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I see no formal proof of the Born rule.

Well, the normalizing condition $\int_\infty|\Psi|^2dx=1$ is because of Born rule if I am not wrong. Does this imply that our reality is a $L_2$ space?

If our reality is a $L_1$ space, will we have $\int_\infty|\Psi|dx=1$?

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    $\begingroup$ Related to your first line: physics.stackexchange.com/q/73329. This is a heuristic explanation, and when you search for formal proofs, you get it "proved" depending on the application it's put to. Possibly MathSE has covered this. $\endgroup$ – user163104 Jul 23 '17 at 1:46
  • $\begingroup$ @Countto10 The link explained why $\Psi^2$ works, but not proved that other rules won't work. $\endgroup$ – High GPA Jul 23 '17 at 2:32
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    $\begingroup$ No, I appreciate that, I just thought there might be a related link to a definite proof (which may not exist). The formal proof, even if I saw it, I would not be able to judge it's validity. I hoped the comment might encourage someone better than me to provide an answer. Anyway, +1, best of luck with it. $\endgroup$ – user163104 Jul 23 '17 at 2:45
  • $\begingroup$ Related: physics.stackexchange.com/q/41719/2451 and links therein. $\endgroup$ – Qmechanic Jul 23 '17 at 3:03
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  1. You see no formal proof of the Born rule because it is not a theorem of quantum mechanics, it is a postulate. See, for instance, this question and its answers.

  2. Normalization has nothing to do with the Born rule, since we can formulate the Born rule for arbitrary non-normalized states $\lvert \psi\rangle,\lvert \phi\rangle$ as $$ P(\lvert \psi\rangle,\lvert \phi\rangle) = \frac{\lvert \langle\psi\vert\phi\rangle \rvert ^2}{\lvert \langle\phi\vert\phi\rangle \rvert \lvert \langle\psi\vert\psi\rangle \rvert }.\tag{1}$$ See also this question about the (non-)necessity of normalization.

  3. $L^2$ space is not a necessary consequence of the Born rule, since eq. (1) makes sense in every Hilbert space, in particular also in finite-dimensional ones such as they are used for qubits and other two-level systems.

  4. Saying "reality is a $L^p$ space" mischaracterizes what quantum mechanics is saying. One of the basic assumptions of QM is that reality is well-modeled by saying that states are vectors (or, more accurately, rays) in a Hilbert space. Which Hilbert space is appropriate for modelling any given system depends on the system. Reality "is" no more a Hilbert space in QM than it is a symplectic manifold in classical Hamiltonian mechanics - these structures (inner product/symplectic structure) exist on the space of states, not "on reality".

  5. If we want a Hilbert space on which the canonical commutation relations $[x_i,p_i] = \mathrm{i}\hbar$ hold, then the Stone-von Neumann theorem guarantees that any such space is equivalent to $L^2(\mathbb{R}^n)$, with position acting as multiplication and momentum acting as differentiation. Even if this were not the case, other $L^p$ spaces cannot occur since $p=2$ is the only choice for which the space is a Hilbert space (i.e. possesses an inner product that induces the $L^p$ norm).

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