# Dirac Projection Postulate

1. Dec 28, 2006

### mlukowski

There is many projection (or measurement) postulates in quantum mechanics axioms: von Neumann measurement, Luders postulate...
But does anybody know sth. about DIRAC POSTULATE?
Thx

2. Dec 29, 2006

### Symbreak

Dirac's best known postulate is his antiparticle postulate; that every particle has an antiparticle with equal mass but opposite charge and spin. This has been proven.

We get a good picture of particles/antiparticles through Dirac's 'antimatter field' - whereby antiparticles are disruptions in a symmetric energy field.

3. Dec 29, 2006

### mlukowski

not this!!!

I didn't mean THIS good-known postulate. I'm working with quantum theory of measurement, and I was asking about Dirac's projection postulate, which is one of Axioms in QM!
THX anyway...

Last edited: Dec 29, 2006
4. Jan 3, 2007

### dextercioby

The projection postulate of QM in the Dirac formulation is due to Dirac and von Neumann. I've seen it attributed to von Neumann rather than to Dirac, but this is less relevant.

It states that for a quantum system found in the pure state $\psi$ one measures an observable and gets the eigenvalue $a$ of the s-adj op. $A$, then, immediately after this measurement the system jumps into $P_{\mathcal{H}}_{a}}\psi$ state.

Daniel.

5. Jan 3, 2007

### mlukowski

OK dextercioby, I know, but look at this:

If we simultaneously measure commuting observables A, B and C, and the outcome connected with A is between $a_1$ and $a_2$, connected with B is between $b_1$ and $b_2$, connected with C is between $c_1$ and $c_2$, then after the measurement statevector is given by equation:

$[E_A(a_2)-E_A(a_1)][E_B(b_2)-E_B(b_1)][E_C(c_2)-E_C(c_1)]|\psi\rangle = |\psi\rangle$

You can find this strange measurement postulate in book ,,Mathematics of Classical and Quantum Physics" By Frederick W. Byron, Robert W. Fuller in chapt. 5.11. I'm not sure, but it may be called ,,Dirac's postulate".

What does this mean? Statevector AFTER the measurement is not connected with statevector BEFORE measurement? How Born's probability rule works here (is it works)?

6. Jan 3, 2007

### dextercioby

This is both new and weird to me. Need to check the book and references listed.

Daniel.

7. Jan 4, 2007

### Staff: Mentor

I don't have Byron & Fuller handy, but this looks to me to be just an unusual way of stating that after the measurement the statevector is an eigenfunction of all three observables.

I don't see this as making any statement about the statevector before the measurement. I certainly wouldn't call it a projection postulate.

Last edited: Jan 4, 2007
8. Jan 4, 2007

### mlukowski

I agree, that it shouldn't be called projection postulate. We just prepare, not project. Pay attention on the interesting feature of this theorem: If measure $E_{A,B,C}$ is dim=1 then we know everything about the state $|\psi\rangle$, but if it is more than dim=2 the equation has free parameters. If so, statevector is not (but may be) one of eigenfunctions of all three observables A, B, C.