4.2.4 Ein Blick auf Quantennetzwerke

A good way to verify that you have a given state $\left|\psi\right\rangle$, is to find a quantum operation that brings $\left|\psi\right\rangle$ back to $\left|0\right\rangle$ and then measure. If you then get $100\%$ for outcome 0, you know that you indeed had $\left|\psi\right\rangle$ (this is because quantum operations are invertible, so whenever you know the output state of an operation there is only one possible input state). This is the same general strategy that we followed in the previous homework problem, and it generalizes to multiple qubits, too! Suppose that you have a quantum operation $U$ that sends some multi-qubit state $\left|\psi\right\rangle$ back to $\left|0\cdots 0\right\rangle$. If you then get outcome $0$ with certainty for each qubit, then the state you had before applying $U$ must have been state $\left|\psi\right\rangle$.

So in our case, we want to verify that the teleported state in the donkey’s and Bob’s qubit is:

To do so, we need to operate the state back to $\left|00\right\rangle$. The state $\left|\Phi^{+}\right\rangle$ was created by

Therefore we need to first invert $\mathrm{CNOT}_{1\to 2}$ and then $(H\otimes I)$. But both of these operations invert themselves! Let us check this. If we first apply a $\mathrm{CNOT}$ gate where Bob’s qubit is the control qubit and the donkey’s qubit is the target qubit, we obtain the following state:

So, if Bob can now apply a Hadamard gate to his qubit, then Bob’s and the donkey’s qubits end up in the state $\left|00\right\rangle$. Measuring both qubits (jointly) will now result in a measurement outcome of $0$ with $100\%$, which we can check with Quirky. This is done by the circuit below, where we added the final $\mathrm{CNOT}$, $H$ and measurements that we talked about above:

Indeed, we see that both measurement outcomes are $0$ with probability 100%, indicating that the state $\left|\Phi^{+}\right\rangle$ was indeed prepared successfully.

Alternatively, you could also look at the probabilities of joint measurement outcomes, as in the following circuit (we swapped the second and the fourth qubit so that the two qubits of interst are next to each other):

We see that the outcome $[00]$ arises with 100% probability.

Note that, a priori, these are two different test, because in the first one we are looking at the probabilities of the individual measurement outcomes, whereas in the second test we are looking at the probability distribution of both outcomes at the same time. However, the two tests are completely equivalent (meaning that you can use either of them to determine if the state is $\left|00\right\rangle$. This is because if $p$ is a probability distribution of two bits such that measuring the first bit of $p$ always gives 0 and measuring the second bit of $p$ also always gives 0, then the probability distribution must be $p=[00]$ (this is very intuitive – as soon as there is another term there would be some probability of outcome 1 in either the first or the second bit).

Is there an easier way to solve this problem? Some of you may have tried to put the measurements and probability boxes directly after the teleportation circuit, without applying any additional quantum operations. For example, you may have tried simply measuring qubits 1 and 4 individually. However, this provides very little information (in particular, this does not even tell you whether there are correlations between the outcomes or not). This is illustrated in the following figure:

Here, qubits 1 and 2 are in the state $\left|+\right\rangle\otimes\left|+\right\rangle=\frac{1}{2}(\left|00\right\rangle+\left|01\right\rangle+\left|10\right\rangle+\left|11\right\rangle)$, while qubits 3 and 4 are in the state $\left|\Phi^{+}\right\rangle=\frac{1}{\sqrt{2}}(\left|00\right\rangle+\left|11\right\rangle)$. These are very different states – one is maximally entangled, while the other is a product state! But the figure shows that if we only look at the probabilities of single-qubit measurement outcomes then in each case we get two $50\%$ boxes. So simply measuring the individual qubits is not a good strategy.

We can do somewhat better by looking at the joint probabilities of measurement outcomes:

In this case, we can see a difference between $\left|+\right\rangle\otimes\left|+\right\rangle$ and $\left|\Phi^{+}\right\rangle$. Indeed, for the first state, all four outcomes $[00]$, $[01]$, $[10]$, and $[11]$ are equally likely, while for the second state we only get $[00]$ or $[11]$ with equal probability. However, this does still not confirm that the state was $\left|\Phi^{+}\right\rangle$! Indeed, the state could just as well have been $\left|\Phi^{-}\right\rangle=\frac{1}{\sqrt{2}}(\left|00\right\rangle-\left|11\right\rangle)$, which gives the same probabilities:

Indeed, all you know after the looking at the joint probabilities is that the state is either $\left|\Phi+\right\rangle$ or $\left|\Phi^{-}\right\rangle$. In order to distinguish between these two cases we first have to apply some unitary, and this is exactly what is achieved by the solution that we discussed above. Note that this is similar to the discussion in Abschnitt 2.5.1 and in particular 2.6.