# The wavefunction of the superconductor A consists of two parts: B and C

I can only remembered that when $H_A=H_B+H_C$, one can write $|\psi_A>=|\psi_B>\otimes |\psi_C>$. I don't know how this is related to the above situation.
My basic understanding is this: if the system $B$ has a different phase of that of $C$, then $B$ and $C$ will interact through josephson effect. The particle number will flow from $B$ to $C$(or $C$ to $B$), depending on their relative phase. However, if their phase is same, they have no interaction, thus the whole system can be considered as two independent part.
However, why can we choose a $N_B$ that the phase $\phi_B=\phi_C$? Even if the initial phase is not same, and the Josephson effect happens; after sometime, their phase will be the same, because the current cannot flow forever. At this time, we count the number of B and C, why can't we write A as a direct product of them.
• If we go to a basis with fixed particle number, a state where the relative superconducting phase between $B$ and $C$ is a highly entangled state. A product state with fixed numbers on $B$ and $C$ does not have a well-defined relative phase. – Meng Cheng Apr 27 '15 at 17:12
• Mathematically speaking, the Fourier transform of a product is not a product. This is already a guess for the answer. You can think of the separation from A to B+C as making a Josephson system as well. Then the phase in A (say zero) can be split in B (say $+\varphi$) and C (say $-\varphi$) with well localised phases. The number of particles (electrons, say) also splits nicely, but tunnelling phenomena takes place, and so the number of particle is not a localised quantity. Say differently, you generate entanglement between the left and right parts, and the wave function can not be factorised. – FraSchelle Apr 28 '15 at 10:41