Consider a maximally mixed state composed of $n$ equal qubits. Its density matrix is then:$$\hat{\rho}^{(n)}=\hat{\rho}^{\otimes n}.$$ The eignevalues of such a matrix, $\lambda$ will all be equal (since it is maximally mixed) and $\lambda=\sigma^{n}$, where $\sigma$ is the eigenvalues of each quibts' density matrix. By considering now that in the system there are infinite qubits, then the eigenvalues' relation becomes: $$lim_{n\to \infty}\sigma^{n}=0,$$ since $0<\lambda<1$. The density matrix by definition must have at least one non-zero eigenvalue (corresponding to the pure state), which means that in the infinite case this expression is no longer a density matrix. Is this an inconsitency of the density matrix formalism? Furthermore, since in QFT theories have infinite degrees of freedom is not possible the generalization of density matrices in such theories?
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I think there are ways, but how do you define the tensor product of infinitely many Hilbert spaces? Is the operator $\rho^{(n)}$ in this limit well-defined, and positive and of trace-class? ... – Tobias Fünke Mar 06 '24 at 15:25
1 Answers
I think the issue here is the assumption here is the assumption that there is always a well defined maximally mixed state. If we take the quantum states in the computational basis as a binary representation of a number, then what you are after here is similar to the quantum equivalent to a uniform probability distribution on the natural numbers. However, no such distribution exists, for essentially the same reason you encountered here, all the probabilities vanish. (In fact what you have is worse as it involves things like an infinite sequence of alternating zeros and ones, which does not correspond to a finite number). This does not mean there is some fundamental flaw in the theory of density matrices. It just means that the particular state you attempted to write down does not exist. Other probability distributions on the naturals are possible, such as an exponential distribution, and similarly there do exist well defined density matrices on the space of infinite qubits.
When it comes to QFT, I would point out that the lack of a uniform distribution on the naturals does not stop us developing probability theory in with continuous variables and indeed we can use density matrices in QFT, give or take some technical subtitles.
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