# Is the trace distance upper bounded by the Euclidean distance?

Suppose we have two pure state $$|\psi\rangle$$ and $$|\phi\rangle$$.

I was wondering whether the statement: $$\||\psi\rangle\!\langle\psi|- |\phi\rangle\!\langle\phi|\|_{\rm tr}$$ is at most the Euclidean distance between $$|\psi\rangle$$ and $$|\phi\rangle$$. I know that under the qubit condition, the trace distance is exactly half of Euclidean distance.

If it is right, then what should the vector representation be corresponding to these two pure states?

• are you asking whether the trace distance between two states is upper bounded by the $L_2$ distance, that is, whether $\|\rho-\sigma\|_1\le \|\rho-\sigma\|_2$?
– glS
Aug 1 at 11:37
• Yes, is that true? Aug 1 at 11:44

This can be inspected following Hölder's inequality with $$p=q=2$$:

$$||AI||_1\leq ||A||_2 ||I||_2$$ for any matrices $$A$$ and $$I$$ of dimension $$d$$. Setting $$I$$ to be the identity matrix, we find $$||A||_1\leq \sqrt{d}||A||_2$$ (also mentioned on Wikipedia). We thus do not have a proof that $$||A||_1\leq ||A||_2$$. For example, choosing any $$d>1$$ and $$A=I$$, we have $$d=||I||_1 > ||I||_2=\sqrt{d}$$.

Now that we know the trace norm is not always smaller than the Euclidean for an arbitrary matrix $$A$$, are there any special properties when $$A$$ is the difference between two states, $$A=\rho-\sigma$$?

I used Mathematica to check whether random density matrices have this special property and readily found examples that they don't: the 2-norm can be smaller than the 1-norm by more than a factor of 2

SeedRandom[dim];
r = With[{U = RandomVariate[CircularRealMatrixDistribution[dim]]},
U.(RandomReal[{0, 1}, dim] U\[Transpose])];
r = r/Tr[r];
s = With[{U = RandomVariate[CircularRealMatrixDistribution[dim]]},
U.(RandomReal[{0, 1}, dim] U\[Transpose])];
s = s/Tr[s];
A = r - s;
svals = SingularValueList[A];
n1 = Total[svals]
n2 = Sqrt[Total[svals^2]]


The outputs are n1=0.841057 and n2=0.295807. If you want to check the matrices yourself, they are $$\rho=\left( \begin{array}{cccccccccc} 0.0559484 & 0.0391899 & -0.00173254 & 0.0150553 & -0.00665287 & 0.0423966 & -0.0237883 & -0.00254245 & 0.00403863 & 0.0141285 \\ 0.0391899 & 0.114607 & -0.0355029 & 0.0449404 & 0.00791358 & 0.0208125 & -0.0166859 & 0.0136876 & -0.00365799 & 0.0577457 \\ -0.00173254 & -0.0355029 & 0.111272 & -0.00530915 & -0.0237596 & -0.00530644 & -0.0144068 & 0.025036 & 0.0104944 & -0.019394 \\ 0.0150553 & 0.0449404 & -0.00530915 & 0.0779406 & 0.0128229 & -0.00717888 & -0.0125049 & -0.00948783 & -0.0199239 & 0.0163034 \\ -0.00665287 & 0.00791358 & -0.0237596 & 0.0128229 & 0.0684224 & -0.00253051 & -0.00578879 & 0.00811702 & -0.00295013 & -0.0421486 \\ 0.0423966 & 0.0208125 & -0.00530644 & -0.00717888 & -0.00253051 & 0.166275 & 0.0078063 & -0.0168575 & 0.0515418 & -0.03558 \\ -0.0237883 & -0.0166859 & -0.0144068 & -0.0125049 & -0.00578879 & 0.0078063 & 0.106101 & -0.00781364 & 0.00546259 & 0.00478611 \\ -0.00254245 & 0.0136876 & 0.025036 & -0.00948783 & 0.00811702 & -0.0168575 & -0.00781364 & 0.078272 & -0.00208764 & 0.00160243 \\ 0.00403863 & -0.00365799 & 0.0104944 & -0.0199239 & -0.00295013 & 0.0515418 & 0.00546259 & -0.00208764 & 0.0919301 & -0.0107645 \\ 0.0141285 & 0.0577457 & -0.019394 & 0.0163034 & -0.0421486 & -0.03558 & 0.00478611 & 0.00160243 & -0.0107645 & 0.129231 \\ \end{array} \right)$$ and $$\left( \begin{array}{cccccccccc} 0.125917 & -0.0348865 & -0.0329077 & -0.00975323 & -0.00550824 & 0.0247898 & -0.00381375 & 0.0177037 & 0.0115744 & -0.00041872 \\ -0.0348865 & 0.0910228 & -0.00273958 & -0.0161907 & 0.010816 & -0.00247926 & 0.0103943 & 0.0055119 & 0.0220808 & 0.00719803 \\ -0.0329077 & -0.00273958 & 0.0671699 & 0.0178862 & -0.0175591 & 0.0186505 & -0.000334635 & -0.00606378 & 0.00894835 & -0.0522578 \\ -0.00975323 & -0.0161907 & 0.0178862 & 0.0976361 & 0.0494538 & 0.0108696 & -0.00845623 & 0.0206701 & -0.0305535 & 0.00555425 \\ -0.00550824 & 0.010816 & -0.0175591 & 0.0494538 & 0.122415 & 0.0255157 & 0.00702227 & -0.0282939 & 0.0271073 & 0.00524606 \\ 0.0247898 & -0.00247926 & 0.0186505 & 0.0108696 & 0.0255157 & 0.10571 & -0.00900667 & -0.0204656 & -0.00510038 & -0.0140047 \\ -0.00381375 & 0.0103943 & -0.000334635 & -0.00845623 & 0.00702227 & -0.00900667 & 0.0731207 & 0.00192831 & 0.0358366 & -0.00855586 \\ 0.0177037 & 0.0055119 & -0.00606378 & 0.0206701 & -0.0282939 & -0.0204656 & 0.00192831 & 0.0696558 & 0.0089128 & 0.0131745 \\ 0.0115744 & 0.0220808 & 0.00894835 & -0.0305535 & 0.0271073 & -0.00510038 & 0.0358366 & 0.0089128 & 0.13509 & -0.00930882 \\ -0.00041872 & 0.00719803 & -0.0522578 & 0.00555425 & 0.00524606 & -0.0140047 & -0.00855586 & 0.0131745 & -0.00930882 & 0.112262 \\ \end{array} \right).$$

• I think you can say something quite a bit stronger: you always have the opposite inequality, $\|A\|_1\ge \|A\|_2$. You see it easily from the fact that $\|A\|_1$ is the sum of the singular values of $A$, while $\|A\|_2^2$ is the sum of the squares of the singular values. Thus $\|A\|_1^2 \ge \|A\|_2^2$, as the square of a sum is always larger than the sum of squares, which is equivalent to $\|A\|_1 \ge \|A\|_2$.
– glS
Aug 1 at 17:46
• You can also strengthen the inequality you cited replacing $d$ with the rank of $A$, which you can prove observing your argument also holds if instead of $I$ you use the projection on the support of $A$ (at least it works for positive semidefinite operators; the argument might need some adjustment for the more general case, I'm not sure). These inequalities are a standard tool discussing t-designs, see eg en.wikipedia.org/wiki/Welch_bounds
– glS
Aug 1 at 17:51
• @gIS well... that's a better answer than mine! Aug 1 at 18:01

You always have the opposite inequality: $$\|A\|_1\ge \|A\|_2$$.

You see it easily from the fact that $$\|A\|_1$$ is the sum of the singular values of $$A$$, while $$\|A\|_2^2$$ is the sum of the squares of the singular values. Thus $$\|A\|_1^2 \ge \|A\|_2^2$$, as the square of a sum is always larger than the sum of squares, and thus $$\|A\|_1 \ge \|A\|_2$$.

For the other direction, as pointed out in the other answer, you can still say something, as long as you introduce the dimension of the space, or better still, the rank. For any matrix $$A$$ you have $$\|A\|_2\le \|A\|_1 \le \sqrt{\operatorname{rank}(A)} \|A\|_2,$$ with the upper bound obtained via Holder's inequality as per the other answer. More generally, you have $$\|A\|_q \le \|A\|_p \le \operatorname{rank}(A)^{1/p-1/q}\|A\|_q,$$ for all $$1\le p\le q \le\infty$$ such that $$1/p+1/q=1$$. This is mentioned e.g. in chapter 1 of Watrous' book. A way to prove this is using the more general inequality for Schatten norms: $$\|ST\|_a\le \|S\|_b \|T\|_c, \qquad \frac1a=\frac1b+\frac1c, \quad a,b,c\in[1,\infty].$$ Applying this with $$S=A$$, $$T=\Pi_{\operatorname{supp}(A)}$$, $$a=p$$, $$b=q$$, we get $$\|A\|_p \le \|A\|_q \|\Pi_{\operatorname{supp}(A)}\|_{(1/p-1/q)^{-1}} = \|A\|_q \operatorname{rank}(A)^{1/p-1/q},$$ and we must have $$1/p-1/q\ge1$$, thus $$q\ge p$$.

In the special case of Hermitian matrices you can also use a simpler proof: given Hermitian $$A$$, let $$U=\Pi_+-\Pi_-$$, with $$\Pi_\pm$$ projections onto the spaces spanned by eigenvectors of $$A$$ with positive and negative eigenvalues, respectively. Then $$\operatorname{tr}(A U)=\operatorname{tr}|A|$$, and thus from Holder we get $$|\operatorname{tr}(AU)|=\operatorname{tr}|A| \le \|A\|_2 \|U\|_2 = \sqrt{\operatorname{rank}(A)} \|A\|_2,$$ because $$\|U\|_2=\|\Pi_{\operatorname{supp}(A)}\|_2=\sqrt{\operatorname{rank}(A)}$$ and $$\operatorname{tr}|A|=\|A\|_1$$.

Specialising to the case of pure states, everything's simple: you have $$\|\mathbb{P}_\psi-\mathbb{P}_\phi\|_1 = 2\sqrt{1-F}$$ with $$F\equiv |\langle\psi|\phi\rangle|^2$$, and I used the notation $$\mathbb{P}_\psi\equiv|\psi\rangle\!\langle\psi|$$. This follows from a direct calculation of the eigenvalues of $$\mathbb{P}_\psi-\mathbb{P}_\phi$$. Also $$\|\mathbb{P}_\psi-\mathbb{P}_\phi\|_2^2 \equiv \operatorname{tr}[(\mathbb{P}_\psi-\mathbb{P}_\phi)^2] = \operatorname{tr}(\mathbb{P}_\psi+\mathbb{P}_\phi-2\mathbb{P}_\psi\mathbb{P}_\phi) = 2 (1-F).$$ Thus $$\|\mathbb{P}_\psi-\mathbb{P}_\phi\|_1 = \sqrt 2\|\mathbb{P}_\psi-\mathbb{P}_\phi\|_2.$$ By contrast, the Euclidean distance between the ket vectors themselves reads $$\||\psi\rangle-|\phi\rangle\|_2^2 = 2 - 2\operatorname{Re}\langle\psi|\phi\rangle \ge 2(1-\sqrt F).$$ Furthermore $$2(1-\sqrt F) \ge \left(\frac{\|\mathbb{P}_\psi-\mathbb{P}_\phi\|_1}{2}\right)^2=1-F$$ always holds, because $$2(1-\sqrt F)-(1-F) = 1 - 2\sqrt F + F = (1-\sqrt F)^2 \ge0.$$ If follows that $$\||\psi\rangle-|\phi\rangle\|_2 \ge \frac{\|\mathbb{P}_\psi-\mathbb{P}_\phi\|_1}{2}.$$

Note that the "trace distance" is usually defined as half of $$\|\mathbb{P}_\psi-\mathbb{P}_\phi\|_1$$.

• In the case of pure state, my understanding is as following: $|| |\psi\rangle, |\phi\rangle ||_{tr} = \sqrt{1 - |<\phi|\psi>|^2}$, and the Euclidean distance is $|| |\psi\rangle - |\phi\rangle || = \sqrt{2 - <\phi|\psi> - <\psi|\phi>}$ and the latter is large and equal to the former because $(<\phi|\psi> - 1)(<\psi|\phi> - 1)\ge 0$. Does that have some problem? Aug 2 at 4:35
• @ZehongFan well, I'm not sure, the comment is hard parse. But the correct calculation is at the end of the post. You should be careful how you define things though. What exactly is "euclidean distance" for you in this context? Are you asking about $\|\rho-\sigma\|_2$ or are you asking about the Euclidean distance between the ket vectors, that is, $\| |\psi\rangle-|\phi\rangle\|_2$?
– glS
Aug 2 at 6:40