# Hilbert space of n qubits always has a decoherence free subspace if n is even?

I've heard that a subspace of an $$n$$ qubit Hilbert space is called decoherence-free if it is not affected by errors of the form $$g^{\otimes n}$$ where $$g \in SU(2)$$.

Let $$\textbf{2}$$ denote the fundamental rep of $$SU(2)$$. Then an $$n$$ qubit Hilbert space transforms as $$\textbf{2}^{\otimes n}$$ with respect to the action of $$g^{\otimes n }$$.

The representation $$\textbf{2}^{\otimes n}$$ contains trivial subreps if and only if $$n$$ is even. And a trivial subrep is exactly one for which the action of errors of the form $$g^{\otimes n}$$ has no effect on the subspace.

In this way we can see that two qubits has a 1 dimensional decoherence free subspace, $$\textbf{2}^{\otimes 2} = \textbf{1} \oplus \textbf{3}$$ four qubits has a 2 dimensional decoherence free subspace, $$\textbf{2}^{\otimes 4} = (2) \textbf{1} \oplus (3) \textbf{3} \oplus \textbf{5}$$ six qubits has a 5 dimensional decoherence free subspace, $$\textbf{2}^{\otimes 6} = (5) \textbf{1} \oplus (9) \textbf{3} \oplus (5) \textbf{5} \oplus \textbf{7}$$ etc...

On the other hand for odd number of qubits you don't get any trivial subreps because $$-1$$ to an odd number is $$-1$$ so all the subreps of $$\textbf{2}^{\otimes n}$$ for $$n$$ odd are faithful, but of course the trivial rep $$\textbf{1}$$ is not faithful so it cannot show up in $$\textbf{2}^{\otimes n}$$ for $$n$$ odd.

So my question is: do I understand decoherence free subspaces correctly? The topic is very new to me. If so is there some more physical explanation for why even numbers of qubits always have decoherence free subspaces? The math is clear but it all just seems kind of weird.

• I don't understand why you are saying 2 qubits have 1-dimensional DFS. 2 qubits have 2-dimensional DFS under a collective noise model like $g^{\otimes n}$, which is permutation-symmetric. Feb 22 at 23:05
• After writing my answer, I am realizing, are you perhaps mixing the concepts of decoherence-free subspaces and noiseless subsystems? Feb 23 at 20:52

## TL;DR

Decoherence-free subspaces (DFS) do exist for both, even and odd numbers of qubits $$(N)$$, for noise of the form $$g^{\otimes N}$$.

Let me give you some examples from a different perspective than representation theory.

Consider the collective dephasing noise model. Under collective dephasing, all qubits collectively undergo the dephasing process in a permutation-symmetric way. So every qubit $$j$$

\begin{align} |0\rangle_j &\mapsto |0\rangle_j\,,\\ |1\rangle_j &\mapsto e^{i\phi}|1\rangle_j\,. \end{align}

The $$N$$ qubits are collectively coupled to the bath. Permutation-symmetric means here that even if you swap two qubits or do any permutations, the effect of the bath on your $$N$$ qubit system will be the same, and this symmetry is exactly the thing which allows us to preserve information.

## Two qubit case

The basis state under this noise would transform as follows:

\begin{align} |00\rangle &\mapsto |00\rangle\,,\\ |01\rangle &\mapsto e^{i \phi}|01\rangle\,,\\ |10\rangle &\mapsto e^{i \phi}|10\rangle\,,\\ |11\rangle &\mapsto e^{2i\phi}|11\rangle\,. \end{align}

Observe that the basis states $$|01\rangle$$ and $$|10\rangle$$ acquire the same phase. This phase is an error in the case of a single qubit as it reduces the purity, but here, this error acts as a global phase if we consider the span of these two bases.

So, we can work in this subspace and be protected from error.

We can do our logical encoding as \begin{align} {|0\rangle}_L &= |01\rangle\,,\\ {|1\rangle}_L &= |10\rangle\,. \end{align}

This gives us a protected subspace of 2 dimensions. Note that we also have two trivial DFS here with one dimension each for basis states $$|00\rangle$$ and $$|11\rangle$$.

So, for two qubits, we have three DFS in total, out of which one encodes a qubit, and the other two are trivial.

\begin{align} \text{DFS}_{2}(0) &= \text{Span}\{ |00\rangle \}\,,\\ \text{DFS}_{2}(1) &= \text{Span}\{ |01\rangle, |10\rangle \}\,,\\ \text{DFS}_{2}(2) &= \text{Span}\{ |11\rangle \}\,.\\ \end{align}

## Notation

The notation $$\text{DFS}_{N}(w)$$ represents a decoherence-free subspace for $$N$$ qubits, which is spanned by all the basis states of weight $$w$$, i.e., the number of $$1$$s in the basis states.

## Three qubit case

Similar to the two-qubit case, here the basis states would transform as follows:

\begin{align} |000\rangle &\mapsto |000\rangle\,,\\ |001\rangle &\mapsto e^{i \phi}|001\rangle\,,\\ |010\rangle &\mapsto e^{i \phi}|010\rangle\,,\\ |100\rangle &\mapsto e^{i \phi}|100\rangle\,,\\ |011\rangle &\mapsto e^{2i \phi}|011\rangle\,,\\ |101\rangle &\mapsto e^{2i \phi}|101\rangle\,,\\ |110\rangle &\mapsto e^{2i \phi}|110\rangle\,,\\ |111\rangle &\mapsto e^{3i \phi}|111\rangle\,, \end{align}

and thus, we get the following DFSs:

\begin{align} \text{DFS}_{3}(0) &= \text{Span}\{ |000\rangle \}\,,\\ \text{DFS}_{3}(1) &= \text{Span}\{ |001\rangle, |010\rangle, |100\rangle \}\,,\\ \text{DFS}_{3}(2) &= \text{Span}\{ |101\rangle,|110\rangle,|011\rangle \}\,,\\ \text{DFS}_{3}(3) &= \text{Span}\{ |111\rangle \}\,. \end{align}

## N qubit case

Consider computational basis states for $$N$$ qubits. Let these basis states have $$w$$ numbers of $$1$$s and $$N-w$$ numbers of $$0$$s. Each such state will transform as follows under the collective dephasing:

$$|w(1\text{s}), N-w (0\text{s})\rangle \mapsto e^{w i \phi}|w(1\text{s}), N-w (0\text{s})\rangle\,.$$

Therefore, you will have DFSs as follows:

$$\text{DFS}_N(w) = \text{Span}\{ |w(1\text{s}), N-w (0\text{s})\rangle \}\,.$$

Your hilbert space $$(\mathcal{H})$$ under this collective type of noise is split as follows

$$\mathcal{H} = \bigoplus_{w=0}^N \text{DFS}_N(w)\,.$$

The dimensions of these DFSs are as follows:

$$\text{dim}\big( \text{DFS}_N(w) \big) = {n \choose w } = \frac{n!}{w!(n-w)!}\,.$$

## Visual representation of DFSs

You can see below the Bratteli diagram showing DFSs for each $$N$$. $$X$$-axis is the number of qubits and the $$Y$$-axis represents the number of $$0$$s minus number of $$1$$s in your basis states.

Each point in the figure represents a DFS. For example, the point $$(2,0)$$ represents the spanned by $$\{ |01\rangle, |10\rangle \}$$, where the number of $$0$$s minus the number of $$1$$s in the basis state is the same. The points above and below $$(2,0)$$ are the other two trivial DFSs we talked about, i.e., $$(2,2)$$ is the one-dimensional subspace spanned by $$\{|00\rangle\}$$ and $$(2,-2)$$ is the other one-dimensional subspace spanned by $$\{|11\rangle\}$$.

(Fun fact: The number of paths from the origin to a given point is exactly the dimension of that DFS.)

## Conclusion

For collective noise of the form $$g^{\otimes N}$$, there exists DFS for odd and even number of qubits. I have shown the example of the collective dephasing type of noise, but this argument can be made general.

Since $$g$$ is a unitary, let it have eigenvalues $$\{ e^{i\phi_1} , e^{i\phi_2} \}$$ for the corrsponding eiongevectors $$\{ |a\rangle, |b\rangle \}$$.

Now you can just work in the basis $$\{ |a\rangle, |b\rangle \}$$, and you will still have the same DFSs regardless of what exactly $$g$$ is. You will just do your encoding as follows for the example of 2 qubit case.

\begin{align}{|0\rangle}_L = |ab\rangle\,,\\ {|1\rangle}_L = |ba\rangle\,. \end{align}

After the action of collective noise $$g^{\otimes 2}$$, these states will acquire the same phase and would be a global phase for the subspace spanned by these two states.