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.

  • 2
    $\begingroup$ 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
  • $\begingroup$ Two comments: First, if you rotate further one matrix becomes [1 0;0 0] and the other [c^2 cs; sc s^2] (c=cos,s=sin), which is even nicer. Second, I don't think it is usually used to bring projectors into this form (at least I never used it that way) - the mere existence of this form can already be used to simplify many things tremendously (e.g., you know that anything depending on the eigenvalues of any f(A,B) will depend only on those angles which can be obtained e.g. from the spectrum of A*B, or otherwise simplified by playing around with the 2 2x2 matrices above. $\endgroup$ Jun 10, 2021 at 8:48

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.