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How is that, in many-body physics, particle creation and annihilations are possible even though it is a non-relativistic theory?

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  • $\begingroup$ Particle creation and annhilation comes about by advancing from a Hilbert space to a Fock space. This doesn't require relativity in principle, it's just that standard QM is inconsistent when used to describe a relativistic theory. $\endgroup$ – Jerry Schirmer Oct 7 '14 at 18:26
  • $\begingroup$ You mean by non-linear effects or inhomogeneities (like lattice defects or dopant atoms)? $\endgroup$ – CuriousOne Oct 7 '14 at 18:26
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    $\begingroup$ The question is pretty vague, and asserts something that sounds false as a general statement. But anyway, trying to guess what might be relevant --eE.g., in low-energy nuclear physics we talk about creating quasiparticle excitations, which are basically particle-hole excitations. I believe this happens in condensed-matter physics as well. A lot of the formalism is similar to what you see with particle-antiparticle excitations, even though the calculations are nonrelativistic. $\endgroup$ – Ben Crowell Oct 7 '14 at 18:33
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Can we have particle creation or annihilation in non-relativistic physics?

The answer is an emphatic yes. It is quite a routine in non-relativistic (NR) many-body physics.

How is particle creation or destruction possible without relativity?

Absence of relativity does not allow creation (and annihilation) of particle-antiparticle pairs (or mass) with disappearance (or appearance) of energy. Non-relativistic physics, therefore, forbids energy turning into mass or vice-versa. Since relativity is not incorporated in non-relativistic many-body theory, mass creation or mass destruction is not possible.

What is happening then?

The energy supplied to a Fermi vacuum (which is simply the Fermi level) "creates" an electron-hole pair in the lattice. This electron-hole pair "creation" is not same as the electron-positron pair creation from the vacuum of a relativistic quantum field theory. Had it been so, one would have to supply at least $2m_ec^2$ amount of energy to the Fermi vacuum to create the electron-hole pair which is the minimum energy needed to create an electron-positron pair in relativistic physics. In metals, energy goes into the electron (already present at the Fermi level), which then absorbs it to get excited at a higher level. You can call this appearance of the electron above the Fermi level and the hole below it as the "creation" of the electron-hole pair. The typical incident energy $h\nu(\ll m_ec^2)$ is sufficient to really create an electron-hole.

Addendum

Moreover, particle creation may not require relativity if the particle is gapless excitation. This is the case for phonons and magnons which are essentially non-relativistic concepts. Any tiniest bit of energy will create them. The energy supplied goes to the lattice vibration and there is nothing like energy being converted to rest mass.

If we look at the "second-quantized" Schrodinger equation, the Hamiltonian doesn't contain the rest energy. This is because, in non-relativistic physics, the mass cannot change to energy and therefore, mass is eternal and can be ignored in the Schrodinger equation. On the other hand, relativistic quantum field theories start with Lagrangians that include the rest mass from the very beginning.

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    $\begingroup$ maybe one should stress that the particles created are not elementary standard model particles, but mathematical definitions of particle states. the creation and annihilation operators on harmonic oscillator states for example. Assigning the name "particle" to the creation does not make it an elementary particle. It is just counting the number of states. $\endgroup$ – anna v Jan 30 '18 at 8:46
  • $\begingroup$ @annav I agree. $\endgroup$ – SRS Jan 30 '18 at 8:47
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I don't know if that is the case, but you may have been misguided by the formalism.

For convenience, the language of second quantization is often utilized in non-relativistic many body physics. So you may have stumbled upon creation and annihilation operators in that context. Nevertheless, even if written in second quantization, the Hamiltonians of many non-relativistic particles preserve the number of particles.

A concrete example: the Hamiltonian of $N$ non-relativistic bosons with pair interaction (symmetric) potential $V$ is usually written on $L^2_s(\mathbb{R}^{3N})$ (the symmetric $L^2$, under permutation of the three dimensional variables $x_i$, $i=1,\dotsc,N$) as: $$H=\sum_{i=1}^N-\frac{\Delta_i}{2M}+\frac{1}{N}\sum_{i<j}V(x_i-x_j)$$

On the $N$-particle sector, $H$ agrees with the second quantized operator on $\Gamma_s(L^2(\mathbb{R}^3))$ (symmetric Fock space over $L^2(\mathbb{R}^3)$): $$\tilde{H}=\int_{\mathbb{R}^3}a^*(x)\Bigl(-\frac{\Delta_x}{2M}\Bigr)a(x)dx+\frac{1}{2}\int_{\mathbb{R}^6}V(x-y)a^*(x)a^*(y)a(x)a(y)$$

So, as you see, the operator $\tilde{H}$ uses the formalism of creation and annihilation operators $a^\#(x)$, but the operator leaves each fixed particle sector invariant, as expected in a non-relativistic theory.

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