# Operators in Bra-Ket notaion

1. Jun 24, 2007

### mrandersdk

I'm reading an article where there are an atom with two states, lets call them |0> and |1>. Then the writer defines an operator by

|0><1|

I know how this operator works in the bra ket notaion, but how does it work, if I want to use it in the position basis?

Someone told me that I just can use (as an approximation)

a_0 x a_1^*

where a_0 and a_1 are the states in the position basis. Of cause |0> means a_0 and <1| means a_1^* in the position basis, but just to use a_0 x a_1^* seems to need some explanation, can someone give me one?

2. Jun 24, 2007

### StatusX

Sorry, I have a tendency to answer questions by starting from the basic concepts, so you might already know most of this.

A state is just a vector in the state space, and we can represent this in different ways by taking different bases of state space, just as you can represent an ordinary vector in euclidean space by different numbers depending on the basis of euclidean space you take. Once we choose a basis, we can write down the numbers corresponding to a state. If this basis is the set of eigenstates of the position operator, ie, the states |x> where the particle is certain to be at the position x, then the corresponding numbers are the wavefunction.

That is $|\Psi>$ is just an abstract quantity, not a function with complex values, but we may write:

$$|\Psi> = \int |x><x|\Psi> dx = \int \Psi(x) |x> dx$$

where we define $\Psi(x)=<x|\Psi>$, which is the ordinary wavefunction. Now we express linear operators in this basis in the same way you represent linear operators in a chosen basis of euclidean space, by matrices, albeit infinite dimensional ones. Specifically, similar to before, we expand:

$$L = \left(\int |x><x| dx\right) L \left(\int |x'><x'| dx'\right)$$

$$= \int \int <x|L|x'> |x><x'| dx dx'$$

The $<x|L|x'>$ are analagous to matrix elements $L_{ij}$. Then we get the representation of L|\Psi> by:

$$L|\Psi> = \int \int <x|L|x'> |x><x'|\Psi> dx dx'$$

$$= \int \left(\int <x|L|x'> <x'|\Psi> dx'\right) |x> dx$$

So the wavefunction $L\Psi(x)$ is given by:

$$L\Psi(x) = \int <x|L|x'> <x'|\Psi> dx'$$

so we need to look at the components:

$$<x'|L|x>$$

In the case of the operator $|0><1|$, this becomes:

$$<x'|0><1|x> = \Psi_0(x') {\Psi_1}^*(x)$$

So finally we get:

$$L\Psi(x) = \int \Psi_0(x') {\Psi_1}^*(x) \Psi(x') dx'$$

which is exact.

3. Jun 24, 2007

### mrandersdk

If I understand you correct you say that the operator L in my case is given by

$$L: \Psi(x) -> \int \Psi_0(x') {\Psi_1}^*(x) \Psi(x') dx'$$

And the one i'm advised to use is

$$L': \Psi(x) -> \Psi_0(x) {\Psi_1}^*(x) \Psi(x)$$

As I can see this only makes sense if $$\Psi(x)$$ is some kind of delta-function. But maybe it's a good approximation ??? Can you explaine why it is in my case?

By the way do you know any webpages, that explaine the mathematics for vector spaces spaned by eigenkets with continuous spectra.

I have had a course in the spectral theorem for bounded normal operators, and think it have something to do with that, but would like to read a little more about the connection. Because to me representing an operator by an integral can only be a notation (as in the course i had), and not something you can actual use to calculate things, but I can see the notation kind of makes sence, because the notation gives a smart way to represent $$L \Psi (x)$$, which you can calculate.

but

$$L= \int \int <x|L|x'> |x><x'| dx dx'$$

Last edited: Jun 24, 2007
4. Jun 24, 2007

### StatusX

I don't know, you'd have to ask the person who gave you that approximation. I don't see why it would be useful.

Also, I don't know any web pages, maybe a professor or someone else here can help you. But I can tell you that a rough way to think about an integral of vectors or operators is in the same way as an integral of real or complex valued functions: as a limit of Riemann sums. So if we write f(x,x')=<x|L|x'>, you'd have:

$$L = \int \int f(x,x') dx dx' = \lim_{\Delta x, \Delta x' \rightarrow 0} \sum_i \sum_j f(x_i,x_j') \Delta x_i \Delta x_j' |x_i><x_j'|$$

The inside of the limit is always well-defined, and the limit is (probably) well-defined since we're assuming the space is complete. I'm not sure about the generality of this definition, but it's a good image to have in mind. In any case, I'm sure someone's defined a measure on these Hilbert spaces and set up a more rigorous theory of Lebesgue integration of the vectors and operators, but I'm not familiar with that theory. If you're worried about the mathematical foundations, I would suggest looking for a book on Hilbert space theory.

Last edited: Jun 24, 2007
5. Jun 24, 2007

### olgranpappy

...exactly wrong. You mixed up your 'x's and 'x-prime's this should read

$$L\Psi(x) = \int \Psi_0(x){\Psi_1}^*(x')\Psi(x') dx'$$

which, is also equal to

$$L\Psi(x) = \Psi_0(x) \left( \int dx' {\Psi_1}^*(x')\Psi(x')\right)$$

it takes any vector and makes it parallel to |0> with a magnitude that depends on how much of |1> is mixed into |Psi>.

6. Jun 24, 2007

### olgranpappy

...even then it doesn't make sense.

7. Jun 24, 2007

### olgranpappy

Maybe what you should be thinking about are the length scale on which
$$\Psi_1(x)$$ varies (call it lambda) compared to the other typical distances in your problem (seperation between arguments of your operator). If lambda is much greater than the other length scale then the approximation

$$L(x,y)\approx L(x,x)$$

is okay

8. Jun 24, 2007

### mrandersdk

ok thanks for your help, I try to ask him why it make sense to do what i'm suppose to do. If it makes sense i post the answer later.

9. Jun 24, 2007

### olgranpappy

no problem.

p.s. just curious--what's the article you're reading?

10. Jun 24, 2007

### StatusX

You're right, I switched them on the third to last line. I should have realized the answer is just |0><1|Psi>, which immediately gives the final equation, but I was focused on showing the general method for looking at linear operators in a certain representation.

Last edited: Jun 24, 2007
11. Jun 25, 2007

### mrandersdk

actually it's a big paper, but it can be found here:

http://www.phys.au.dk/ltc/publications/Specialerymann.pdf

the operator is defined on page 39, where $$\sigma^j_{\my v}$$ is the slowly varying (in time) part of the operator defined on page 13.

I know the operator here is a sum, but thought that if i could understand it for the simple one I could for the sum, but maybe not. Today i'm gonna try a little more.