Particle number in Schrodinger equation (Hamiltonian method)

Question

Can it be proved using Hamiltonian mechanics that the number of particle for a given system remains constant after evolution in time? For example for a single particle S.Eq. is $$-\frac{\hbar^2}{2m}\frac{\partial^2 \psi(x,t)}{\partial x^2}+V(x)\psi(x,t)=i\hbar\frac{\partial \psi(x,t)}{\partial t}$$ so how can I prove that number of particle i.e $1$ remains same as the system evolves further in time using concepts of Hamiltonian mechanics.

I know that evaluating $\frac{\partial}{\partial t}\int |\psi(x,t)|^2 dx$ gives me $0$ assuming $V(x)$ is real so one can say that only single particle is there neither it decays nor any other particle pops out of nowhere.

Is this correct reasoning or is there any other way to look at total number of particle in a system using some exotic tools of Hamiltonian.

Answer 1

Yes, basically this is correct for first quantization. As the integral over $|\psi(x)|^2$ is interpreted as the number of particles, and this is conserved, then there is always one particle. Note that for your Hamiltonian to be valid (a hermitian operator) then $V$ must be real.

The full picture is a bit more complicated, though: let us use Dirac's notation and write the TDSE as $H|\psi\rangle = i\hbar \partial_t |\psi\rangle$. Now we can formally write the time evolution operator $U(t,t')$ that takes the state from time $t'$ to time $t$, which is an exponential function of the Hamiltonian. In this respect, the integral over $|\psi(x)|^2$ represents the expression $\langle \psi | \psi \rangle$ which we demand to be constant under time evolution. This means that $U$ is a unitary operator - which it should be as it is an exponential of $i$ times a Hermitian operator. This is related to conversation of probabilities and the unitarity of time evolution.

This, however, doesn't tell us a lot about the number of particles in the system :) The reason is that we didn't define what is the number of particles. Indeed, $|\psi\rangle$ might describe a state with a lot of particles, yet $\langle \psi | \psi \rangle = 1$. In first quantization, which is the way you wrote the TDSE, there is no mechanism that allows the changing of the number of particles, and its conservation is 'trivial' in a sense. It is just given as we describe it as a property of the system when we write the wave function and the equation, and it cannot change.

There are ways to describe a change in the number of particles, and they require what's called 'second quantization' -- which is exactly the quantization of the number of particles (this is the difference between first and second quantization, by definition). Then, we can define operators that count the number of particles of certain types, and they are similar to the creation and annihilation operators of harmonic oscillator, which you might be familiar with. The operator $\hat{n} = a^{\dagger}a$ is called a 'number' operator because it counts how many particles of type $a$ are in the system. Then, we can formally check if the number of particles is conserved by calculating the commutation relation of $\hat{n}$ with the Hamiltonian, just like any other operator. If $[\hat{n},H]=0$ it is conserved, and if it is nonzero it can change during time evolution.

Answer 2

The lecture notes for 221B on Quantum Field Theory by Hitoshi Murayama give another explanation, as follows. Another limitation of the multi-body wave functions is that it is incapable of describing processes where the number of particles changes. For instance, think about the emission of a photon from the excited state of an atom. The wave function would contain coordinates for the electrons in the atom and the nucleus in the initial state. The final state contains yet another particle, photon in this case. But the Schrodinger equation is a differential equation acting on the arguments of the Schrodinger wave function, and can never change the number of arguments.