# Picture of Entanglement

1. Jul 23, 2008

### wawenspop

I understand a wavepacket as a better description of a particle than the classical 'solid object' but is there a similar simple picture for an entangled pair?

Is it a double wave packet? Is it two joined wavepackets?
Is it one wave packet with two objects in it? Is it actually
one complex wave?

Pls you clever people have a shot at a description for me. It can't be
that difficult? Even google is failing me here.

2. Jul 23, 2008

### Staff: Mentor

Recall that the wave function for a single particle is a function of the three position variables plus time, ignoring spin for simplicity. The wave function for two entangled particles (again ignoring spin) is a single function of six position variables (three for each particle) plus time. I don't know a way to visualize easily a six-dimensional function. :uhh:

3. Jul 23, 2008

### wawenspop

So lets just consider the x direction and think about two entangled particles. I can picture two overlapping wave functions (analysed by product 2 Hilbert spaces) but is that correct and how does it look over a large separation? How does it look even when they 'overlap'. No obfuscating please!

4. Jul 23, 2008

### ZapperZ

Staff Emeritus
Have you looked at the wavefunction written for a bipartite system? If you have, then maybe you can describe what part of it that you don't understand. If you haven't, why not?

Zz.

5. Jul 23, 2008

### wawenspop

Yes, I have done the bra ket math and Hilbert space tensor multiplications to produce the entanglement equation and the energy density equation for entangled particles. I wish you would let someone else answer than you!

6. Jul 23, 2008

### vanesch

Staff Emeritus
If you are talking about some mental picture (not a formal description), then here's my personal one:
Imagine you have the state |x1> |x2> + |x3> |x4>
I see it as "two parallel worlds", which can still interfere with each other. In one world, the first particle is in point x1 and the second in point x2, in the second world, the first particle is in point x3 and the second in point x4.

7. Jul 23, 2008

### wawenspop

hmmm parallel worlds eh? Let me think for a while and get back to you on this, its very interesting.

8. Jul 23, 2008

### vanesch

Staff Emeritus
Of course it is :tongue2:

The reason why one can do this, is that the time evolution operator is a linear operator. So evolving the state |x1>|x2> + |x3>|x4> from time t1 to time t2 (using U(t1,t2) ) brings us to:

U(t1,t2) { |x1>|x2> + |x3>|x4> } = U(t1,t2) |x1>|x2> + U(t1,t2) |x3>|x4>

So these states evolve "independently". Of course, that's only true as long as we remain with unitary evolution. When a measurement is applied (projection on a certain measurement basis), then we have to project of course the entire state and not just the individual terms: that's where "interference" can be noticed: certain measurement basis terms which are present in each "world" individually add up algebraically, and can hence interfere constructively or destructively.

But I find it a good way to visualize unitary evolution.

That's BTW already good enough for "simple" superposition. For instance, a photon state that falls in onto a beamsplitter can be seen as going "one way" in one world, and "the other way" in the other world. In each world, the photon then undergoes whatever it has to undergo (going through lenses, mirrors, whatever), but it might meet again with "itself" in the other world upon an interference experiment.

9. Jul 24, 2008

### wawenspop

The tensor product for the two Hilberts spaces $$H_{A} \otimes H_{B}$$ for two entangled particles (consirering one state only),

Gives: $$\frac{1}{\sqrt{2}}(|0>_{A }\otimes |1>_{B}-|1>_{A}\otimes|0>_{B})$$

I assume there is no euclidean distance separation between particles (photons) in this equation? Or have we made some assumptions here about convergence etc?

If not, then the separation does not enter into the validity of this equation and we are left with a conceptual issue of a composite wave over a large separation where the particles at each end appear to 'know' the state of the other?

10. Jul 24, 2008

### LaserMind

Imagine a 1000 mile long wave packet that has detectors
at each end of its wavefront. The photon
must somehow 'know' that it is being observed and thus
must collapse its wave function all along its length instantly.
Its position along the wavefront is a probability, not the fact
that it 'bumps' into a detector.
At the instant it is detected, then the other part of the wavefront
must 'know' that its in fact gone because it was trying to detect it
at exactly the same instant.

To collapse it must know its being observed
at all parts of the front. Otherwise how is the choice made? If it
is a probabilty, the position of the detector is irrelevant. The order
of events is 1) I am being observed 2) I will only collapse my observable
statistically (not by the position of a detector)

11. Jul 25, 2008

### vanesch

Staff Emeritus
If each hilbert space is the full statespace of a particle, and if we assume that we entangle "classically-looking states (wave packets, say)", then in each term, the two particles are at more or less well-defined positions (and hence have a more or less well-defined distance between them).
In each term individually, the particles appear in a product state (state_of_1) x (state_of_2), and hence can be considered "independent" from one another. If the time evolution operator makes things only interact locally, then *within each term* the distant wave packets do not interact. (state_of_1) will evolve independently if (state_of_2) if they are spatially separated, and if the dynamics is "local".