4 ms·

Yeah, the relevant snippit is here: However, we never observe collapse to a single eigenstate of a continuous-spectrum operator (e.g. position, momentum, o

Yeah, the relevant snippit is here: However, we never observe collapse to a single eigenstate of a continuous-spectrum operator (e.g. position, momentum, or a scattering Hamiltonian), because such eigenfunctions are non-normalizable. In these cases, the wave function will partially collapse to a linear combination of "close" eigenstates (necessarily involving a spread in eigenvalues) that embodies the imprecision of the measurement apparatus. I suppose I just took away that the particle 'point' idea is likely just an asymptotic approximation of what the thing could be modeled as, but in reality - deep down at the atomic scales - it's really a 'linear combination of "close" eigenstates'.

tjradcliffe 12y ago
Speculation: the person who wrote that has never heard of the Dirac delta function? Regardless of what happens in reality, the Dirac delta is precisely a normalized point-like eigenstate (infinite height, zero width, unit area.)
- HCIdivision17 12y ago
  Reminds me now of the wonder I felt as after a few classes that had us doing integrals of various tedious geometries for EM fields, we come to QM - the really hard course - and we just pop a Dirac delta in to trivialize some integrals. It's like suddenly being able to say, "Enough! Just tell me the value right there!" Which turns out to reduce to your friendly neighborhood point particle model. (Pretty much by design, if I'm not forgetting my history correctly - it's sort of a made up tool that sort of abuses notation, like the Heaviside step function.) But deep down I never trusted that a particle was actually a basically zero dimensional-like point, though. It was just useful to think of it as one.
millstone 12y ago
The key word there is continuous-spectrum. A free electron can have any energy, and therefore the energy eigenstates are not physically realizable, due to the normalization requirements. But a bound electron, like in a quantum harmonic oscillator, has quantized energies and therefore can assume those eigenstates exactly. Think about electron orbitals in molecules: very precise energies.

tjradcliffe 12y ago

Speculation: the person who wrote that has never heard of the Dirac delta function? Regardless of what happens in reality, the Dirac delta is precisely a normalized point-like eigenstate (infinite height, zero width, unit area.)

HCIdivision17 12y ago
Reminds me now of the wonder I felt as after a few classes that had us doing integrals of various tedious geometries for EM fields, we come to QM - the really hard course - and we just pop a Dirac delta in to trivialize some integrals. It's like suddenly being able to say, "Enough! Just tell me the value right there!" Which turns out to reduce to your friendly neighborhood point particle model. (Pretty much by design, if I'm not forgetting my history correctly - it's sort of a made up tool that sort of abuses notation, like the Heaviside step function.) But deep down I never trusted that a particle was actually a basically zero dimensional-like point, though. It was just useful to think of it as one.

millstone 12y ago

The key word there is continuous-spectrum. A free electron can have any energy, and therefore the energy eigenstates are not physically realizable, due to the normalization requirements. But a bound electron, like in a quantum harmonic oscillator, has quantized energies and therefore can assume those eigenstates exactly. Think about electron orbitals in molecules: very precise energies.