4.2.5 The uncertainty principle

For the last phenomenon that we want to discuss, we will only need a single quantum bit. Recall the rules for measuring a single quantum bit from Eq. 2.6, as in the following picture:

Given a quantum state $\left|\psi\right\rangle=\psi_{0}\left|0\right\rangle+\psi_{1}\left|1\right\rangle$ , we obtain the two possible outcomes with probabilities

p_{0}=\psi_{0}^{2},\qquad p_{1}=\psi_{1}^{2}.

(4.19)

What are the deterministic states for which we get one of the outcomes with certainty? These are states for which one probability is 100% and the other is 0%. In other words, states where one amplitude is $\pm 1$ and the other is zero. Up to possibly an overall minus sign, which we know from 2.7 is irrelevant, there are only two such states:

\left|0\right\rangle=\begin{pmatrix}1\\ 0\end{pmatrix},\qquad\left|1\right\rangle=\begin{pmatrix}0\\ 1\end{pmatrix},

(4.20)

i.e., the basis states. These are the only states for which we can predict the measurement outcome perfectly well, so we will say that there is no uncertainty in the measurement outcome.

What happens if we first perform an operation on the qubit and then perform a measurement? For example, suppose that we first apply a Hadamard operation and then a measurement, as in the following picture:

Here, the only states for which we are completely certain about the measurement outcome are

\left|+\right\rangle=\frac{1}{\sqrt{2}}\begin{pmatrix}1\\ 1\end{pmatrix},\qquad\left|-\right\rangle=\frac{1}{\sqrt{2}}\begin{pmatrix}1\\ -1\end{pmatrix}

(4.21)

(up to overall sign). This is because the Hadamard operation maps $\left|+\right\rangle$ , $\left|-\right\rangle$ back to the basis states $\left|0\right\rangle$ , $\left|1\right\rangle$ (you showed this in 4.5); and the latter states are precisely the states with complete certainty about the final measurement, as we discussed above. We can also see this by computing the probabilities of the two measurement outcomes

q_{0}=\frac{(\psi_{0}+\psi_{1})^{2}}{2},\qquad q_{1}=\frac{(\psi_{0}-\psi_{1})% ^{2}}{2}.

(4.22)

If $q_{1}=0$ then the state is $\left|+\right\rangle$ , while if $q_{0}=0$ then the state is $\left|-\right\rangle$ (up to overall sign).

These are all calculations that we have seen several times – but there is an interesting observation that we have not made before. Since no state appears in both Eqs. 4.20 and 4.21, this means that, for every state, at least one of the two procedures will have some uncertainty in the measurement outcome. This result is nothing but the famous Heisenberg uncertainty principle!

Can we make this observation more quantitative? We first need a way to quantify the uncertainty or ‘randomness’ given by a probability distribution $p=\bigl{(}\begin{smallmatrix}p_{0}\\ p_{1}\end{smallmatrix}\bigr{)}$ . The following function is a good choice:

\displaystyle\operatorname{uncertainty}(p)=p_{0}(1-p_{0})=p_{0}p_{1}.

It maps values to $[0,1/4]$ , is minimal if one of the outcomes is zero (i.e., if $p_{0}=0$ or $p_{0}=1$ ) and it is maximal if both outcomes are equally likely (i.e., if $p_{0}=p_{1}=1/2$ ). Here is a plot which confirms these properties:

Now suppose that we start with a state $\left|\psi\right\rangle$ and perform either of the two procedures described above. That is, we either measure the state directly or we first apply a Hadamard operation and then measure. The corresponding uncertainties are $\operatorname{uncertainty}(p)$ and $\operatorname{uncertainty}(q)$ , where the distributions $p=\bigl{(}\begin{smallmatrix}p_{0}\\ p_{1}\end{smallmatrix}\bigr{)}$ and $q=\bigl{(}\begin{smallmatrix}q_{0}\\ q_{1}\end{smallmatrix}\bigr{)}$ are given in Eqs. 4.19 and 4.22 above. Then:

\displaystyle\operatorname{uncertainty}(p)+\operatorname{uncertainty}(q)>0.

Indeed, this inequality means precisely that there exists no state for which both uncertainties are simultaneously zero. In the following homework problem, you will show a stronger result:

Homework 4.4 (Uncertainty tradeoff).

Show that, for every qubit state $\left|\psi\right\rangle$ :

\operatorname{uncertainty}(p)+\operatorname{uncertainty}(q)=\frac{1}{4}.

(4.23)

Moreover, find a qubit state $\left|\psi\right\rangle$ with $\operatorname{uncertainty}(p)=\operatorname{uncertainty}(q)$ .

Bonus question: Construct this state using Quirky, and confirm that $\operatorname{uncertainty}(p)=\operatorname{uncertainty}(q)$ by using Quirky.

Hack.

Using Eqs. 4.19 and 4.22 and the fact that $\psi_{0}^{2}+\psi_{1}^{2}=1$ , we confirm the equality:

	$\displaystyle\operatorname{uncertainty}(p)+\operatorname{uncertainty}(q)$	$\displaystyle=\psi_{0}^{2}\psi_{1}^{2}+\frac{(\psi_{0}+\psi_{1})^{2}}{2}\frac{% (\psi_{0}-\psi_{1})^{2}}{2}$
		$\displaystyle=\psi_{0}^{2}\psi_{1}^{2}+\frac{(\psi_{0}^{2}+2\psi_{0}\psi_{1}+% \psi_{1}^{2})(\psi_{0}^{2}-2\psi_{0}\psi_{1}+\psi_{1}^{2})}{4}$
		$\displaystyle=\psi_{0}^{2}\psi_{1}^{2}+\frac{(1+2\psi_{0}\psi_{1})(1-2\psi_{0}% \psi_{1})}{4}$
		$\displaystyle=\psi_{0}^{2}\psi_{1}^{2}+\frac{1-4\psi^{2}_{0}\psi^{2}_{1}}{4}=% \frac{1}{4}.$

Alternatively, you could also parametrize the qubit state by $\left|\psi(\theta)\right\rangle$ and compute that

	$\displaystyle\operatorname{uncertainty}(p)$	$\displaystyle=\sin^{2}(\theta)\cos^{2}(\theta)=\frac{1}{4}\sin^{2}(2\theta),$
	$\displaystyle\operatorname{uncertainty}(q)$	$\displaystyle=\dots=\frac{1}{4}\cos^{2}(2\theta),$

which again shows that $\operatorname{uncertainty}(p)+\operatorname{uncertainty}(q)=\frac{1}{4}$ .

For the second part of the problem, an intutive guess is that $\left|\psi(\pi/8)\right\rangle$ should work, since it lies right in the middle of the two states $\left|0\right\rangle=\left|\psi(0)\right\rangle$ and $\left|+\right\rangle=\left|\psi(\pi/4)\right\rangle$ . Indeed:

\displaystyle\operatorname{uncertainty}(p)=\frac{1}{4}\sin^{2}(\pi/4)=\frac{1}% {4}\cdot\frac{1}{2}=\frac{1}{8},

so also $\operatorname{uncertainty}(q)=\frac{1}{4}-\operatorname{uncertainty}(p)=\frac{% 1}{8}$ .

We can test this as follows using Quirky:

The picture confirms that $p_{1}=q_{1}$ , hence also $\operatorname{uncertainty}(p)=\operatorname{uncertainty}(q)$ .

The formula that you just proved in 4.4 is quite remarkable: it shows that there is a simple tradeoff between the uncertainties of the two procedures. In particular, if one procedure has zero uncertainty then the other produces a uniformly random outcome! ¹³¹³ 13 We can also see this directly. E.g., if we directly measure $\left|+\right\rangle$ (a state that has zero uncertainty for the second procedure) without first doing a Hadamard, we get outcomes $0$ and $1$ with equal probability.