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.

  • 4
    $\begingroup$ 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. $\endgroup$
    – Rammus
    Jun 9, 2021 at 13:49

1 Answer 1


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.

  • $\begingroup$ 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? $\endgroup$ Jun 9, 2021 at 18:31
  • $\begingroup$ 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? $\endgroup$
    – glS
    Jun 9, 2021 at 19:16
  • $\begingroup$ Shouldn't these be hermitian projectors? $\endgroup$ Jun 9, 2021 at 20:55
  • $\begingroup$ @NorbertSchuch you are right, I didn't really read how the method worked that carefully. Should be correct now $\endgroup$
    – glS
    Jun 10, 2021 at 6:31
  • 2
    $\begingroup$ 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. $\endgroup$
    – Max
    Dec 23, 2022 at 0:32

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