Inner products on a Hilbert space

Join the discussion
Registration is free. Ask a follow-up in this thread, or start your own.
2 replies · 2K views
Decimal
Messages
73
Reaction score
7
Hello,

I am taking a quantum mechanics course using the Griffiths textbook and encountering some confusion on the definition of inner products on eigenfunctions of hermitian operators. In chapter 3 the definition of inner products is explained as follows: $$ \langle f(x)| g(x) \rangle = \int f(x)^{*} g(x)\, dx $$ Should you need to express some function ##f(x)## as a linear combination of functions ## f_n(x)## then the appropriate constants ##c_n(x)## can be found using Fouriers trick: $$c_n(x) = \langle f_n(x)| f(x) \rangle$$ This I understand. In chapter 4 this idea is applied to find the probability of measuring a certain spin of spin 1/2 particle in the x direction. The spinor ##\chi## will need to be expressed in the eigenfunctions of ##\textbf{Sx}##: ##\chi_{x+}## and ##\chi_{x-}##. So to find the appropriate coefficients one can apply fouriers trick again. $$c_+ = \langle \chi_{x+}| \chi\rangle$$ However when this inner product is calculated according to Griffiths: $$c_+ = \langle \chi_{x+}| \chi\rangle = (\chi_{x+})^{\dagger} \chi$$ It seems like the integral from the original definition has disappeared? Why is this? I understand you are working with vectors here instead of scalar valued functions, but does that change the definition of the inner product on the Hilbert space? This result looks a lot more like the standard definition of the inner product, yet the first vector is also conjugated. Can someone explain this difference to me? Thanks!
 
Physics news on Phys.org
These belong to a different space. It is finite dimensional, so it is the standard inner product. And it is complex, hence the conjugate.
 
Decimal said:
It seems like the integral from the original definition has disappeared?

There is a sum here (as is present when you write out the individual terms of the standard inner product for vectors represented in column form) instead of an integral. The sum and integral are analogous, except that one is for finite dimensional spaces and the other is for an infinite dimensional space.