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$\begingroup$ I would like to point out that the concept of the wave vector in lattice vibrations (or vibrations in general) has nothing to do with Quantum theory, it is a purely classical concept. If you simply want to enumerate the eigenmodes of vibration for a lattice of harmonic springs, the wave vector picture is the natural one.
In special relativity the four-vector k k is defined: k = (k, ω c) k = (k, ω c) We can then write: k ⋅x − ωt = k ⋅ x k ⋅ x − ω t = k ⋅ x. Where of course: x = (x, ct) x = (x, c t) My question is how do we know k k is indeed a four vector? I'm asking because before using its property of transforming by Lorentz Transformation from ...
The wave vector is normal to the surfaces of constant phase ('wave fronts') with a length inversely proportional to wavelength. It corresponds to wave momentum, and is one way to describe wave propagation (another one is energy flux, which, in case of electromagnetic waves, is given by the Poynting vector; depending on the medium, wave and ...
Hence we have 1 λ2 = 1 λ2x + 1 λ2y. We can now define the components and magnitude of the wave-vector k in the usaual way (kx = 2π / λx, ky = 2π / λy, k = 2π / λ), and get the Pythagorean relation for the wave-vector k2 = k2x + k2y. The extension from 2-dimensional to 3-dimensional space is easy.
Wave functions, ψ(x) ψ (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 ...
The basic classical interpretation of this expression is that EM waves, or any wave described by a null 'wave-4-vector' like gravitational waves, travel at the speed of light. You can easily see this by considering the classical definition of the wave-4-vector k = (ω c,k) k = (ω c, k), where k k is the usual wavevector in 3-space.
2. Think about what the wave vector represents, and what kind of wave your equation describes. The most general equation for the spatial variation of the electric field for a plane wave is. E(r) =E0e−ik∙r E (r) = E 0 e − i k ∙ r. where E0 E 0 is some vector with no dependence on r r.
The "direction of wave propagation" is the direction of a wave's energy flow, and the direction that a small wave packet will move, i.e. the direction of the group velocity. For light waves, this is also the direction of the Poynting vector. On the other hand, the wave vector points in the direction of phase velocity.
From [1], I understand that, "The four-wavevector is a wave four-vector that is defined, in Minkowski coordinates, as $$\mathbf{K}^\mu = \left(\frac{\omega}{c}, \mathbf{k}\right).\tag{1}"$$ I am studying special relativity using [2]. In [2] Gray, there is derivation of the four-wavevector.
The question is, how would we define the direction of the wave or rather how would we define the unit vector of the complex wave vector (I am thinking of something like $\hat{k}$). Because in the case of a real wave vector, there is a physically conceivable concept of a norm and we, therefore, divide the vector by its norm to arrive at a unit ...