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A complex scalar field that describes a quantum mechanical system. The square of the modulus of the wave function gives the probability of the system to be found in a particular state. DO NOT USE THIS TAG for classical waves.

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First of all, remember we are talking physics here, and in physics zero and so small we cannot measure it are just the same thing. Having said that, it is not true that one always assume that the "el …
answered Dec 27 '14 by glS
13
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Where did you get that formula? The correct normalization does not involve the time integral. Denoting with $| \psi(t) \rangle$ the state at time $t$, the normalization condition reads $$ \tag{A} \la …
answered Jan 21 '15 by glS
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No, it is just a matter of convention, which heavily depends on the context. As an example, in basic quantum mechanics, when dealing with single- and many-particle states, $\psi$ tends to be used for …
answered Dec 12 '17 by glS
4
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The way you wrote it, they are distinguishable (unless $a=b$ of course). For the particles to be indistinguishable their wavefunction must be of the form $$ \psi(r_1,r_2) = \frac{1}{\sqrt{2 … distinguishable: you can measure one without affecting the other. A separable wavefunction like the one you wrote can describe a couple of non-interacting, distinguishable particles. If …
answered Dec 21 '14 by glS
4
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Are you asking why taking the squared modulus of a superposition of (eigen)states turns out to be considerably more complicated than the squared modulus of a single eigenstates? It this is so, I'd say …
answered Nov 14 '14 by glS
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If you are referring to a global phase factor $e^{i\phi_0}$ (with $\phi_0 \in \mathbb{R}$ a real number, NOT the azimuthal angle $\phi$ upon which the hydrogenic wavefunction depends) multiplying the … whole wavefunction, then that has probably been simply taken to be 1, as this choice does not affect any of the physics. This is because the probability density of finding a particle in some location …
answered Jan 24 '15 by glS
9
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Those factors usually come out when separating the wave-function for a 3-dimensional problem (like the Hydrogen atom) in its radial and angular parts: $$ \psi(r, \theta, \phi) = R(r) Y(\theta, \phi). …
answered Oct 17 '16 by glS
46
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that they tend to also not be in similar states. This amounts to an effective repulsive effect between particles. You can see this by remembering that to get a physical two-fermion wavefunction you … have to antisymmetrize it. This means that if the two single wavefunctions are similar in a region, the total two-fermion wavefunction will have nearly zero probability amplitude in that region, thus …
answered Oct 25 '16 by glS