'Rectilinear' and 'Diagonal' Basis in BB84 Protocol



A lot of the tutorials on BB84 protocol talks about these two measurement bases, 'Rectilinear' or 'Vertical-Horizontal' and 'Diagonal'. I understand that it is possible to create a physical device that would be able to measure a qubit in both vertical and horizontal direction, or in other words, in 'Rectilinear' basis, but what would be the matrix representation of it?

For example, we can use $\lvert 0 \rangle \langle 0 \lvert$ to measure a qubit in the $\lvert 0 \rangle$ basis and $\lvert 1 \rangle \langle 1 \lvert$ to measure in the $\lvert 1 \rangle$ basis. But what would be the combined measurement basis which we could call 'rectilinear' or 'vertical-horizontal'?

Hasan Iqbal

Posted 2018-05-16T17:42:16.183

Reputation: 1 270


I think the terminology comes from photon polarisations which according to wikipedia at least is how the protocol was originally described. https://en.wikipedia.org/wiki/Quantum_key_distribution#BB84_protocol:_Charles_H._Bennett_and_Gilles_Brassard_(1984)

– snulty – 2018-05-16T19:48:22.390

1As written right now, the question seems to (wrongly) state that |0> is a basis and |1> is a different basis. Instead, they are the two possible results in the same basis. – agaitaarino – 2018-05-17T04:25:50.430



Talking about bases such as $\left|0\rangle\langle0\right|$ and $\left|1\rangle\langle1\right|$ (or the equivalent vector notation $\left|0\right>$ and $\left|1\right>$, which I'll use in this answer) at the same time as 'horizontal' and 'vertical' are, to a fair extent (pardon the pun) orthogonal concepts.

On a Bloch sphere, there are 3 different orthonormal bases - we generally consider $\left|0\right\rangle$ and $\left|1\right\rangle$; $\frac{1}{\sqrt 2}\left(\left|0\right\rangle+\left|1\right\rangle\right)$ and $\frac{1}{\sqrt 2}\left(\left|0\right\rangle-\left|1\right\rangle\right)$; $\frac{1}{\sqrt 2}\left(\left|0\right\rangle+i\left|1\right\rangle\right)$ and $\frac{1}{\sqrt 2}\left(\left|0\right\rangle-i\left|1\right\rangle\right)$. I'll refer to these as the 'quantum information bases' as this is the notation generally used in quantum information.

That looks a bit of a mess, so we can also write this as $\left|\uparrow_z\right\rangle$, $\left|\downarrow_z\right\rangle$; $\left|\uparrow_x\right\rangle$, $\left|\downarrow_x\right\rangle$; $\left|\uparrow_y\right\rangle$, $\left|\downarrow_y\right\rangle$, where the different bases are now labelled as $x, y$ and $z$. In terms of spin-half particles, this has a natural definition of up/down spin in each of those directions. However, there is freedom in choosing which direction (in the lab) these axes are in (unless otherwise constrained).

Photons (used in the BB84 protocol) aren't spin-half particles (they have a spin of one), but nevertheless have similarities to this - the 'axes' are the possible directions of the polarisation of a photon, only instead of labelling these as $x, y$ and $z$, they're labelled as horizontal/vertical, diagonal/antidiagonal and left-/right-circular, or in vector notation, this is shortened to $\left|H\right>$, $\left|V\right>$; $\left|D\right>$, $\left|A\right>$; $\left|L\right>$ and $\left|R\right>$. These can then be mapped on to the 'quantum information' bases above, although which basis gets labelled as $\left|0\right>$ and $\left|1\right>$ is somewhat arbitrary.

For the BB84 protocol (and indeed, frequently used in other applications), the rectilinear (vertical/horizontal) basis is the one labelled using $\left|0\right>$ and $\left|1\right>$.

That is: $$\left|H\right>=\left|0\right>$$ $$\left|V\right>=\left|1\right>$$ $$\left|D\right>=\frac{1}{\sqrt 2}\left(\left|H\right\rangle+\left|V\right\rangle\right)=\frac{1}{\sqrt 2}\left(\left|0\right\rangle+\left|1\right\rangle\right)$$ $$\left|A\right>=\frac{1}{\sqrt 2}\left(\left|H\right\rangle-\left|V\right\rangle\right)=\frac{1}{\sqrt 2}\left(\left|0\right\rangle-\left|1\right\rangle\right)$$ $$\left|R\right>=\frac{1}{\sqrt 2}\left(\left|H\right\rangle+i\left|V\right\rangle\right)=\frac{1}{\sqrt 2}\left(\left|0\right\rangle+i\left|1\right\rangle\right)$$ $$\left|L\right>=\frac{1}{\sqrt 2}\left(\left|H\right\rangle-i\left|V\right\rangle\right)=\frac{1}{\sqrt 2}\left(\left|0\right\rangle-i\left|1\right\rangle\right)$$

If you want to measure in any of these bases, use the 'projectors' of that basis. That is, if you want to measure in the rectilinear basis, the projectors are $\left|H\rangle\langle H\right|$ and $\left|V\rangle\langle V\right|$. Similarly, in the diagonal basis, $\left|D\rangle\langle D\right|$ and $\left|A\rangle\langle A\right|$; and in the circularly polarised basis, $\left|L\rangle\langle L\right|$ and $\left|R\rangle\langle R\right|$


Posted 2018-05-16T17:42:16.183

Reputation: 3 226


I think something like Figure 11 in What we can learn about quantum physics from a single qubit would be a perfect illustration for this answer!

– agaitaarino – 2018-05-17T04:31:42.740


For the diagonal basis, the measurement operators are the $|0\rangle\langle 0|$ and $|1\rangle\langle 1|$, as stated in the question. For the other basis, any mutually unbiased basis will do, but people usually go for the two operators $(|0\rangle+|1\rangle)(\langle 0|+\langle 1|)/2$ and $(|0\rangle-|1\rangle)(\langle 0|-\langle 1|)/2$.

The labels of which basis you call what are fairly arbitrary, but I think that the rectilinear basis is usually the one that corresponds with horizontal/vertical polarisation and is labelled 0/1. The diagonal basis is then the other one.


Posted 2018-05-16T17:42:16.183

Reputation: 35 722