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Quantum mechanics starts with wave functions living in Hilbert space. But later for Born's interpretation, the wave function need to be of unit energy (I mean total probability = 1, $\int_{-\infty}^{\infty}|\psi(x)|^2dx = 1$). But for two elements of Hilbert space when summed to get a third element in the same space, the third need not be of unit energy. So these two assumptions mathematically inconsistent. Either you have to throw away Hilbert space or the Born's interpretation.

If we throw away Hilbert space, assume $\psi$ to be of unit energy and is an element of a metric space (rather than a vector space) with suitable metric defined on it. Would this kind of an idea which makes more mathematical sense would lead to a physical theory. Right now the mathematics of QM is simply ad-hoc.

Rajesh D
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  • I don't understand exactly what do you mean when you say that the mathematics of QM is ad hoc. The mathematics of QM is robust, and it is such that it agrees with observation. Are you suggesting a different mathematical formulation that would leave its results/predictions unchanged (i.e. maintain concordance with experiments?) – QuantumDot Dec 31 '14 at 08:42
  • @QuantumDot : Yes,precisely. I want the normalization of the wave function to gel with other mathematical aspects in a more fundamental way. – Rajesh D Dec 31 '14 at 08:45
  • I should remind you that when you form superpositions of states, the sum should be appropriately normalized by an overall constant to give unit probabilities. There is nothing mathematically inconsistent with this. – QuantumDot Dec 31 '14 at 08:48
  • @QuantumDot : Oaky, in that sense, inconsistent may not be appropriate, but the word necessary is 'mathematical beauty' which is needed. – Rajesh D Dec 31 '14 at 08:51
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    http://en.wikipedia.org/wiki/Projective_Hilbert_space – Robin Ekman Dec 31 '14 at 08:56
  • The Born rule is not an interpretation. It simply tells you how quantum mechanical systems connect to classical variables trough measurements. Energy has nothing to do with probability. Units of energy are [J], units of probability are [1]. I would suggest that you get a modicum of an understanding of a physical theory before you make claims about its consistency. – CuriousOne Dec 31 '14 at 10:15
  • And, once more again, one of my favourite answers by Urs Schreiber shows the premise of the question to be false - quantum mechanics is not ad hoc, it is a natural theory to consider when looking closely at ordinary classical Hamiltonian mechanics. – ACuriousMind Dec 31 '14 at 14:56

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The mathematics of QM seems ad-hoc when one sees it in physics courses. In von Neumann's approach it becomes clear that a good mathematical candidate for the description of quantum phenomena are operator algebras, specifically C*-algebras and von Neumann algebras.

Their representation theory then leads to the usual Dirac/Schrödinger/Heisenberg picture of quantum mechanics (it comes from the fact that, thanks to von Neumann's theorem, any system with finitely many degrees of freedom has only one irreducible representation, namely Schrödinger's, up to unitary equivalence). States are, in this picture, positive and normalised linear maps from the C*-algebra of observables to the field $\mathbb C$. Using the GNS construction, every pure state gives an irreducible representation where such state is represented by a cyclic vector.

Now the projective Hilbert space of this representation is in one-to-one correspondence with all the other states in the same superselection sector, and this justifies rigorously the use of "unit energy" vectors. Thus one can do superposition of states through normalised linear combinations, which shows the existence of the phenomenon of "quantum interference", which is absent in classical mechanics because of the commutativity of the C*-algebra of observables.

ACuriousMind
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Phoenix87
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We find that the description of quantum systems with the concept of superposition yields predictions highly consistent with experiment. Linearity seems to be central and essential to this highly successful description, so, until someone works out another way of describing the phenomena we measure, the notion of vector space i.e. linear, Hilbert space is here to stay. A Banach space alone will not do at all.

"But for two elements of Hilbert space when summed to get a third element in the same space, the third need not be of unit energy."

This is absolutely true. But, and sometimes this is not emphasised enough, the quantum state space is a projective space. This means that the fundamental objects are rays within the Hilbert space, not points. A quantum state is actually the whole linear set $\{\alpha\,\psi_0:\,\alpha\in\mathbb{C}\}$ where $\psi_0$ is a unit vector in the Hilbert space. We can do all the linear operations (scaling, addition) on these rays. When we come to the Born interpretation and want to work out probabilities, then our procedure is to choose the normalised vector in the particular ray in question. The jargon for this kind of state space where we think of coset rays instead of points is a projective Hilbert space.

This is how the Born interpretation lives harmoniously with linear superposition notions.

Selene Routley
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Quantum mechanics starts with wave functions living in Hilbert space. But later for Born's interpretation, the wave function need to be of unit energy

No, Hilbert space is a Hilbert space knowing nothing about external constructions to itself like "probabilities as observable values". Normalization of state vectors is an example of such external construction. Vectors $|\psi\rangle$ and $\mathrm{const}|\psi\rangle$ are different vectors in Hilbert's space. But they are not different wrt equivalence relation (multiplication by a const) which, in turn, comes from observalbe probabilities. Word "probability" concerns only numeric representatives of vectors (coordinates). The space itself, operators and their spectra are not taking place this game.

maximav
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