quantum mechanics - Can two bosons be called identical although their momenta are different?

In my introductory book of Quantum Field Theory says that because of the conmutation of the creation operators for a two particles sistem, bosons can be called identical. They give an example using a two-particle state $$|k_1,k_2\gt$$ and argue that this state is symmetric under the interchange of $k_1$and $ k_2$ so there is no possible way of distinguishing one quantum of energy from another.

But for me you can do it because $k_1$ is not necessary equal to $k_2$ so the quanta, or in QFT, the particles would have diferent momenta. So, can two bosons be called identical although their momenta are different?

Answer

The two bosons are identical as species, and their being identical bosons means that the wave function is a symmetric function of all the internal quantities that distinguish them such as the momenta.

So the wave function for two bosons may be written as a function of momenta, $\psi(k_1,k_2)$, and their being identical bosons means that

$$\psi(k_2,k_1) = +\psi(k_1,k_2)$$

If they have some other independent (commuting) pieces of information, such as the polarizations of the spin, those must also be exchanged for the equation above to hold.

So the momenta (or similarly positions) may be used to effectively distinguish two particles of the same type, e.g. two Higgs bosons. That's no contradiction. The word "identical" doesn't mean that they should be thought of as the "same object". Two Higgs bosons aren't the same object i.e. one object. They're two objects – with identical internal properties ("abilities") so the internal properties can't be used to distinguish them.

But they may occupy different states of various momenta or spin polarizations and those may be used to say "which is which".

Their being identical has consequences. In particular, if you have two identical bosons with momenta $k_1,k_2$ in the initial state, and two with $k_3,k_4$ in the final state, you can't say whether the particle with $k_1$ is "the same" as the particle with $k_3$, and $k_2$ is the same as $k_4$; or whether $k_1$ is the same as $k_4$ and $k_2$ is the same as $k_3$. Instead, you must sum over both histories (or corresponding Feynman diagrams) separated by the semicolons and only the total amplitudes gives you the right scattering probability amplitude.

On top of momenta (or positions, only one wave function is possible, they are Fourier transforms of each other) and spin polarizations, two identical particles may carry different colors (if quarks of 3 colors are said to be "the same particle" with some extra internal quantum number), or different isospin polarization (if neutron-and-proton or electron-and-neutrino are considered the same particle, a nucleon or a lepton doublet, respectively).

For example, if you have two nucleons, the wave function of these two identical fermions is antisymmetric – analogously to the symmetric one above. If one of them is known to be a neutron and one of them is known to be a proton, the particles may effectively be considered distinguishable. You may write the wave function of momenta (and spin polarizations) as a function where the neutron's momentum is written as the first argument, and the proton's momentum is the second. No extra symmetry constraints exist for this wave function. Alternatively, you may describe them as a wave function of the two momenta, two spin polarizations, and two isospin polarizations (the latter distinguishes protons from neutrons, one qubit with two possible values), and that function has to be antisymmetric under the simultaneous exchange of all properties of the two particles.