Prove that the density matrix of $|+\rangle$ with respect to basis $\{|+\rangle, |-\rangle\}$ is given by

$$\rho = \begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix}.$$


1 Answer 1


You may compute the density matrix for a given pure state $|\psi\rangle$, with \begin{align*} \rho = |\psi\rangle \langle \psi|. \end{align*} Thus, when you define $\{|+\rangle,|-\rangle\}$ to be your standard basis vectors, you have, \begin{align*} \rho = |+\rangle \langle +| = \begin{pmatrix}1 \\ 0 \end{pmatrix}\begin{pmatrix}1 & 0 \end{pmatrix} = \begin{pmatrix}1 & 0 \\0 & 0\end{pmatrix}. \end{align*}

Note that in general, when considering some matrix in a given basis, you are considering the definition of that matrix given that the standard basis vectors are the given basis. Thus, when considering the above density matrix, we are operating under the definition that, \begin{align*} |+\rangle \equiv \begin{pmatrix}1\\0\end{pmatrix}, |-\rangle \equiv \begin{pmatrix}0\\1\end{pmatrix}. \end{align*}

It may help to consider what would happen if we defined the standard basis to be $\{|0\rangle, |1\rangle\}$. In our computation we would have, \begin{align*} \rho = |+\rangle \langle +| = \frac{1}{2}\begin{pmatrix}1 \\ 1 \end{pmatrix}\begin{pmatrix}1 & 1 \end{pmatrix} = \frac{1}{2}\begin{pmatrix}1 & 1 \\1 & 1\end{pmatrix}. \end{align*} Recall that for an invertible matrix $A$, we may diagonalize it as $A = PDP^{-1}$, where $D$ is a diagonal matrix containing the eigenvalues of $A$, and $P$ consists of its eigenvectors. Diagonalizing the above matrix yields, \begin{align*} \rho = \frac{1}{2}\begin{pmatrix}1 & 1 \\1 & 1\end{pmatrix} &= \frac{1}{2}\begin{pmatrix}1 & -1 \\1 & 1\end{pmatrix}\begin{pmatrix}2 & 0 \\0 & 0\end{pmatrix}\begin{pmatrix}1 & 1 \\-1 & 1\end{pmatrix}\\ &= \begin{pmatrix}1 & -1 \\1 & 1\end{pmatrix}\begin{pmatrix}1 & 0 \\0 & 0\end{pmatrix}\begin{pmatrix}1 & 1 \\-1 & 1\end{pmatrix}. \end{align*} And so you can see that the diagonal matrix recovered is indeed the same as the matrix we originally computed. Thus, it may be clear that in $A = PDP^{-1}$, $P^{-1}$ brings us into the basis where its columns are the standard basis vectors, and $P$ brings us out of it.

Please let me know if this doesn't answer your question.

  • $\begingroup$ Can you explain the reason why you are considering $|+\rangle$ as [1 0]^transpose instead of standard [1 -1]^transpose ? $\endgroup$ Apr 19, 2020 at 15:48
  • $\begingroup$ I have amended my answer, does it help? $\endgroup$
    – Arthur-1
    Apr 19, 2020 at 16:43
  • $\begingroup$ I think that there is a missing $\frac{1}{2}$ before matrix $\rho$ in fourth and fifth expression. In fifth expression should be $\frac{1}{4}$ in second step (one $\frac{1}{2}$ from $\rho$ and second from inverse of $P$). Moreover, in sentence ...where $D$ is a diagonal matrix containing the eigenvalues of $D$..., the second matrix should be $A$. Right? $\endgroup$ Apr 19, 2020 at 21:30
  • $\begingroup$ Agreed! Thanks for the corrections. $\endgroup$
    – Arthur-1
    Apr 20, 2020 at 2:49

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