# How do we understand Jordan's Lemma?

In quantum computing protocols, jordan's lemma keeps cropping up. See, for example, here: https://cims.nyu.edu/~regev/teaching/quantum_fall_2005/ln/qma.pdf

For any two projectors $$\phi_1$$, $$\phi_2$$, there exists an orthogonal decomposition of the Hilbert space into one-dimensional and two-dimensional subspaces that are invariant under both $$\phi_1$$ and $$\phi_2$$. Moreover, inside each two-dimensional subspace, $$\phi_1$$ and $$\phi_2$$ are rank-one projectors.

What does a decomposition of a hilbert space mean? Can anyone give an example? I sort of understand the proof but I don't understand what the theorem is trying to say or how it is useful.

• You decompose the Hilbert space into a direct sum of subspaces, with each one being two or one dimensional. One of the reasons that this is useful is that each subspace then looks like a qubit system and as the projectors leave the subspaces invariant they also look like single qubit projectors when you restrict them to a subspace. This can sometimes then allow you to reduce the analysis of a problem to these subspaces and hence to qubit systems. I don't know anything about QMA but this is also useful in device-independence as it allows one to reduce an arbitrary dimension problem to qubits. Commented Jun 9, 2021 at 13:49

It means to find a basis for the Hilbert space that can be partitioned into singles and pairs of vectors which form one- and two-dimensional subspaces left invariant by the action of the projectors $$\Pi_i$$. More formally, a decomposition for the space means to write the (here Hilbert) vector space $$\mathcal H$$ as a direct sum of subspaces which are invariant under the action of the given operators, that is: $$\mathcal H=\bigoplus_i \mathcal H_i, \qquad \text{with}\,\, \Pi_i \mathcal H_j\subseteq\mathcal H_j\forall i,j.$$

The idea seems pretty simple: given any pair of (ortho)projectors $$\Pi_1$$ and $$\Pi_2$$, if we use as basis for the vector space the basis of (orthonormal) eigenvectors of $$\Pi_1+\Pi_2$$, then automatically both $$\Pi_1$$ and $$\Pi_2$$ assume a particularly simple matrix form. More specifically, they are simultaneously block-diagonalisable, with blocks of dimensions 1 or 2.

As an example, suppose $$\dim(\mathcal H)=3$$, with $$\mathcal H={\rm span}(\{e_1,e_2,e_3\})$$, and consider the projectors $$\Pi_1\equiv e_+ e_+^\dagger$$ with $$e_+\equiv \frac{1}{\sqrt2}(e_1+e_2)$$ and $$\Pi_2\equiv e_2 e_2^\dagger+ e_3 e_3^\dagger$$.

Then, the basis of eigenvectors of $$\Pi_1+\Pi_2$$ is $$\{\frac{1}{\sqrt2}(e_1+e_2),\frac{1}{\sqrt2}(e_1-e_2),e_3\}$$. Representing the projectors in this basis, we get $$\Pi_1\doteq\begin{pmatrix}1&0&0\\ 0&0 & 0 \\0&0&0\end{pmatrix}, \qquad \Pi_2\doteq\begin{pmatrix}0&0&0\\ 0&0 & 0 \\0&0&1\end{pmatrix}.$$

For another example, consider $$\Pi_1=e_{123}e_{123}^\dagger, \qquad \Pi_2=e_{23}e_{23}^\dagger,\\ e_{123}\equiv\frac{1}{\sqrt3}(e_1+e_2+e_3), \qquad e_{23}\equiv\frac{1}{\sqrt2}(e_2+e_3).$$ Following the same procedure we get (calculations are not as nice here so I just got the result via software): $$\Pi_1\doteq\begin{pmatrix} \frac{1}{2}+\frac{1}{\sqrt{6}} & -\frac{1}{2 \sqrt{3}} & 0 \\ -\frac{1}{2 \sqrt{3}} & \frac{1}{2}-\frac{1}{\sqrt{6}} & 0 \\ 0 & 0 & 0 \end{pmatrix}, \qquad \Pi_2\doteq\begin{pmatrix} \frac{1}{2}+\frac{1}{\sqrt{6}} & \frac{1}{2 \sqrt{3}} & 0 \\ \frac{1}{2 \sqrt{3}} & \frac{1}{2}-\frac{1}{\sqrt{6}} & 0 \\ 0 & 0 & 0 \end{pmatrix}.$$

The simultaneous block-diagonal structure is again clearly visible.

• Hi! What confuses me is that if the hilbert space is dimension 3, how do we get 2 dimensional and 1 dimensional subspaces? Is it the block structure that characterizes the dimension? Commented Jun 9, 2021 at 18:31
• I don't know that I understand your problem here. I'm tempted to just answer: because $3=2+1$, but maybe I'm missing your point?
– glS
Commented Jun 9, 2021 at 19:16
• Shouldn't these be hermitian projectors? Commented Jun 9, 2021 at 20:55
• @NorbertSchuch you are right, I didn't really read how the method worked that carefully. Should be correct now
– glS
Commented Jun 10, 2021 at 6:31
• You have a typo in your first example that threw me off for a while: Π2≡𝑒2𝑒†2+𝑒3𝑒†3 should not have the 𝑒2𝑒†2 term. Your own representation of Π2 confirms this.
– Max
Commented Dec 23, 2022 at 0:32