# For an event that can occur in many ways, why is the wavefunction of the event the sum of wavevfunction for each way separately?

The wavefunction of identical particles is given as: $$\psi_{1,2} (x_1,x_2) = \psi_1(x_1)\psi_2(x_2) + \psi_2(x_1)\psi_1(x_2)$$ . Why is it so? Why is it the sum of the two states? What is the explanation behind this? Yes, I know that wavefunction being a linear combination of the solutions of Schrodinger's equation is a solution itself. But that does answer only mathematically.

In order to get a deep insight, I again read Feynman's introductory lectures where he wrote:

When an event can occur in several ways, the probability amplitude for the event is the sum of the probability amplitudes for each way considered separately. There is interference. $$\phi = \phi_1 +\phi_2$$ .

Why is it so?

• This choice reflects the fact that (in this case) microscopic "particles" are indistinguishable. If you do an experiment, there is no way to tell if you are looking at particle 1 or particle 2. All you can tell is whether you measured "a particle" or not. We basically have to undo the fact that we are using two independent functions that are distinguishable in our calculation. The mathematical symbols $\psi_1$ and $\psi_2$ are classical objects, but they are describing quantum mechanical objects which have fewer properties than the symbols, so we need to perform a symmetrization. Jun 26, 2015 at 19:34
• @CuriousOne: You are right, sir; we don't know which state it is at now & so it is a superposition of states having equal probability;either former or later. I'll only ask you: does this wavefunction collapses into one state when we measure like other quantum measurement problems? I am having some doubt whether we can actually measure which particle is at which state but hope that it will also follow the measurement-problem & wavefunction-collapse. Just can you clarify it?
– user36790
Jun 27, 2015 at 2:15
• The "collapse of the wavefunction" is a figure of speech for the Born rule. No such thing exists. What we mean by that is that for a measurement we have to couple the quantum system strongly to a classical measurement apparatus, which means that we have many thermodynamic degrees of freedom in play which have nothing to do with the wavefunction that we are trying to measure. There are multiple ways to "prove" that after the coupling the resulting wavefunction has the same properties as a "collapsed" one would have, see "decoherence": en.wikipedia.org/wiki/Quantum_decoherence Jun 27, 2015 at 4:30
• A cat in a closed box is a dead cat, classically, quantum mechanically, biologically, medically, philosophically and theologically. :-) Jun 27, 2015 at 4:46
• A pure state represents the system's actual quantum mechanical state, a mixed state in addition includes the level of our statistical knowledge of that state. The equivalent of mixed states in classical mechanics would be a probability distribution due to measurement errors or to measure average properties of a system. Jun 28, 2015 at 19:52

As per the first question: given $\mathcal{H}_1$ and $\mathcal{H}_2$ as the Hilbert spaces of states for the solutions of the equations of motions for the particles $1$ and $2$, the Hilbert space of the system composed by both identical particles is the symmetric (or anti-symmetric) tensor product thereof, and therefore its states can be expressed as any symmetric (or anti-symmetric) linear combinations of the initial states $\psi_1,\psi_2$ of the two particles. This is the so called Pauli principle which is proven true experimentally, no other reason than that.
As per the second question: given $A$ and $B$ as two different events, the probability of either of the two is simply, by definition, $P(A \cup B) = P(A) + P(B)$. If you then interpret the wave function modulus square as a probability density function then Feynman's formula automatically follows.