# Second quantization with qubits

Is "second quantization" means system wich can contain variable, unknown, superposed and otherwise uncertain number of qubits?

Can "second quantized" system contain 0.5% of 1 qubit and 95% of 2 qubits?

Does this mean that quantum field state cannot be described with quantum computer with fixed number of qubits?

Or may be this is wrong and the "power" of qubit is sufficient?

• In Fourier modes, states can be represented by creation operators, labelled by momentum and polarization, acting on vacuum. So, you can write a state like : $|\Phi> = 0.005~a~^+(\vec k, \lambda)|0\rangle + 0.95~a^+(\vec {k'}, \lambda')~a^+(\vec {k''}, \lambda'')|0\rangle$ that is : $|\Phi> = 0.005~|\vec k, \lambda\rangle + 0.95~|k', \lambda', k", \lambda"\rangle$ A state like $|k', \lambda', k", \lambda"\rangle$ is automatically symmetrical for bosonic fields and antisymmetrical for fermionic fields, because of the properties of the creation operators. Commented Aug 17, 2013 at 17:31

The next difficulty is mapping the fermion or boson operators to q-bit operators. Boson operators are easiest. If you fix the number of particles on a site to $$N=2^m$$, then you need m q-bits per site. The binary number given by this q-bit register tells you how many number of bosons are on that site. The creation operator/annihilation operator will have to be implemented with a sequence of gates acting on this q-bit register. The nice thing is that bosons commute, so adding sites, is as simple as adding registers.