# How to perform a basis change on a 2x2 density operator?

I have an ensemble described by following density operator: $$P=3/8 |+\rangle\langle+| + 5/8 |-\rangle\langle-|$$

I am trying to write this operator in $$\{|0\rangle, |1\rangle\}$$ basis. I know that to change basis, we should multiply it's matrix representation to U matrix: $$U=\frac{1}{\sqrt{2}}\begin{pmatrix} 1 & 1\\1&-1 \end{pmatrix}$$

The problem is the final matrix I achieve does not make sense! I am new to quantum computing, can any one help me and point me in the right direction? I am guessing the main problem is finding matrix representation of this operator!

• Can you detail what you've tried and what you got? It'd be easier to help you to understand what's wrong Oct 29, 2022 at 15:29

I'm not exactly sure what you mean, but I'll try my best. $$\newcommand{\ket}[1]{|#1\rangle}\newcommand{\bra}[1]{\langle#1|}$$The way I would do this is by first noting that $$\ket{+} = \frac{1}{\sqrt{2}}(\ket{0} + \ket{1})$$ $$\ket{-} = \frac{1}{\sqrt{2}}(\ket{0} - \ket{1})$$ Basically, we can just substitute these identities into the expression: $$\begin{split} \frac{3}{8}\ket{+}\bra{+} + \frac{5}{8}\ket{-}\bra{-} &= \frac{3}{8}\cdot\frac{1}{2} (\ket{0} + \ket{1})(\bra{1} + \bra{0}) + \frac{5}{8}\cdot\frac{1}{2}(\ket{0} - \ket{1})(-\bra{1} + \bra{0})\\ &= \frac{3}{16}(\ket{0}\bra{0} + \ket{0}\bra{1} + \ket{1}\bra{0} + \ket{1}\bra{1}) \\ & \quad\qquad + \frac{5}{16}(\ket{0}\bra{0} - \ket{1}\bra{0} - \ket{0}\bra{1} + \ket{1}\bra{1}) \\ &= \frac{1}{2}\ket{0}\bra{0} - \frac{1}{8}\ket{0}\bra{1} - \frac{1}{8}\ket{1}\bra{0} + \frac{1}{2}\ket{1}\bra{1} \end{split}$$ which is $$P$$ in the $$\{\ket{0}, \ket{1}\}$$ basis.

To maybe clarify why what you were doing was wrong: first of all, in case you don't know, if we wanted to apply some gate $$O$$ to a density matrix like $$P$$, the new expression would be $$UPU$$ This is because density matrices can be seen as a weighted sum of pure statevectors. To apply a gate to a statevector $$\ket{\psi}$$, we just do: $$\ket{\psi} \mapsto U\ket{\psi}$$ Then, the density matrix for $$U\ket{\psi}$$ would be $$U\ket{\psi}(U\ket{\psi})^\dagger = U\ket{\psi}\bra{\psi}U^\dagger$$ Then, since density matrices are just a sum of statevectors, we can apply $$U$$ like this to each statevector outer product in the sum, and then factor each $$U$$ on the left and on the right to get $$UPU$$.

So, applying that gate you were using (which is the Hadamard gate, and is commonly denoted $$H$$), we get $$\begin{split} HPH &= \frac{3}{8}H\ket{+}\bra{+}H + \frac{5}{8}H\ket{-}\bra{-}H \\ &= \frac{3}{8}\ket{0}\bra{0} + \frac{5}{8}\ket{1}\bra{1} \end{split}$$ which is a different answer than earlier. We can still do what we did earlier where we substituted in expressions for $$\ket{+}$$ and $$\ket{-}$$, but now for $$\ket{0}$$ and $$\ket{1}$$. Doing that, we get that this is equal to $$\frac{1}{2}\ket{+}\bra{+} - \frac{1}{8}\ket{+}\bra{-} - \frac{1}{8}\ket{-}\bra{+} + \frac{1}{2}\ket{-}\bra{-}$$ which is parallel to earlier. To be honest, I'm pretty hazy on how change of basis works, but what I perceive to be the reason for this is that $$H$$ maps a vector in the $$\{\ket{+}, \ket{-}\}$$ basis to the coefficients it would have in the $$\{\ket{0}, \ket{1}\}$$ basis, but still stays in the $$\{\ket{+}, \ket{-}\}$$ basis. Normally you would do change of basis in matrix form, and I did it in bra-ket notation here, so I guess that's why it's weird?

Anyways, I hope that helps!

You can perform a basis change in one of two ways. Use the identity operator, which I prefer, or use a Unitary matrix, which can get messier if you for get that you are not performing a transformation/ rotation in hilbert space.

You have $$\rho =\frac{3}{8}|+\rangle\langle+|+\frac{5}{8}|-\rangle\langle-|$$ so if you want to perform a change of basis, from hadmard to computational, you can use the matrix you provided $$U=\begin{bmatrix} \frac{1}{\sqrt{2}}&\frac{1}{\sqrt{2}}\\ \frac{1}{\sqrt{2}} &-\frac{1}{\sqrt{2}} \end{bmatrix}$$

Now here we need to remember, that, as opposed to a unitary rotation, the index for the columns and rows represent a different basis, not the same.

So acting on $$\rho$$ with $$U$$, we get $$U\rho U =\frac{3}{8}U|+\rangle\langle+|U+\frac{5}{8}U|-\rangle\langle-|U$$ where I have omitted the conjugate transpose as it's a hermitian matrix.

From here, we simply perform the matrix multiplication of the vectors, take the outer product, and finally the sum, giving you

$$U\rho U=\frac{3}{16}\begin{bmatrix} 1&1\\ 1 &1 \end{bmatrix}+\frac{5}{16}\begin{bmatrix} 1&-1\\ -1 &1 \end{bmatrix}=\begin{bmatrix} \frac{1}{2}&-\frac{1}{8}\\ -\frac{1}{8} &\frac{1}{2} \end{bmatrix}$$