Limit as distinguishable->indistinguishable

1. Jun 5, 2006

mikeu

I have a question that I sort of feel should be easy to answer, but I haven't figured out how yet... Hopefully someone can either show me the easy answer or tell me that it's a little more subtle :)

Consider a system of two particles, distinguishable by some continuous parameter, which can each be in one of two places. Suppose each particle has an equal probability of being in either place, so the system is in the state
$$|\psi\rangle=\frac{1}{2}\left(|ab;0\rangle+|a;b\rangle + |b;a\rangle+|0;ab\rangle\right)$$
where a and b represent the value of the parameter and the left- and right-hand sides of the semi-colon represent the two locations. This state is normalized for all distinct a and b, but has norm squared of 3/2 for b=a (since |a;b>+|b;a> becomes 2|a;a> when b=a). How does the norm of the state vector change discontinuously like that?

Next look at the probability of finding one particle in each location. Initially we have
$$p=|\langle a;b|\psi\rangle|^2+|\langle b;a|\psi\rangle|^2=\frac{1}{2}$$
but when b=a we get instead
$$p=\frac{|\langle a;a|\psi\rangle|^2}{|\langle\psi|\psi\rangle|} =\frac{2}{3}.$$
Again, how does it make sense for the probability to jump discontinuously like that? And why would it do that, it seems that if the particles are indistinguishable (bosons) the probability should still be 1/2 of finding one in each place.

I think I'm just overlooking something simple, but I've been hitting my head against this for a little while with no new thoughts, so hoped I'd find some help here.

Thanks,
Mike

Last edited: Jun 5, 2006
2. Jun 5, 2006

CarlB

Your calculation for norm squared = 1 makes the assumption that |a> and |b> are orthonormal. That is, you are implicitly assuming that <a|b> = 0. But in this case, you cannot make a continuous transformation that takes b to a.

To see this more explicitly, use an orthonormal basis like, for example, |L> and |R>. Set |a> = |L>, and let $$|b\rangle = \cos(\theta) |R\rangle + \sin(\theta)|L\rangle$$. Now compute the normalization of the state as a function of theta.

Carl

Last edited: Jun 5, 2006
3. Jun 6, 2006

vanesch

Staff Emeritus
I think the problem lies within this "distinguishable by some continuous parameter". Isn't this what leads to the Gibbs paradox ?
I remember reading in Reif that what saves us is that we cannot have, quantum-mechanically, systems which are "just an epsilon distinguishable".

4. Jun 6, 2006

Hans de Vries

For instance you can have:

indistinguishable: particles with identical spin.
distinguishable: particles with opposite spin.

You might go continously from distinguishable to indistinguishable by
rotating the spin of one of the particles. You will have to split the
amplitudes for spin-up and down cases, do the calculations for both
the distinguishable and the indistinguishable case and finally add the
propabilities together.

Regards, Hans

5. Jun 6, 2006

vanesch

Staff Emeritus

"particles of almost identical spin" :grumpy:

Do not forget that this distinguishable/indistinguishable comes into play when counting (orthogonal) microstates. As such, you can work in any quantized spin basis. It might be that you can neglect the spin degree of freedom, or not, but you cannot consider it to be *different* orthogonal microstates to have spin |n>, |n'>, |n"> ... where n, n' and n" are slightly different axes.
You have to count the relevant number of BASIS STATES.
So if you have to take into account "slightly different spins" you use the full set of spin basis states ; if the spins are of no matter, you don't count the degree of freedom.

It's more like "electrons of mass 511KeV, electrons of mass 511.3 KeV and so on, and that's not something that is bound to occur.

6. Jun 6, 2006

Hans de Vries

There's a whole bunch of distinguishable/indistinguishable examples at the
beginning of Feynman's"The theory of Fundamental Processes".

Particles with identical spin are indistinguishable.
Particles with opposite spin are distinguishable.

A system with one particle having a z+ spin and another having an x+ spin:
There is a 50% chance that the latter has a z+ spin (indistinguishable)
and 50% chance that it has a z- spin (distinguishable) The angle of
rotation is a continuous parameter which results in a continuous shifting
percentage from a 100% chance of being distinguishable to 0% chance
of being distinguishable.

My understanding of mikeu's question was that he feared there would
be a discontinuous jump in propabilities somewere:

But there is no discontinous jump in probabilities anywhere when going
from distinguishable->indistinguishable, since the total propability is
simply a continously shifting weighted sum of the distinguishable and
indistinguisable cases.

Regards, Hans

7. Apr 23, 2008

bilha nissenson

you are mentioning only spin, but in classical mechanics particles are distinguishable, also by position and time, could that be continuous? or almost? is the limit the uncertainty principle? So it is only mathematics? and we can say nothing more about it?

8. Apr 24, 2008

Hans de Vries

Feynman in his book "The theory of Fundamental Processes" talks in chapter one
about distinguishable and indistinguishable particles in the context of scattering
experiments, so, when they are in the same place at the same time.

indistinguishable particles do not interact (partly) as distinguishable particles
because they are not exactly in the same place at the same time.

Regards, Hans

Last edited: Apr 24, 2008
9. Apr 25, 2008

bilha nissenson

Thank you for the answer, but the fundamental question seems still open, what is actually the difference between distinguishable and indistinguishable particles, and when do they become one or the other? Are those circumstances some sort of an empiric mathematical definition?

[indistinguishable particles do not interact (partly) as distinguishable particles
because they are not exactly in the same place at the same time.
/QUOTE=Hans de Vries;1704177]

Last edited: Apr 25, 2008
10. Apr 25, 2008

Hans de Vries

For photons this is determined by the 90 degrees polarization angle. Horizontal polarized
photons are distinguishable from vertical polarized photons and they do not interfere.
Equally polarized photons are indistinguishable and they do interfere. Each angle
in between leads to partial interference.

For spin half particles this angle becomes 180 degrees. Spin up and spin down electrons
do not interfere. (for instance in atomic orbitals where they pair up that way), while
fermions with the same spin direction do interfere.

Regards, Hans

11. Jan 28, 2012

Degeneration

You say they do not interact the same because they are not in the same place at the same time, but if they have the same spatial wavefunctions and different spins, won't they still be in the same place at the same time? Do we still need to consider a superposition of the two (instead of psi_tot=psi(1)psi(2))?

12. Jan 28, 2012

DrDu

In my oppinion, the question has perfectly been answered in post #2. The states considered are position eigenstates which are orthogonal as long as a not equal b. So the expression is really discontinuous.