We know that a quantum state can be represented by a matrix $\rho$, where $\rho$ is positive semi-definite and trace is $1$.

So, what is the definition of $\sqrt{\rho}$ and how can I calculate it?


1 Answer 1


According to the spectral theorem, every $d$-dimensional density operator $\rho$ has a unique expression in terms of its eigenvectors and eigenvalues like the following: \begin{equation} \rho = \sum_{i=1}^d \lambda_i |\lambda_i\rangle \langle \lambda_i | \tag{1} \end{equation}

where $\lambda_1, \dots, \lambda_d$ are eigenvalues of $\rho$, where for simplicity we just set $\lambda_k = 0$ for $k > \text{rank}(H)$, and $\{|\lambda_1\rangle, \dots, |\lambda_d\rangle\}$ are orthonormal eigenvectors of $\rho$. Then, the definition of $\sqrt{\rho}$ is given straightforwardly as \begin{equation} \rho^{1/2} = \sum_{i=1}^d \lambda_i^{1/2} |\lambda_i\rangle \langle \lambda_i | \tag{2} \end{equation}

where $\sqrt{\lambda_i}$ is real since a density matrix must be positive semidefinite (nonnegative eigenvalues) by definition. Note that this is just another way of writing a diagonalization of $\rho$, for example: \begin{align} \rho &= U \Lambda U^\dagger \tag{3a} \\ \rho^{1/2} &= U \Lambda^{1/2} U^\dagger \tag{3b} \end{align}

where the columns of $U$ are the eigenvectors $|\lambda_i\rangle$ and $\Lambda$ contains the eigenvalues of $\rho$ on its diagonal. This means that one way to calculate $\rho^{1/2}$ is by finding the eigenvectors and eigenvalues $\rho$ and then reconstructing $\rho^{1/2}$ using either Eq. 2 or Eq. 3b.

  • $\begingroup$ Thank you for your detailed answer! $\endgroup$
    – Lancdorr
    Feb 25 at 4:44

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