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I have a qubit. The probabality $p$ to find it in either of the two possible states $\{|0\rangle,|+\rangle\}$ is $p = 1/2$. I now want to measure it to maximize the probability that the measurement is correct.

From this presentation (page 5), I know that I can determine the probability as $$\frac{1}{2}+\frac{1}{4}\left\||0\rangle-|+\rangle\right\|_{\rm tr}.$$

But how do I find the associated projection?

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  • $\begingroup$ The initial conditions are ambiguous. Are saying it is either an incoherent mixture of 50% $|a\rangle$ where $a$ is $0$ or $+$, and the remaining 50% of the density matrix is an arbitrary mixture of any number of pure states that are not eigenstate in the given basis? $\endgroup$
    – JEB
    Sep 17, 2021 at 3:45
  • $\begingroup$ There are only two states $|0\rangle$ and $|+\rangle$. The probability to find one of them is 50%. Is that clearer? $\endgroup$
    – Johny Dow
    Sep 17, 2021 at 3:48
  • $\begingroup$ No, any density matrix of the form $\rho = \begin{pmatrix} 1/2 & b^* \\ b & 1/2\end{pmatrix}$ with $b$ a complex number such that $|b|\leq 1$ satisfies your condition, so this is still ambiguous. $\endgroup$
    – SolubleFish
    Sep 17, 2021 at 8:02
  • $\begingroup$ @JohnyDow It's not clear, because if you $p=\frac 1 2$ for one state, then it's 1/2 for the other in a two state system, so one would not say "the probability of finding one if 1/2", even though it's true, one would say "the probability of finding either state is 1/2". $\endgroup$
    – JEB
    Sep 17, 2021 at 12:51
  • $\begingroup$ @JEB I see. I tried to edit the question. Is it clear now? $\endgroup$
    – Johny Dow
    Sep 17, 2021 at 13:09

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I made a silly mistake in the comments. The measurement you are looking for consists of the two projectors unto the positive and negative subspaces of $|0\rangle\langle 0 | - |+\rangle \langle + |$. In particular, the matrix $|0\rangle\langle 0 | - |+\rangle \langle + |$ can be diagonalised as $\sum_i \lambda_i |\psi_i\rangle \langle \psi_i |$, where the $|\psi_i\rangle $ form an orthonormal basis and the $\lambda_i$ need not be positive. Let's assume that all the $\lambda_i$ are non-zero, you can check later that this will not really matter.

Now write $$|0\rangle\langle 0 | - |+\rangle \langle + | = \sum_i \lambda_i |\psi_i\rangle \langle \psi_i |\\ = \sum_{\lambda_i>0} \lambda_i |\psi_i\rangle \langle \psi_i | + \sum_{\lambda_i<0} \lambda_i |\psi_i\rangle \langle \psi_i |$$

where I have split up the sum into those where the $\lambda_i$ are positive and negative, respectively.

The projectors unto the so-called positive and negative subspaces are then $\Pi_+ = \sum_{\lambda_i>0} |\psi_i\rangle \langle \psi_i|$ and $\Pi_- = \sum_{\lambda_i<0} |\psi_i\rangle \langle \psi_i|$, and give you the projectors you seek for your measurement.

The above can be generalised to the case where the two states $\rho_1$ and $\rho_2$ are given with unequal probabilities $p_1 \neq p_2$. Then one can repeat the above game, but calculating the positive and negative subspaces of $p_1\rho_1 -p_2\rho_2$ instead.

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  • $\begingroup$ Thank you for this! So specifically in this example I get the following matrix: $|0\rangle\langle 0|-|+\rangle\langle+|=\left[\begin{array}{cc}\frac{1}{2} & -\frac{1}{2} \\ -\frac{1}{2} & -\frac{1}{2}\end{array}\right]$. Which has the eigenvectors: $\left[\begin{array}{ll}0.38 & 0.92\end{array}\right]^{T}$ and $\left[\begin{array}{ll}-0.92 & 0.38\end{array}\right]^{T}$. Therefore $\Pi_+ = \left[\begin{array}{ll}0.38 & 0.92\end{array}\right] * \left[\begin{array}{ll}-0.92 & 0.38\end{array}\right]^{T}$ . Correct? $\endgroup$
    – Johny Dow
    Sep 17, 2021 at 15:24
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    $\begingroup$ Your expression for $\Pi_+$ is not of the form $|\psi\rangle \langle \psi|$. There is only a single term since the subspace is one-dimensional, since there is only one positive eigenvalue. Furthermore, you should also be careful with writing down only the rounded, numerical values. $\endgroup$
    – Kenneth Goodenough
    Sep 17, 2021 at 15:34
  • $\begingroup$ Term $|\psi_i\rangle \langle \psi_i|$ corresponds to the outer product, right? So what is the problem with the way I wrote $\Pi_+$? $\endgroup$
    – Johny Dow
    Sep 17, 2021 at 15:41
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    $\begingroup$ What you wrote down is of the form $A = |\psi_i\rangle \langle \psi_j|$ for two different $|\psi_i\rangle$ and $|\psi_j\rangle$. Note that unlike $\Pi_+ = |\psi_i\rangle \langle \psi_i|$ this cannot be a projector, since a projector $B$ satisfies $B^2=B$ by definition, while $A^2 \neq A$ and $\left(\Pi_+\right)^2 = \Pi_+$, as you can check. $\endgroup$
    – Kenneth Goodenough
    Sep 17, 2021 at 15:45
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    $\begingroup$ If $A$ is any matrix of the form $\sum_i |\psi_i\rangle \langle \psi_i|$ with the $|\psi_i\rangle$ orthonormal needs to be a projector. Proof: $\left(\sum_i |\psi_i\rangle \langle \psi_i|\right)\left(\sum_j |\psi_j\rangle \langle \psi_j|\right) = \sum_{i, j} |\psi_i\rangle \underbrace{\langle \psi_i| |\psi_j\rangle}_{=\delta_{ij}} \langle \psi_j| = \sum_i |\psi_i\rangle \langle \psi_i|$. We have the simple case where we only have a single summand, so I suspect you've made a calculational mistake somewhere. $\endgroup$
    – Kenneth Goodenough
    Sep 17, 2021 at 15:56

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