# Prove that for any state $\rho$ there is a basis $|e_k\rangle$ corresponding to uniformly distributed probabilities

I am new to quantum mechanics, and I'm trying to learn about Majorization in quantum states. These are the notes I am following http://michaelnielsen.org/blog/talks/2002/maj/book.ps by Michael Nielsen. I am stuck in an exercise that already has an answer, but I still can't understand it. It is the following:

Exercise 1.3.1: Let $$\rho$$ be an arbitrary state of a $$d$$-dimensional quantum system. Prove that there always exists an orthonormal basis $$|e_k >$$ such that the probabilities for a measurement in that basis are uniformly distributed. Given $$\rho$$ can you explicitly construct a basis $$|e_k >$$ such that this is true?

At the end of chapter 1, it says that the answer is: The basis $$|e_k >$$ to measure in is the Fourier transform of the eigenbasis of $$\rho$$.

But I can't see how this is proved. Or even what is the intuition of why this is true?

So it might help to just start with one state. We diagonalize $$\rho$$ and we find out that it is
$$\rho =\rho_1 |\psi_1\rangle\langle\psi_1| +\dots+\rho_d|\psi_d\rangle\langle\psi_d|$$. Define $$|e_0\rangle = \sqrt{\frac1d} \big(|\psi_1\rangle+ |\psi_2\rangle + \dots + |\psi_d\rangle\big).$$ Now observe, the occupation in this state is $$\operatorname{Tr}\big(|e_0\rangle\langle e_0|~\rho\big)=\langle e_0|\rho|e_0\rangle = \frac1d \sum_{k=1}^d \rho_d.$$ That sum is conveniently 1, so we have our first vector.
What about our next vector? Well suppose $$d$$ is even, consider $$|e_1\rangle = \sqrt{\frac1d} \big(|\psi_1\rangle -|\psi_2\rangle + \dots - |\psi_d\rangle\big).$$ This is clearly orthogonal to $$|e_1\rangle$$, but it has the exact same sum when we form $$\operatorname{Tr}\big(|e_1\rangle\langle e_1|~\rho\big)$$ so we have our second vector.
Can also try going $$1,i,-1,-i$$ and if $$d$$ is divisible by 4, this will be orthogonal to both of the above and will generate the same $$1/d$$.
In fact, any phase factors leading the terms will allow us to get the same $$1/d$$ and so the only thing we have to focus on is orthogonality. So the basic idea there is just that this discrete Fourier transform arrangement does the job well. Indeed both of the above are special cases of this. If you have a sum of nth roots of unity, the finite geometric series formula automatically forces the sum to be zero, which you can kind of intuit because that sum calculates the center of mass of a bunch of masses evenly spread out on a ring of radius one around an origin. This basic rule gives you your Kronecker delta for your overlap integrals $$\langle e_i | e_k\rangle =\delta_{ik}$$, so you have orthogonality and you have the trivial probability $$1/d$$ and everything is good.
• Thank you so much, this was really useful. Just a quick question, since the discrete Fourier transform is defined as $\sum_{n=0}^{N-1} x_{n} \cdot e^{-\frac{i 2 \pi}{N} k n}$ , the idea would be that each $x_{i}$ is $\left|\psi_{i}\right\rangle$ and each of the basis elements $\left| e_{k} \right\rangle$ are constructed by varying $k \in \{0,...,N-1 \}$. It seems that this $N$ should be prime for the argument to work? Commented Sep 26, 2020 at 15:54
• No $N$ does not need to be prime, for example I gave you the solution for $N=4$ explicitly above (well, 3 of the vectors but the fourth was $1,-i,-1,i$) and 4 is not prime. You just need $N=d$ to get orthogonality, because orthogonality gives you $$\sum_{k=0}^{d-1} \frac1d\exp\left(\frac{2\pi i}{d}(m-n)k\right)$$ and this is either a sum of ones or it is a sum of points evenly spaced around the unit circle in the complex plane. Doesn't matter if they're spaced by exactly $2\pi/d$, like with $u=(1,i,-1,-i),v=(1,-i,-1,i)$ you get $u^\dagger v= 1-1+1-1=0$ with no problem. Commented Sep 26, 2020 at 16:10