I often see both terms used in textbooks, but I am not sure whether I understand the difference between them. Both describe the state of a system, however, they seem different in some ways. From what I have found, what is important in the wavefunction is its direction. Wavevectors, on the other hand, require a magnitude and direction. What is their difference? And how are they related?
[Physics] Difference between a wavevector and wavefunction
quantum mechanicssolid-state-physicswavefunctionwaves
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Assuming that the incoming "first" particle is prepared in a pure state, interaction with another particle does seem necessary. Such an interaction might simply be the spontaneous emission of a photon or other particle by the original incoming particle, however.
Most importantly, such an interaction is not itself sufficient. For a measurement event to occur (wave function collapse in the Von Neumann formalism) we must also "physically lose track" of the some of the information of the interacting particle after the interaction has taken place, so that we must replace the entangled state description of the second particle after the interaction with a probabilistic mixture of such states, forcing a description of the first particle after the interaction in terms of a real valued probability density matrix rather than as the complex valued pure state amplitude we started with. This change of description automatically includes an increase in entropy, which also occurs physically.
Unless the second, interacting, particle either escapes the apparatus or interacts with a third particle which so escapes, i.e. "interacts with the environment", no measurement has yet occurred, the entire interaction is in principle reversible, and the complex amplitude description remains appropriate. Measurement requires "loss" (via decoherence) of the entangling information by further entangling with the environment and dissipation.
The escaping third particle is often an emitted photon or phonon. See the reference in the linked answer What is the difference between a measurement and any other interaction in quantum mechanics?, particularly the 1939 article by London and Bauer (but avoid their metaphysics) for details. More recently, see this book on quantum measurement theory, particularly page 102 referring to the view of Zeh.
You may have noticed that some ambiguity remains in this descriptipn. This has been analyzed in great detail and resolved by Zurek, but it gets a little tricky. See e.g. http://arxiv.org/abs/1001.3419 and references therein.
Dear Josh, the wave functions are only perfectly symmetric and/or perfectly antisymmetric - each of the factors is - in the case of two particles. As Fabian correctly says, for more than 2 particles, the wave function isn't perfectly symmetric and isn't perfectly antisymmetric with respect to particular transpositions of the two particles. The general character of the wave function's behavior is given by a Young diagram.
That's why what you want to be proved can only be proved for two particles. In the text below, I will therefore assume that there are just two electrons. What you want to be proved is that if the total $L=L_1+L_2$ of the two electrons is even, the wave function is even under the exchange of the two electrons, and it's odd if $L$ is odd.
It's not hard to see. By rotational symmetry, the states with a given value of $L$ form a multiplet with $2L+1$ components because $L_z$ may go from $-L$ to $+L$ with the spacing one. Without the loss of generality, you may focus on the wave function with the maximum value of $L_3$, namely $L_z=L$.
To get this maximum value, you need $L_z=L_{z,1}+L_{z,2}$ to be composed of two equally maximal, equally oriented terms. Because $L_{z,1}$ and $L_{z,2}$ also go between $-L_1$ and $+L_1$, or similarly between $-L_2$ and $+L_2$, it's clear that the only way to get $L_z=L$ is to have $L_{z,1}=L_1$ and $L_{z,2}=L_2$: both $z$-components must be maximum, too. But if it is so, then the wave function is simply $$ \psi_{L_3=L} = \psi_{L_{z,1}=L_1} \otimes \psi_{L_{z,2}=L_2} $$ For $L_1=L_2$, you can easily see that this wave function has to be symmetric under the permutation of $1$ and $2$. After all, it is the tensor product of the two equal pieces. If you antisymmetrized it, you would get zero. And in fact, the total wave function cannot have any complete symmetry or complete antisymmetry under the exchange of the particles $1,2$ if $L_1\neq L_2$. It's because $L_1^2$ acting on the total wave function gives you $L_1(L_1+1)$ times the wave function, while $L_2^2$ acting on the total wave function gives you $L_2(L_2+1)$. Because the two eigenvalues are not equal for $L_1\neq L_2$, the total wave function can't be symmetric under the exchange of $1,2$.
Again, the question about the symmetry of the orbital wave function only has a sharp answer is there are two particles and if they have the same $L_1=L_2$ - and in that case, the function is symmetric under the permutation.
Similarly, one can prove that the exchange of the two particles with the same half-integer spins $S_1=S_2$ produces a minus sign - assuming that $S_1=S_2$ differs from an integer by $1/2$. To do such things, it's useful to imagine that the components of the multiplet with a given $J$ are organized into a symmetric spintensor.
All the $2J+1$ components of the multiplet with the angular momentum $J$ may be expressed as a completely symmetric "tensor" with 2-valued spinor indices, $$ T_{abc\dots z} $$ Each index is either $0$ or $1$. The number of ones goes from $0$ to the number of indices $N$ - so there are $N+1$ components of this tensor.
Because each index brings $1/2$ to the total angular momentum, it's clear that the number of indices is $N=2J$. Indeed, then we have $N+1=2J+1$ components.
If there were pairs of indices that are antisymmetric with respect to the exchange of the two indices, one could factorize $\epsilon_{ab}$, a totally antisymmetric object. So only the total symmetry is relevant for our simplest case. Now, the spintensor for an angular momentum $J=J_1+J_2$ object may be written as the symmetrization of $$ T_{abc\dots z} = T^{1}_{(abc\dots m} T^{2}_{nop\dots z)} $$ where the parentheses represent the complete symmetrization - because the multiplet is represented by the totally symmetric tensor, as we said. The tensors $T^1$ and $T^2$ correspond to the $J_1$ and $J_2$ pieces.
However, if the total angular momentum $J$ differs by an odd number from its maximum value $J_1+J_2$, like in the case of the singlet where $J_1=J_2=1/2$ but $J=0$, then we must factorize those epsilons. $$ T_{abc\dots z, \,\,{\rm missing\,\,}{mn}} = T^{1}_{(abc\dots l} T^{2}_{op\dots z)} \epsilon^{mn} $$ The epsilon was added to the right hand side to reduce the number of indices by two - i.e. the total spin by one. Because the epsilon is antisymmetric, it changes the symmetry of the whole $T$ under the exchange of the two groups of indices. Each time you reduce the total $J$ by one, the symmetry property changes from symmetry to antisymmetry or vice versa. So the sign obtained from the permutation is given, for $J_1=J_2$, by $(-1)^{J-J_1-J_2}$.
It's also useful to know that the orbital wave functions of a single particle that can be expressed as spherical harmonics $Y_{lm}$ pick the factor of $(-1)^{l}$ under parity i.e. the factor of $(-1)^{l+m}$ under $\theta\to\pi-\theta$.
Best Answer
Wave functions, $\psi(x)$, describe the state of a quantum mechanical system, say a valance electron in an atom, in the position basis. An important set of states in any system are the eigenstates of the system's Hamiltonian. Another way to say this, is that these states have a well-defined (as in "not probabilistic") energy associated with them.
In a confined system such as an atom these energy eigenstates are discrete -- these are the ubiquitous shells of Bohr's modell or the orbitals of chemistry. In the case the electron moves arbitrarily far away as time passes, such as the case of a free particle moving through space, the spectrum is continuous, i.e. the energies of 'neighboring states' are infinitely close together. (This may not be 100% sound mathematically.)
For convenience, we seek to label these energies and their associated states by some other quantities than just the energy. (One reason is, that sometimes energies are degenerate, i.e. there is more than one different state with the same energy and we'd like to discern them.) These labels are referred to as quantum numbers. Now, with a valance electron in an atom, the quantum numbers are $n, l, m$ and spin -- you may have seen these in chemistry class to label orbitals. They arise from the mathematical treatment of the Schrödinger equation. In contrast, for a free particle, the energy eigenstates can be labelled by the particles momentum, which is proportional to a three-dimensional vector, $\vec k$, the so-called wave vector: $\vec p=\hbar\vec k$. People tend to set $\hbar=1$ and use momentum and wave vector synonymously.
In a solid-state environment -- the case you are interested in -- the quantum numbers turn out to be the band index (which is typically the same as the atomic quantum numbers) and a momentum or wave vector. This arises from Bloch's theorem.
Long story short: the wave vector (+ the band index) uniquely identifies an energy eigenstate of the solid state system which is itself described by a wave function.