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In arXiv:quant-ph/0504102v1, A.J. Bracken says

if we think of the phase space formulation of QM as more fundamental, arising directly from a deformation of classical mechanics in phase space [12] we can think of the formulation of QM in Hilbert space and the associated introduction of complex numbers as a computational device to make calculations easier.

Reference [12] is Bayen F, Flato M, Fronsdal C, Lichnerowicz A, and Sterheimer, D, Annals of Physics 111 (1978) 61-110, 111-151. This seems to answer the question of why complex variables. Can anyone explain this? Thanks.

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2 Answers 2

This paper is talking about the Wigner function, which is the Fourier transform of the density matrix $\rho(x,x')$ to be a function of $\rho(x,p)$. The density matrix formulation is arguably more fundamental, but the Wigner function is just a bad set of coordinates for the more fundamental density matrix, see my answer to this question: an example of a quantum system for which wigner function transitions to negative values

The density matrix is real valued on diagonal, and only the complex parts of the matrix are off diagonal. But the mystery (if there is one) of complex numbers in quantum mechanics arises in physical circumstances, when pure states diffract. They diffract in the density matrix formulation too, because the pure states are just special density matrices. The arguments in the paper are overly formal and free of content, and this paper need not be read.

To explain further--- the formulas "28" and "34" in the paper express the complex off diagonal density matrix elements with some other notation, as the author states somewhere at the end. This paper just gives a few new symbols for old things.

Why QM is Complex

If you want to get rid of complex numbers in QM, take the real and imaginary parts of the wavefunction. That sounds glib, and it is: the complex nature of amplitudes means that you have states pared up in twos, $|x_r\rangle$ and $|x_i\rangle$, and these two states are indistinguishable, and if $x_r$ turns into $y_i$ with time, then $x_i$ turns into $-y_r$.

The statement that there is a complex structure to quantum mechanics is simply the statement that there is such a doubling of states. To see why this needs to be so, consider the Eigenstates of H. These have to preserve length so they must rotate into each other, so they must come in these pairs (if you insist on using real numbers, unitary time evolution has only complex eigenvalues, so you can't diagonalize it, but you can block-diagoalize it into 2 by 2 blocks, which have this doubling). Now there is a manifest symmetry of multiplying by "i" (where you introduce an operator "i" which takes the top component of each two-by-two block to the bottom component, and the bottom component to minus the top).

This works for all states except possibly the vacuum (because the vacuum has zero energy, it doesn't have to move at all under time evolution, so it doesn't need a partner). The statement is that there is always an "i" symmetry which squares to -1. This "i" commutes with almost everything else, but doesn't commute with time reversal, which switches the sign of i. This shows you that it's not really complex numbers in QM, but the "i" symmetry above, because time reversal doesn't respect i.

I believe this argument appears somewhere in the literature.

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The concise and intuitive answer (if you know a little bit of linear algebra) is the following: although $\mathcal H$ is a complex Hilbert space, the set of states and measurements are the Hermitian operators on $\mathcal H$ and these form a real vector space. Most natural representations of these operators (from a linear algebra point of view) contain complex numbers (standard matrix representations, for example). As phase space representations (and other quasi-probability representations) demonstrate, this is not necessary -- the reason being already stated: the objects we are ultimately interested in for calculations are real vector spaces, not complex ones.

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This is repeating the wrong point made in the paper. We are not interested in representations of any algebra of operators--- the vectors themselves are physical--- you can't substitute some other representation of the same algebra, when it exists. Each representation is different. The "real vector spaces" are red herrings, the quantities involved are still complex (in real terms, the operator "i" still is in there). These arguments are sophisticated sounding but profoundly ignorant mathematical gibberish. But this mistake is the paper's: you are repeating the author's nonsense accurately. –  Ron Maimon Oct 27 '11 at 16:15
    
@Ron Maimon: If you assume vectors in $\mathcal H$ are physical, then you have far bigger worries than complex numbers... –  Chris Ferrie Oct 27 '11 at 17:56
    
I should add that the first instance I have found of this argument is H. Jeffreys, Philosophical Magazine, Vol. 33 (1942), pp. 815-831. –  Chris Ferrie Oct 27 '11 at 17:57
    
they are physical in the sense that every quantity they contain can be measured in an identically prepared ensemble, and they determine the future. The issue of whether they are physical in the interpretational sense is irrelevant. Going to a density matrix does not stop there from being a pairing of states corresponding to multiplication by i, so it doesn't stop quantum mechanics from being complex, regardless of the mathematical jibberish about "real representations". A spin 1/2 and spin 1 particles are two representations of the same abstract algebra, but have different physics. –  Ron Maimon Oct 28 '11 at 4:57

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