
#1
Nov1412, 08:08 AM

P: 34

Hello
I have a problem in understanding wave functions Let [itex]q \mapsto \phi(q)[/itex] a position wave function for a single particle of mass [itex]m[/itex] The equivalent momentum wave function is said to be computable with using fourier transform: [tex]\psi : p \mapsto \int_q \phi(q) \cdot e^{i/\hbar \cdot \langle p, q\rangle} \delta^3 q[/tex] with [itex]\langle , \rangle[/itex] the duality bracket or inner product (depending if you consider duality or not). But I feel unconfortable because technically (if you consider strictly galilean spacetime) [itex]q[/itex] is not a vector but a point in an affine space. Turning it into a vector is equivalent to choosing a origin (= injecting nonphysical data into my modelisation). Similarly the set of all possible [itex]p[/itex] for my particle is also a affine space. Turning it into a vector is equivalent to choosing a inertial frame (= which again corresponds to nonphysical data in my modelisation). By making explicit those two origins, I can compute: [tex]\psi_0 : p \mapsto \int_q \phi(q) \cdot e^{i/\hbar \cdot \langle p  p_0, q  q_0\rangle} \delta^3 q[/tex] [tex]\psi_1 : p \mapsto \int_q \phi(q) \cdot e^{i/\hbar \cdot \langle p  p_1, q  q_1\rangle} \delta^3 q[/tex] And of course [itex]\psi_0 \neq \psi_1[/itex]. But by defining the equivalence relation : [tex]\psi_A \sim \psi_B \Leftrightarrow \left( \exists p_*, \exists \vec p, \exists \vec q: \psi_A(p) = e^{i/\hbar \cdot \langle p  p_*, \vec q\rangle} \cdot \psi_B(p + \vec p) \right)[/tex] we have [itex]\psi_0 \sim \psi_1[/itex]. It seems that [itex]\sim[/itex]equivalence on (position or momentum) wave functions is compatible with vector space structure and with Fourier transform (though I've not checked for the hermitian product). So my question is: am I right to consider that two equivalent (position or momentum) wave functions are physically equivalent and that one should not consider the classical Hilbert spaces of position and momentum wave functions but rather their respective quotient spaces by relation [itex]\sim[/itex] ? If true, should I consider an even weaker equivalence relation ? Thanks 



#2
Nov1412, 10:50 AM

Sci Advisor
HW Helper
P: 11,866

Of course, all you've 'discovered' is the projective Hilbert space of a quantum system, i.e. the space of (unit) rays on a (complex, separable) Hilbert space, i.e. the space of all pure quantum states. A pure quantum state is thus an equivalence class of vectors.




#3
Nov1512, 05:28 AM

P: 34

Actually I was wrong on one point: my relation does not conserve the vector space structure and I feel that it is not a very good candidate. But even if we don't care about the vector space structure, the [itex]\mathbb{C}[/itex]colinearity you're suggesting is still too strong. It separates different wavefunctions that should not be separated for the reason I've explained in my previous message: Fourier transform will match a single position wavelength to non[itex]\mathbb{C}[/itex]colinear momentun wavelength depending on how your choose a spacial origin and an inertial frame. 



#4
Nov1612, 02:41 AM

P: 34

Equivalence of wave functions
So this means nobody has a clue on how to make quantum mechanics consistant with change of spatial origin or change of inertial frame ?




#5
Nov1612, 10:23 AM

Sci Advisor
HW Helper
P: 11,866

Hmm, I'm thinking that the affine structure on the parameter space (R, R^2, R^3 for Quantum mechanics done on noncurved manifolds) is implemented in the formalism through the linear unitary (continuous in the norm topology) representations of the translation group, i.e. psi(x) and psi(xx_0) as ordinary vectors of unit modulus from the respective rays are linked through a unitary operator, if the quantum system admits pure space translations along Ox axis as a symmetry (consequence of theorems by Wigner and Bargmann).
The Fourier transformation takes the ray of psi(x) into the ray of psi(p), inasmuch the ray of psi(p) is mapped into the ray of psi(pp_0). 



#6
Nov1912, 07:47 AM

P: 34




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