jul 5, 2009

### snoopies622

A question from a beginner:

If the position operator changes a state vector into a wave function, how does one change a wave function back into a state vector? I have read that the position operator for a one-dimensional vector is simply multiplying by x. Does this mean that what amounts to the inverse position operator (or position inverse operator, perhaps) is putting a

$$\frac {1}{x}$$

in front of the wave function? For example, a particle in a one-dimensional "box" of length L has a wave function

$$\psi _n = \sqrt { \frac {2}{L} } sin ( \frac {n \pi x } {L })$$

Is the corresponding ket

$$| \psi _n \rangle = \frac {1}{x} \sqrt { \frac {2}{L} } sin ( \frac {n \pi x } {L })$$

This seems logical but something tells me that it is too easy to be correct.

2. Jul 5, 2009

### dextercioby

I'm afraid your question doesn't make 2 much sense, as the passage from <bra| and |ket> to wavefunctions can be reversed only at a conceptual level. It's going from generalities to particular cases and viceversa.
The "bra-ket" corresponds to an abstract mathematical setting (a rigged HS in which the HS is not specified and is not relevant). Going to wavefunctions from "bra-ket" is just choosing the Hilbert space to be $L^2$ over some set and the associated Lebesgue measure.
This choise means going from general to particular.

3. Jul 5, 2009

### Fredrik

Staff Emeritus
Who says that a position operator changes a state vector into a wave function? The wave functions are state vectors. The position operator (I'll call it Q) can be defined by Qf(x)=xf(x), and if we define an operator A by Af(x)=f(x)/x, we see that AQf(x)=(1/x)Qf(x)=(1/x)xf(x)=f(x), so AQ is the identity operator, and that means that A=Q-1, as expected.

Edit: A clarification on what I and bigbau said. You can postulate that the states are represented by equivalence classes of vectors in some separable Hilbert space H, or you can postulate that they are represented by equivalence classes of vectors in the Hilbert space $L^2(\mathbb R^3)$ of square-integrable functions. When we choose the former option, the vectors are usually called state vectors, and when we choose the latter option, they are usually called wavefunctions. But $L^2(\mathbb R^3)$ is a separable Hilbert space, so the wavefunctions are state vectors. There's actually no significant difference between these two options since there's a theorem that says that all infinite-dimensional separable Hilbert spaces are isomorphic to each other.

Last edited: Jul 5, 2009
4. Jul 5, 2009

### Edgardo

First you have to understand that there are different ways to represent the quantum
state for example of an electron in a quantum well.

1) The ket $| \psi \rangle$ is an abstract representation of the state.
It really is just abstract and (simplified) it doesn't really tell you anything about how the
state looks like.

2a) Now, depending on what you are interested in, you can choose the position representation.
For this you write: $\langle x | \psi \rangle = \psi(x)$
What we have done here is we projected the ket $| \psi \rangle$ onto $| x \rangle$.
(As you may know the scalar product can be viewed as a projection of one vector onto another one.)

2b) If you want the momentum representation you just project the abstract ket
vector $| \psi \rangle$ onto $| p \rangle$ and get $\langle p | \psi \rangle = \psi(p)$.

Why is this important to have different representations? Suppose you want to know what
the probability is to find your electron in the position intervall [x1,x2]. Then you need to
calculate the probability density $| \psi (x) |^2$, i.e. you need the position representation $\psi(x)$.

Suppose you want to know what the probability is that your electron has momentum in the
intervall [p1,p2]. Then you calculate the probability density $| \psi (p) |^2$, i.e.
you need the momentum representation $\psi(p)$.

I recommend reading the pdf Elements of Dirac Bracket Notation by Prof. Frank Rioux from here.

5. Jul 5, 2009

### snoopies622

I was thinking of the bra / ket notation like it was just another basis. OK, I will read that Rioux pdf; it looks very promising. Thanks all.

6. Jul 7, 2009

### Edgardo

I noticed that what I wrote above is misleading. It has nothing to do with position or momentum representation. The above should be:

2a) The relationship between $| \psi \rangle$ and $\psi(x)$ is given by:
$\langle x | \psi \rangle = \psi(x)$
What we have done here is we projected the ket $| \psi \rangle$ onto $| x \rangle$. (As you may know
the scalar product can be viewed as a projection of one vector onto another one.)

2b) Similarly, you get $\psi(p)$ by projecting the vector $| \psi \rangle$ onto $| p \rangle$:
$\langle p | \psi \rangle = \psi(p)$.