# Help needed pls

1. Sep 21, 2005

### Neitrino

Gents pls help me on the below since im confused and getting frustrated

suppose QM system is characterized by this expression $$\int dp \ e^{-ipx} |p>$$
So how do i interpret it - one the one hand since that expression is |x>, I guess I say that my system is in ket state |x> and the observable which describes this system (position in this case) has eigenvalue x. Briefly if thinking about particle I say this particle is at position x.

on the other hand it is a superposition of plane waves exp(-ipx) and superposition of some |p> ket states right? I can no attach these two ideas - on the one hand particle at position x, on the other hand superposition of plane waves (free particle which has any momentum SOMEWHERE in space)

And what additional role |p> play here? (except with exp(-ipx) it gives me |x> )

Any response much appreciated

2. Sep 21, 2005

### George Jones

Staff Emeritus
Roughly, the set of all $\left| p \right>$ is a complete set of states, so

$$1 = \int dp \ \left|p \right> \left< p \right|,$$

and

$$\left|x \right> = \int dp \ \left|p \right> \left< p \right| \left x \right>.$$

Regards,
George

3. Sep 23, 2005

### Neitrino

Besides the above questions i have another question: if $$\phi(x)|0>$$ describes the creation of a particle at position x being in a superposition state of momentum and then its propagation/spread in all over space and if $$<0|\phi(y)$$ describes the same (particle birth at position y and its propagation in all over space) how/why does $$<0|\phi(y)\phi(x)|0>$$describe particle propagation from x to y ?

4. Sep 23, 2005

### dextercioby

Nope, it doesn't.

$$\left\langle 0 | \hat{T}\left(\hat{\phi} (x_{1})\hat{\phi} (x_{2})\right) |0\right\rangle =\mathcal{G}^{(2)}_{0} (x_{1},x_{2})=i\Delta_{F} (x_{1}-x_{2})$$

This is the propagator.

Daniel.

5. Sep 23, 2005

### Neitrino

ou sorry, seems I was no quite precise. I wanted to say if this is correct
how/why $$<0|\phi(y)\phi(x)|0>$$
is the expresion for apmlitude for a particle to prapagate from x to y. Because if partcicle emerged at position x ( ($$\phi(x)|0>$$ )and having momentum superposition state it just "flown" in oll over space and the same didi particle at position y. Why/how in this amplitude expresion $$<0|\phi(y)\phi(x)|0>$$ it "runs" from x to y? (emerged at position x it spread in all over space not particulary to position y)In Peskin Schroeder in the chapter Casuality is said - "the amplitude for a particle to prapagete from x to y is $$<0|\phi(y)\phi(x)|0>$$"

6. Sep 23, 2005

### vanesch

Staff Emeritus
Let me try to help you. The difficulty resides partly in the fact that we are working in the Heisenberg picture which allows for a less intuitive "flow of time".

Let us first look, in the Schroedinger picture, at a simple case in non-relativistic quantum mechanics. Suppose that at t0, our system is known to be in state |A(t0)> and we want to know what is the amplitude, at t1, for our system to be in state |B(t1)>. What we do in this case, is to evolve state |A> forward in time from t0 to t1, so this is |A(t1)> = U(t0,t1) |A>, and then find the "component" of this state in the direction of |B(t1)>, namely:

amplitude = <B(t1) | A(t1) >
or:

amplitude = <B(t1)|U(t1,t0)|A(t0)>

Now, if we consider A and B to be some fixed states (say, position states), we have of course that |B(t1)> = |B> and |A(t0) > = |A>, and we can write:

amplitude = <B| U(t1,t0) |A>

Let us now look at the Heisenberg picture, and let us take t0 as "reference time" where Schroedinger and Heisenberg states coincide. This time, our system which was in state A at t0 remains of course in state A in the Heisenberg picture, so:

|A> = |A-heisenberg>

however, our system which we wanted to see in state B AT TIME t1 has now to be calculated backwards to t0 to have its corresponding Heisenberg state:
|B(t1)> ---> |B-heisenberg> = U(t0,t1) |B>

Now this is great, because our same amplitude can now be written as:

amplitude = <B_heisenberg | A_heisenberg>

We see that the "amplitude to go from one state to another in the Heisenberg picture" is simply given by the inproduct of the two states, and this independent of the "time" which correspond to these states, because the Heisenberg picture has reduced all times to t0 in the states.

Now, in QFT, the HEISENBERG state of a particle being at a position (x,y,z) at a moment t is given by: |x,y,z,t> = phi(x,y,z,t) |0>

So it should then (naively) be clear that the amplitude to go from a state where the particle is at (x,y,z,t) to a state where the particle is at (u,v,w,s), is given by:
<u,v,w,s|x,y,z,t> = <0| phi(u,v,w,s) phi(x,y,z,t) |0>, or in 4-vector notation:

<0| phi(y) phi(x) |0>

However, as dexter pointed out, this is not entirely correct, because we cannot in fact know whether the particle was created at x and went to y, or whether the particle was created at y and went to x.

In practice this is not exactly what is done. One can make a choice between different propagators ; you can use the commutator [phi(y),phi(x)], or the time ordered product T{phi(y) phi(x)}.
The last one is understanable in that we would like to have event x BEFORE event y (we create before we destroy).
There are some subtleties here which are not very clear to myself either so some enlightment could be useful.

cheers,
Patrick.

Last edited: Sep 23, 2005
7. Sep 24, 2005

### Neitrino

Mr. Vanesch, dextercioby, George thks to u all.