# Momentum in two different k-points

1. Sep 6, 2015

### hokhani

For two different points $k$ and $k\prime$ in the Brillouin zone, can we have the same momentum? (By momentum I mean the real momentum, not the crystal momentum $\hbar k$ or $\hbar k\prime$).

2. Sep 11, 2015

### DrDu

Sure, e.g. phonons all have true momentum=0, independently of the k value. For other particles, like electrons, doesn't have a well defined value in crystals, i.e. the electron is not in a momentum eigenstate. Hence, the momentum distribution of electrons with different k values will in general overlap, so that it is possible to obtain the same momentum when measuring it.

3. Sep 11, 2015

### MisterX

I think maybe not for Bloch wave states, although I am not sure.
If there is crystal momentum at some real momentum p would be shown in $<n, k| p>$
$$\psi_{nk}(x) \propto e^{ik\cdot x} u_n(x)$$
$$<n, k| p> \propto \int dx e^{i(p/\hbar - k)x}u_{n}(x)$$
This can be interpreted as a Fourier transform. Because $u_n$ is periodic it will only have Fourier components at certain frequencies.
$$p/\hbar - k = G$$
for any
$$G = g_1\mathbf{b}_1 + g_2\mathbf{b}_2 + g_3\mathbf{b}_3,\, g_i \in \mathbb{z}$$
with $b_i$ being reciprocal lattice vectors.
Note that this does not mean that there is a momentum component at that value of $k$, it just means there could be. Whether or not it does depends on $u_n$.
So, although each $k$ couples to many different values of momentum, it seems those values of momentum are not shared with others in the same B.Z. But you will not get the values of $p$to overlap for two $k$ in the same BZ. I am trying to illustrate this idea with the following crude picture.

You would have to go out of the BZ to get another value of k that corresponded to the same p. Once outside the BZ you could add a reciprocal lattice vector.
$$p_0/\hbar = k_1 + \underbrace{0}_{G_1} = \underbrace{k_1 + \mathbf{b}_1}_{k_2} - \mathbf{b}_1 = k_2 + G_2$$
So
$$k_2 =k_1 + \mathbf{b}_1$$ would also correspond to the same momentum $p_0$. This was by using a different value of $G$, $G_2 = - \mathbf{b}_1$. But by adding $b_1$, we left the BZ of $k_1$.

Last edited: Sep 11, 2015
4. Sep 14, 2015

### hokhani

Yes, I think so but if we take the expectation value of momentum, namely p=m vgroup , where vgroup is proportional to the slope of energy band, it seems reasonable to have the same momentum at different points of BZ.

5. Sep 16, 2015

### zhanhai

I am confused. When an electron absorb a phonon, it receives both the energy and momentum of that phonon; and a phonon mode is characterized by its energy and momentum (wavevector), and all modes in one branch have their energy-wavevector characteristics face, correct?

Second, as an electron is not in a momentum eigenstate, it would not be associated with any specific k value. But I agree with that the electron will take a k value once it is measured.

6. Sep 17, 2015

### DrDu

You are confusing true momentum p and crystal momentum (or wavevector) k. When an electron absorbs a phonon, the wave vector is conserved up to a lattice vector. True momentum is also conserved, but it is not carried by the phonon, but by the whole crystal.