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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 $a_{i,j}$?

Sanchayan Dutta
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1 Answers1

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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:

glS
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