# How to calculate the spectral norm of the density operator used in Molina et al. 2012 paper?

In Molina et al (2012)'s article on quantum money, the proof of security of Wiesner's quantum money scheme depends on the fact that the density operator

$$Q = \frac{1}{4}\sum_{k \in \{0, 1, +, -\}}\left|kkk\right>\left

has spectral norm $$\|Q\|_{spectral}=3/8$$.

The article presents this without proof, but I was able to show it by noticing that

$$\left|v_0 \right>=2 \left|000\right> + \sqrt{2}\left(\left|+++\right> + \left|---\right>\right)$$ $$\left|v_1 \right>=2 \left|111\right> + \sqrt{2}\left(\left|+++\right> - \left|---\right>\right)$$

are both eigenvectors of $$Q$$ with eigenvalues $$3/8$$, so since $$Q$$ is a density operator its eigenvalues must all be in $$[0, 1]$$ and add to one and therefore this is its maximum eigenvalue.

However, I only found these eigenvectors using a tedious numerical search, and since $$\|Q\|_{spectral}=3/8$$ was presented without proof I wonder - is there a more obvious way to show that this is true?

The eigenvalues of a matrix are independent of the choice of basis in which we represent it. This remains true for choices of bases that are not orthogonal. Consider then a matrix $$A=\sum_k P_k$$, where $$P_k\equiv\lvert u_k\rangle\!\langle u_k|$$, and $$\{|u_k\rangle\}$$ is a set of normalized (not necessarily orthogonal) vectors. Observe that $$A=UU^\dagger$$, where $$U\equiv \sum_k \lvert u_k\rangle\!\langle k|$$. Furthermore, observe that $$U^{-1}AU=(U^\dagger U)^{-1}U^\dagger AU=(U^\dagger U)^{-1}(U^\dagger U)^2=U^\dagger U,$$ where I used the formula for the pseudo-inverse to write: $$U^{-1}=U^+=(U^\dagger U)^{-1}U^\dagger$$. Here I should point out that I'm abusing notation, as $$U$$ might not be invertible. This is however not a problem, because I'm only interested in the restriction of the inverse of $$U$$ on its range, which equals the domain of $$A$$. This is why I can safely understand $$U^{-1}$$ as the pseudoinverse $$U^+$$. Moreover, $$U^\dagger U$$ is invertible, provided the vectors $$|u_k\rangle$$ are linearly independent.

You can actually simplify the above by observing that $$BC$$ and $$CB$$ have the same characteristic polynomial (and thus the same spectral radius) for any $$B$$ and $$C$$.

The gist of the above observations is that, whenever a matrix $$A$$ is a sum of unit-trace projections like the case at hand, one can replace computing the eigenvalues of $$A$$ with computing the eigenvalues of the matrix of overlaps of the component vectors. This can significantly simplify the calculations.

In the case at hand, we use the basis vectors $$\{|kkk\rangle\}_{k\in\{0,1,+,-\}}$$, and thus $$U^\dagger U=\frac{1}{2\sqrt2}\begin{pmatrix}2\sqrt2 & 0 & 1 & 1 \\0 & 2\sqrt2 & 1 & -1 \\ 1 & 1 & 2\sqrt2 & 0\\ 1 & -1 & 0 & 2\sqrt2\end{pmatrix}.$$ Finding the largest eigenvalue of $$U$$ amounts to maximising the quantity $$\langle v|U|v\rangle$$ over all $$\langle v|v\rangle=1$$. This calculation can be simplified observing that (1) $$U$$ is real symmetric, thus we can stick to real numbers, (2) being $$U$$ symmetric, we only need to take into account elements in the upper-right components, and just multiply by $$2$$ the results. We thus easily find $$\langle v|U^\dagger U|v\rangle = 1 + \frac{2}{2\sqrt2}\left( v_1(v_3+v_4) + v_2 (v_3 - v_4) \right).$$ Maximising this quantity thus reduces to maximising the quantity $$C(\vec v)\equiv v_1(v_3+v_4) + v_2 (v_3 - v_4)$$ under the constraint $$v_1^2+v_2^2+v_3^2+v_4^2=1$$. It's not hard from this to guess that the maximum is achieved with $$v_2=0$$ and $$v_3=v_4$$, or $$v_1=0$$ and $$v_3=-v_4$$. But if one wants to make sure of it, you can apply standard Lagrange multipliers, which in this case amounts to imposing that the condition on the gradients $$\nabla C(\vec v)=\lambda \nabla \vec v$$, which gives $$v_3+v_4= 2\lambda v_1, \qquad v_3-v_4= 2\lambda v_2, \\ v_1 + v_2 = 2\lambda v_3, \qquad v_1 - v_2 = 2\lambda v_4,$$ which has solutions for $$\lambda=1/\sqrt2$$, with solution space $$(\sqrt2 v_1,\sqrt2 v_2,v_1+v_2,v_1-v_2)$$ (you don't actually need this for the largest eigenvalue, but you might verify that this matches with your finding).

In conclusion, the largest eigenvalue of $$U^\dagger U$$ is $$1+\frac{1}{\sqrt2}\times \frac{1}{\sqrt2}=3/2$$. Dividing by $$4$$, you again find that the largest eigenvalue of $$Q$$ is $$3/8$$.

Of course, this whole last paragraph about finding the largest eigenvalue of $$U^\dagger U$$ is completely equivalent to the standard ways of diagonalising a Hermitian matrix. I just used this one because I enjoy reinventing wheels.

Personally, I'd jump straight to Mathematica. It took me all of a minute:

zero = ({
{1, 0, 0, 0, 0, 0, 0, 0},
{0, 0, 0, 0, 0, 0, 0, 0},
{0, 0, 0, 0, 0, 0, 0, 0},
{0, 0, 0, 0, 0, 0, 0, 0},
{0, 0, 0, 0, 0, 0, 0, 0},
{0, 0, 0, 0, 0, 0, 0, 0},
{0, 0, 0, 0, 0, 0, 0, 0},
{0, 0, 0, 0, 0, 0, 0, 0}
});
one = ({
{0, 0, 0, 0, 0, 0, 0, 0},
{0, 0, 0, 0, 0, 0, 0, 0},
{0, 0, 0, 0, 0, 0, 0, 0},
{0, 0, 0, 0, 0, 0, 0, 0},
{0, 0, 0, 0, 0, 0, 0, 0},
{0, 0, 0, 0, 0, 0, 0, 0},
{0, 0, 0, 0, 0, 0, 0, 0},
{0, 0, 0, 0, 0, 0, 0, 1}
});
plus = Table[1, {8}, {8}]/8;
minus = KroneckerProduct[{{1, -1}, {-1, 1}}, {{1, -1}, {-1, 1}}, {{1, -1}, {-1, 1}}]/8;
Q = (zero + one + plus + minus)/4;
Eigenvalues[Q]


{3/8, 3/8, 1/8, 1/8, 0, 0, 0, 0}