I Operations regarding tensor-product states

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Summary
Confusion regarding the inner product of tensor-product states
I have a question on tensor-product states that I'd like to ask, thanks in advance!

1. The basis vector of a two-particle state can be written as ##|\mu_i \rangle |\nu_j \rangle## for orthonormal vectors ##|\mu_i \rangle, |\nu_j \rangle## spanning their single-particle Hilbert spaces. The inner product of basis vectors can be written as ##\langle \mu_{i'}|\langle \nu_{j'}|\mu_i \rangle |\nu_j \rangle = \delta_{i,i'} \delta_{j,j'}##

Does this hold for kets with continuous degrees of freedom? Eg momentum ##\langle p',q' |p,q \rangle = \delta(p-p') \delta(q-q')##. After reading Blundell's book on QFT however (screenshot provided below), I attempted to use creation and annihilation operators to obtain the above identity, but what I kept obtaining was ##\langle p',q' |p,q \rangle = \langle 0|\hat{a_{p'}}\hat{a_{q'}}\hat{a_{p}}^\dagger\hat{a_{q}}^\dagger|0\rangle = \delta(p-p') \delta(q-q') + \delta(q-p')\delta(p-q')##.


Screenshot 2019-06-19 at 5.44.00 AM.png
 

vanhees71

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It's all consistent! The point is that in your example from the book you deal with identical particles, and the QFT formalism includes the appropriate symmetrization (bosons) or antisymmetrization (fermions), i.e., you do not deal with the full product space but only with the totally symmetrized/antisymmetrized Fock states for bosons and fermions. A convenient basis for the Fock space are the occupation-number states, which is very natural, because you cannot know in any way which particle is which due to the indistinguishability of identical particles (with identical I mean that the particles carry exactly the same instrinsic quantum numbers, mass and various conserved charges of the Standard Model).

What's wrong in the book is to omit the spin indices. Only for scalar bosons there are of course none. Fermions necessarily have half-integer spin and thus you must include the spin degree of freedom in you single-particle basis.
 
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Ah yes, I completely forgot about the need to satisfy exchange symmetry, thanks for pointing that out!
 

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