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The maximally mixed state on m qubits is defined to be the quantum state with associated density operator $\rho_m = \frac{1}{2^m} I$. Examples are

  • On one qubit this is $\rho_1 = \frac{1}{2}(|0\rangle\langle0|+|1\rangle\langle1|) = \frac{1}{2}I$
  • On two qubits we have $\rho_2 = \frac{1}{4} (|00\rangle\langle00| +|01\rangle\langle01|+|10\rangle\langle10|+|11\rangle\langle11|$.

My question is the following: how can the corresponding state vector $|\phi_m\rangle$ be expressed in terms of the standard basis elements, eg. $|\phi_2\rangle = \sum_{i,j} a_{i,j}|ij\rangle$? What are the values of $a_{i,j}$?

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If by "corresponding state vector" you mean a pure state $\lvert\psi\rangle$ such that $\lvert\psi\rangle\!\langle\psi\rvert$ is maximally mixed, then the answer is that there isn't one.

A density matrix $\rho$ can be written as $\rho=\lvert\psi\rangle\!\langle\psi\rvert$ for some ket state $\lvert\psi\rangle$ if and only if it is pure. One easy way to check for this is to compute $\mathrm{Tr}(\rho^2)$, which is the so-called purity of the state, and equals $1$ if and only if the state is pure.

A maximally mixed state is as far away as possible from this situation, in the sense that it is the state (or one of the states) which corresponds to the minimal possible value of the purity.

You might also want to have a look at these related questions:

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  • $\begingroup$ Well what I meant by corresponding is this: a one qubit system can be described either via it's density matrix or just as an element of the Hilbert space in question ie as $|\phi\rangle = \sum_i |\phi_i\rangle$ for basis vectors $|\phi_i\rangle$. Hmm, on a second look I think I might be confusing mixed states with superposition... So I am guessing when I say $|\phi\rangle$ can be written as linear combination of basis vectors, that's only true for pure states, right? That would explain my confusion. $\endgroup$
    – gen
    Jun 9, 2019 at 14:53
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    $\begingroup$ @gen yes, when you say that a state can be described either via its density matrix $\rho$ or its ket state $\lvert\psi\rangle$, the correspondence between the two is that $\rho=\lvert\psi\rangle\!\langle\psi\rvert$. Not all density matrices correspond to a ket (pure) state. They only do when the state is pure. Otherwise, the density matrices are mixtures of DMs corresponding to pure states, that is, objects of the form $\rho=\sum_k p_k\lvert\psi_k\rangle\!\langle\psi_k\rvert$ for some ensemble of pure states $\{\lvert\psi_k\rangle\}$ $\endgroup$
    – glS ♦
    Jun 9, 2019 at 14:55
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    $\begingroup$ if the question is what mixture gives a maximally mixed state, then the answer is that you can get via any mixture with equal probabilities of an orthonormal set for the space $\endgroup$
    – glS ♦
    Jun 9, 2019 at 14:58

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