Direct products in quantum mechanics

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SUMMARY

The discussion centers on the properties of states in quantum mechanics that can be represented as direct products. Such states exhibit independence between individual components, as demonstrated by the example of two protons with angular momentum zero, where the wavefunction is expressed as a direct product of position and antisymmetric spin states. This independence leads to the conclusion that measurement outcomes for position and spin are uncorrelated. The ground state of identical fermions is shown to be antisymmetric in position while remaining uncorrelated in spin, highlighting the unique characteristics of direct product states.

PREREQUISITES
  • Quantum mechanics fundamentals
  • Understanding of wavefunctions and measurement in quantum systems
  • Fermionic and bosonic statistics
  • Linear algebra concepts related to tensor products
NEXT STEPS
  • Explore the implications of direct product states in quantum entanglement
  • Study the role of symmetry in quantum mechanics, particularly in identical particles
  • Investigate the measurement problem in quantum mechanics
  • Learn about the mathematical formulation of quantum states using Hilbert spaces
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Quantum physicists, advanced students of quantum mechanics, and researchers studying the properties of multi-particle systems will benefit from this discussion.

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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:

\psi(r_1,r_2) (|+->-|-+>)

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:

<f(r_1,r_2)g(s_1,s_2)>=<f(r_1,r_2)><g(s_1,s_2)>

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


\psi_S(r_1,r_2) (|+->-|-+>)

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:

\psi_A(r_1,r_2) (spin \;1\; triplet)

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?
 
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Yeah so this seems right. Take one particle in 2-dimensions, and say you want the two dimensions to be uncorrelated, say:

(|x_1>+\pi|x_2>)\otimes (|y_1>+\lambda|y_2>)

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:

(|x_1 y_1>+\lambda|x_1 y_2>+\pi |x_2 y_1>+ \pi \lambda |x_2 y_2>

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:

<\psi|X_1|\psi> <\psi|X_2|\psi> \neq <\psi|X_1X_2|\psi>

where |\psi> is say |x_1>\otimes|x_2>+ |x_2>\otimes|x_1>

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: |x_1>\otimes|x_2>, or in general |\psi_1>\otimes|\psi_2>
 

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