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Physics
Quantum Physics
Inner products on a Hilbert space
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[QUOTE="Decimal, post: 6053673, member: 625746"] 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 defintion 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! [/QUOTE]
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Inner products on a Hilbert space
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