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I was watching this PBS Spacetime video. He mentions that quantum particles / wavefunctions (he uses electrons as his example) all have their own set of separate 3D coordinates in a coupled system. Does this refer to each particle being a coordinate in quantum configuration space, or is this an incorrect interpretation? How would this encode the interactions between particles?

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The short answer is that, yes, if you have $N$ particles, and each particle can be in $m$ states individually, then $m^N$ complex numbers are required to fully specify the wavefunction of the system (not just $m\times N$). This is because the probability of measuring a particle in a particular state may depend on the state of all other particles. So particle 1 may have probability $x$ to be in state $|\uparrow\rangle$ if the other particles are in states $|\uparrow\rangle$$|\downarrow\rangle$$|\downarrow\rangle$$|\uparrow\rangle$$|\downarrow\rangle$$|\uparrow\rangle$, but probability $y$ if the other particles are in states $|\downarrow\rangle$$|\uparrow\rangle$$|\downarrow\rangle$$|\uparrow\rangle$$|\uparrow\rangle$$|\uparrow\rangle$, etc. and the same for all other particles and possible state combinations. This is known as entanglement and it can be generated by interactions between particles in indefinite states.

DFT and other quantum simulation approaches work when most of this complicated entanglement can be neglected. Ground states of most locally-interacting systems tend not to be too crazily spatially entangled, so you might get away with just storing $1000\times m\times N$ or $1000\times m\times N^2$ complex numbers in some smart way, dependent on the specific structure of entanglement.

"Each particle being a coordinate in quantum configuration space" is not quite right, depending on how you mean it. Remember a point in classical configuration space specifies the state of the whole system. Quantum wavefunctions are spread all through that configuration space. Rather, each particle adds new dimensions to configuration space for the wavefunction to inhabit.

That's a great video by the way, especially the first five minutes.

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  • $\begingroup$ Thank you so much! $\endgroup$ Commented Jan 13, 2022 at 5:58

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