# The philosophy behind the mathematics of quantum mechanics

My field of study is computer science, and I recently had some readings on quantum physics and computation.

This is surely a basic question for the physics researcher, but the answer helps me a lot to get a better understanding of the formulas, rather than regarding them "as is."

Whenever I read an introductory text on quantum mechanics, it says that the states are demonstrated by vectors, and the operators are Hermitian matrices. It then describes the algebra of vector and matrix spaces, and proceeds.

I don't have any problem with the mathematics of quantum mechanics, but I don't understand the philosophy behind this math. To be more clear, I have the following questions (and the like) in my mind (all related to quantum mechanics):

• Why vector/Hilbert spaces?
• Why Hermitian matrices?
• Why tensor products?
• Why complex numbers?

(and a different question):

• When we talk of an n-dimensional space, what is "n" in the nature? For instance, when measuring the spin of an electron, n is 2. Why 2 and not 3? What does it mean?

Is the answer just "because the nature behaves this way," or there's a more profound explanation?

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These are the axioms of quantum mechanics, and axioms are by definition not deducible from "more profound explanation". Any interpretation of these things is just a change of words. For example, the vector space notion is roughly equivalent to superposition principle. –  Siyuan Ren Jul 30 '11 at 13:09
The "philosphy" behind this Math is called Physics, usually. You try to ride the horse backwards, which is understandable for someone coming from math. –  Georg Jul 30 '11 at 15:09
Your line of thinking is (unfortunately) not popular with the practising physicist (though with good reason!). The problem is one of (almost) philosophy, but can be well-posed and one can have coherent thoughts on it. You should be able to find some food for thought in the works of Christopher Fuchs (perimeterinstitute.ca/personal/cfuchs, especially the paper titled "Quantum Mechanics as Quantum Information (and only a little more)"). –  genneth Jul 30 '11 at 22:05
Obligatory answer to all "philosophy" questions: xkcd.com/54 –  Kostya Jul 31 '11 at 12:40

Vector spaces because we need superposition. Tensor product because this is how one combines smaller systems to obtain a bigger system when the systems are represented by vector space. Hermitation operator because this allows for the possibility of having discrete-valued observables. Hilbert space because we need scalar products to get probability amplitudes. Complex numbers because we need interference (look up double slit experiment).

The dimension of the vector space corresponds to the size of the phase space, so to speak. Spin of an electron can be either up or down and these are all the possibilities there are, therefore the dimension is 2. If you have $k$ electrons then each of them can be up or down and consequently the phase space is $2^k$-dimensional (this relates to the fact that the space of the total system is obtained as a tensor product of the subsystems). If one is instead dealing with particle with position that can be any $x \in \mathbb R^3$ then the vector space must be infinite-dimensional to encode all the independent possibilities.

Edit concerning Hermitation operators and eigenvalues.

This is actually where the term quantum comes from: classically all observables are commutative functions on the phase space, so there is no way to get purely discrete energy levels (i.e. with gaps in-between the neighboring values) that are required to produce e.g. atomic absorption/emission lines. To get this kind of behavior, some kind of generalization of observable is required and it turns out that representing the energy levels of a system with a spectrum of an operator is the right way to do it. This also falls in neatly with rest of the story, e.g. the Heisenberg's uncertainty principle more or less forces one to have non-commutative observables and for this again operator algebra is required. This procedure of replacing commutative algebra of classical continuous functions with the non-commutative algebra of quantum operators is called quantization. [Note that even on quantum level operators can still have continuous spectrum, which is e.g. required for an operator representing position. So the word "quantum" doesn't really imply that everything is discrete. It just refers the fact that the quantum theory is able to incorportate this possibility.]

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You can get interference without complex numbers. Sound waves interfere, but they're real-valued. In a bound system of spin-1/2 particles such as a nucleus, you can actually use real-valued wavefunctions; it works fine, and wave interference effects do occur. What you can't have without complex numbers is a traveling wave representing a spin-1/2 particle. –  Ben Crowell Jul 30 '11 at 17:48
@Ben: I don't agree. Waves live naturally in the complex domain: they carry both amplitude and phase. You can pretend that these are two real numbers but this is actually not the case: the phase is $2\pi$-periodic. I.e. this is nothing else than polar decomposition of a complex number. –  Marek Jul 30 '11 at 17:56
Thanks for the good answer. Just one point: "eigenvalues of operators represent physically possible values that can be measured." That's another fact that I don't have an intuition for. Eigenvalues is a mathematical concept, while observables and measurement are physical notions. How do they relate? –  Sadeq Dousti Jul 30 '11 at 19:13
Your first paragraph is a great summary of something I've been annoyed with ever since I opened my first QFT book. I think I have 5 of those books and not in a single one of them do they actually write why they need the math, they just STATE the equations needed and go along. But for us who are not mathematicians but for example programmers or engineers we want to know the why's :) Someone should write a QFT book that concentrates on the physical intuition, sort of like Feynmans "strange theory of light and matter" but more ambitious and math is allowed if it is explained :) –  BjornW Jul 30 '11 at 21:21

Scott Aaronson, himself a (quantum) computer scientist, thinks and writes about a number of these subjects in his paper Is Quantum Mechanics An Island In Theoryspace? - at least the "why complex numbers and not the reals or the quaternions?", and I'm pretty sure he mentions it in his 'Democritus' lectures as well.

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The lectures were fantastic, thanks for the link. –  recipriversexclusion Aug 1 '11 at 19:13
@recipriversexclusion: I have a CS background, and those lectures really made it click what quantum computing was about, in terms of computation (as opposed to the physics aspect). I hope it's the same for people coming from physics. Glad you liked them, in any case! –  yatima2975 Aug 1 '11 at 21:25

Is the answer just "because the nature behaves this way," or there's a more profound explanation

I would say , yes to "because nature behaves this way". It is the most economical description of experimental data using mathematics, to date.

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First of all, the philosophy of quantum mechanics is hardly straightforward to most physicists. This has spawned an entire industry of "quantum interpretations" to the "quantum measurement problem".

Quantum mechanics is probably one of the best solutions to the problem of making mysticism mathematically precise. Quantum mechanics puts mysticism on a firm mathematically footing. Quantum mechanics is about Cosmic Consciousness and Reality.

Really, you can read about it straight from the horse's mouth in the book Quantum Questions, which is a compilation of mystical writings by the founders of quantum mechanics themselves. People like Werner Heisenberg, Erwin Schroedinger, Albert Einstein, Louis de Broglie, Jeans, Max Planck, Wolfgang Pauli and Arthur Eddington. They're not mere "fringe" physicists, even though "fringe" physicists like Jack Sarfatti, David Bohm, Amit Goswami, John Hagelin and Frank Tipler often do have a point...

Another book you can read is Quantum Enigma.

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It is strange that you are not very comfortable with Hilbert spaces because actually continuous functions form Hilbert space and physics is all about functions. The general difference with classical mechanics is in the fact that classical mechanics was formulated long before people who used and learned (and even developed) it could understand manifolds and Lie algebras while for the time of quantum mechanics the idea of Hilbert space and all that stuff became more or less natural.

The same with the rest. You could formulate QM without complex numbers (in principle), but this would be the same as formulation of Maxwell's equations without vectors. People say that Maxwell's equations actually were formulated before physicist were familiar enough with the vectors and it was a nightmare.

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Hilbert spaces are not necessarily composed of continuous functions. For example, all Euclean spaces are Hilbert spaces, but the vectors inside are not continuous functions. –  Siyuan Ren Jul 30 '11 at 15:41
@Karsus Ren, If you read my answer carefully you'd mention that I did not say that Hilbert spaces are necessarily composed of continuous functions. Probably I should had used 'e.g.' instead of 'actually'. –  Misha Jul 30 '11 at 15:49
@Misha: "e.g." wouldn't be much better. Continuous functions are just bad example. Prime examples of Hilbert spaces come from measure theory ($L^2(X, \Omega, \mu)$ spaces) and measurable functions (built on the Borel sigma algebra) are in general much worse behaved than continuous ones; but one sacrifices this in order to obtain better topological properties of the space -- one can take limits and stay in the same space. This is in fact the key difference between Riemann and Lebesgue integral. –  Marek Jul 30 '11 at 16:30
@Marec, when one says "actually sheep are mammal" it is obvious that not every mammal is sheep. Surprisingly, it does not work when you speak of Hilbert spaces. I was trying to point out that Hilbert space is not something complex and/or artificial. It is something which is studied at school but at the different level of abstraction. –  Misha Jul 30 '11 at 16:41
@Misha: but you are not saying continuous functions are measurable functions. You are explicitly saying that they form a Hilbert space (implying there's nothing else in it). This is true about as much as saying that rational numbers form the real line $\mathbb R$, while in fact in both cases these are only dense subspaces. I agree with your sentiment that Hilbert space is nothing artificial. I just wanted you to be bit more precise. –  Marek Jul 30 '11 at 18:00

I asked the same questions some time back.