In these lecture notes, Scott Aaronson gives three arguments for why complex Hilbert spaces are more natural than real ones for formulating quantum mechanics. I don't understand his second argument, which begins at "how many independent real parameters ..." and ends at "... juuuuust right!" It goes as follows:

  1. (Mixed) states are described by positive-semidefinite Hermitian operators, which have $N^2$ real degrees of freedom if the underlying field is the complex numbers and $N(N+1)/2$ real degrees of freedom if the underlying field is the real numbers, where $N$ is the dimension of the Hilbert space over its field. (He also considers the case where the "field" is the division ring of quaternions so that the Hilbert space is actually a module instead of a vector space.)


Intuitively, it seems like the number of parameters needed to describe [a composite system] should equal the product of the number of parameters needed to describe [the individual constituent systems] ... If amplitudes are complex numbers, then happily this is true!

His point is that in order for the real degrees of freedom to compound in the "intuitive" way - multiplicatively - the number of real degrees of freedom for a generic density matrix must be a power of the Hilbert space dimension, which is the case for the complex numbers, but not the reals or the quaternions.

My objection is to his first point. In a given basis, a mixed state isn't characterized by a generic positive-semidefinite Hermitian matrix, but instead by a trace-1 positive-semidefinite Hermitian matrix. The trace-1 requirement imposes an additional constraint on the matrix and removes one real degree of freedom, so that the number of real degrees of freedom is actually $N^2 - 1$, not $N^2$, and his argument falls apart. The number of physical degrees of freedom actually compound supermultiplicatively: if the constituent systems have $M$- and $N$-dimensional Hilbert spaces, then the composite system has $(MN)^2 - 1$ real degrees of freedom, which is strictly greater than the $(M^2 - 1)(N^2 - 1)$ real degrees of freedom that we'd expect if they compounded multiplicatively.

(Sure, you can naturally associate a mixed state to an arbitrary positive-semidefinite Hermitian matrix by renormalizing it to have trace 1, but then this map becomes many-to-one, and the trace is a completely unphysical extra degree of freedom. Unlike in the state-vector formalism for pure states, in the density-matrix formalism there is a natural canonical representation for any state, with no unphysical degrees of freedom like the state vector's global phase.)

Aaronson addresses this point only by saying "we assume, for convenience, that the state doesn't have to be normalized (i.e., that the probabilities can add up to less than 1)." This seems absurd to me - the entire argument hinges on a careful count of a state's actual number of physical degrees of freedom, so you can't just pad it out "for convenience" to the number that you want. You could use the exact same reasoning to show that the real numbers form the "natural" field by introducing $N(N-1)/2$ dummy degrees of freedom "for convenience" that pad the total number out to $N^2$.

Is there something I'm missing that salvages this argument?


1 Answer 1


Scott Aaronson has fleshed out his argument in a recent blog post. As far as I can tell, his initial "physical degrees of freedom should multiply across composite systems" argument was incorrect. The correct argument is as follows: the fact that real degrees of freedom of non-normalized density matrices (not mixed states) multiply across tensor products if the density matrices act on a complex (but not real or quaternionic) Hilbert space leads to the “local tomography” property, which states that an arbitrary state of a composite system can be fully determined from the joint statistics of product measurements only. How appealing you find this property is to some extent a matter of personal taste.


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