I am doing a term project on the BB84 Protocol and it makes use of the Heisenberg Uncertainty Principle. I think I understand the principle in theory. If we have two non-commuting observables, then we cannot simultaneously measure a state with these observables, since after the measurement the state collapses to one of the eigenstates of the observable, but non-commuting observables do not have common eigenstates. This is what I understood.

I also found that the Paulis spin matrices for x and z, which are used in the BB84 protocol do not commute. So they should (and no dot) have common eigenstates. I tried to find the Heisenberg inequality using the formula for the Heisenberg inequality where we use average values and each time for the 0 ket I find the inequality side 0. Shouldn't it give something other than 0?

Then I tried to justify the 0, thinking that maybe for one of the spin matrices the uncertainty is 0 since we are using the right operator, and that is why when we multiply the uncertainties we get 0?

Am I doing something wrong in my calculations, or have I completely misunderstood the subject.

Thanks for any kind of help.


1 Answer 1


First of all, Heisenberg principle is usually invoked to say that two operators don't commute and therefore if your qubit is in $|0\rangle$ which is an eigenstate of $\sigma_z$, you will have uncertainty in the measurement of $\sigma_x$. However, let's check that Heisenberg principle is still holding even in this situation!

Let's see Heisenberg's principle in this case. Heisenberg principle can be derived using schwarz inequality. Usually, some of the terms are omitted, but I will report the full principle. The Heisenberg principle for two operators $\hat A$ and $\hat B$ states that:

$$\langle\Delta^2A\rangle\langle \Delta^2B\rangle \geq \frac{1}{4}\left|\langle\{\hat A, \hat B\}\rangle - 2\langle \hat A \rangle \langle\hat B \rangle\right|^2 + \frac{1}{4}\left|\langle[\hat A, \hat B]\rangle\right|^2$$

You can read this stated in [1].

In our case we have $\hat A = \sigma_x$ and $\hat B = \sigma_y$. As you noticed if our state is in the $|0\rangle$ eigenstate o $\sigma_z$, then the expectation value $\langle\Delta^2A\rangle = 0$, hence the left hand side is 0!

However, for this particular state, also the right hand side is 0:

  • $\{\sigma_z, \sigma_x\}=0$ (property of the Pauli matrices)
  • $2\langle \sigma_z \rangle\langle \sigma_x \rangle = 0$ since $\langle \sigma_x \rangle = \langle 0 | \sigma_x | 0 \rangle = 0$ (you can check this one doing the matrix multiplication!)
  • $\left|\langle[\sigma_z, \sigma_x]\rangle\right|^2 = 0$ because $[\sigma_z, \sigma_x]=-2i\sigma_y$ and the expectation value $\langle 0 | \sigma_y | 0 \rangle=0$.

Hence, the right hand side is 0 as well and heisenberg is saved (note that the inequality allows for the left and right to be equal)!

A last point is that $\sigma_z$, $\sigma_x$ and $\sigma_y$ all have different eigenstates. However, they have the same eigen values $\pm 1$.

[1] https://arxiv.org/pdf/2003.08705.pdf


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