Direct products in quantum mechanics

  • Thread starter RedX
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969
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Not every state can be represented by a direct product. Do states that can be written as a direct product have anything special about them?

It seems that states that can be written as a direct product lose correlation between between each individual state. More specifically, stronger than losing correlation, they become independent.

As an example, take two protons with angular momentum zero:

[tex]\psi(r_1,r_2) (|+->-|-+>) [/tex]

which is a direct product of a function of position, and an antisymmetric spin state. By writing the wavefunction as a direct product, you seem to have lost information on whether the spin up proton is near r1 or r2.

Indeed, measurement of position is independent of measurement of spin, and:

[tex]<f(r_1,r_2)g(s_1,s_2)>=<f(r_1,r_2)><g(s_1,s_2)>[/tex]

Going back to the two protons with angular momentum zero, why is it that the ground state is:


[tex]\psi_S(r_1,r_2) (|+->-|-+>) [/tex]

where S stands for symmetric in position. This total expression is definitely antisymmetric which is required for identical fermions. But does the ground state have to be written in a way such that the spin and positions are uncorrelated?

And why is the next excited state:

[tex]\psi_A(r_1,r_2) (spin \;1\; triplet) [/tex]

again uncorrelated?

Does this have to do with the fact that naturally, position doesn't care about spin, that it's not more likely for spin to like the negative x axis than the positive x axis?

Anyways, are there anymore special properties about states that can be written as direct products?
 
969
3
Yeah so this seems right. Take one particle in 2-dimensions, and say you want the two dimensions to be uncorrelated, say:

[tex](|x_1>+\pi|x_2>)\otimes (|y_1>+\lambda|y_2>)[/tex]

The amplitude to be in x2 is pi times more likely than to be in x1, and y2 lambda times more likely than to be in y1.

Multiplying it out:

[tex](|x_1 y_1>+\lambda|x_1 y_2>+\pi |x_2 y_1>+ \pi \lambda |x_2 y_2>[/tex]

So it doesn't matter if you measure x or y first. If you measure x then the wavefunction collapses and finding y2 is still lambda times more likely than finding y1 no matter what you got for x, and if you measure y first then finding x2 is still pi times more likely than x1.

Of course such a direct product state is silly, because usually in an experiment if you know where the particle is in the x-direction, then you have some idea of where it is in the y-direction.

What's interesting is that multi-particle states of identical particles are in general never totally uncorrelated, so you can get weirdness, such as:

[tex]<\psi|X_1|\psi> <\psi|X_2|\psi> \neq <\psi|X_1X_2|\psi> [/tex]

where [tex]|\psi> [/tex] is say [tex]|x_1>\otimes|x_2>+ |x_2>\otimes|x_1>[/tex]

where here X1 measures the x-position of the first particle, and X2 measures the x-position of the second particle. Intuitively they should multiply, and they do, but only if they're distinct particles so that the state can be written as a direct product that is not symmetrized: [tex]|x_1>\otimes|x_2>[/tex], or in general [tex]|\psi_1>\otimes|\psi_2>[/tex]
 

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