# Unbiased real vector with respect to arbitrary orthonormal basis for a finite Hilbert space

Here I use Dirac notation to denote vectors. I would like to show that for an arbitrary orthonormal basis $$\{ |\psi_k\rangle \}_{k=1}^n \subset \mathbb C^n$$, $$\langle \psi_i | \psi_j \rangle = \begin{cases} 1 & i = j \\ 0 & i \neq j \end{cases}$$ there exists phases $$\{\theta_k\}_{k=1}^n \subset [0, 2\pi]$$ such that $$\sum_{k=1}^n e^{i\theta_k}|\psi_k\rangle \in \mathbb R^n$$. Or equivalently, there exists a real vector $$|v\rangle \in \mathbb R^n (|v\rangle \neq 0)$$ such that $$|\langle v|\psi_1 \rangle| = \cdots = |\langle v|\psi_n \rangle|$$

The case $$n=2$$ is easy. Let $$| \psi_1 \rangle = \begin{pmatrix} \alpha_1 \\ \beta_1 \end{pmatrix} \ \ \ \ | \psi_2 \rangle = \begin{pmatrix} \alpha_2 \\ \beta_2 \end{pmatrix} \ \ \ \ | v \rangle = \begin{pmatrix} x \\ y \end{pmatrix}$$

It is easy to check that the quadratic equation of $$x, y$$ $$|\langle v|\psi_1 \rangle|^2-|\langle v|\psi_2 \rangle|^2 = (|\alpha_1|^2 - |\alpha_2|^2)x^2 + (|\beta_1|^2 - |\beta_2|^2)y^2 + 2\operatorname{Re}(\alpha_1\bar\beta_1 - \alpha_2\bar\beta_2)xy= 0$$ has non-trivial real roots.

However, it seems that the high dimensional cases are more serious. Are there any suggestions to me?

• For finite dimensional spaces I'd recon a proof by induction might work. Or a base transform to the standard unit vectors. – Floris Claassens Feb 14 at 10:30
• @Floris Claassens, I think transforming to computational basis(standard unit vectors) does not change the problem much. Could you elaborate on the proof by induction? Personally, I believe topological method might work. – Zixuan Liu Feb 14 at 11:57
• @Dap, I am wondering how to use the possible duplicate problem to solve my problem? – Zixuan Liu Feb 14 at 14:38
• @ZixuanLiu: yes I didn't read that question carefully enough and have retracted my comment/duplicate vote. – Dap Feb 14 at 14:42

The case $$n=3$$ holds.

Given orthonormal $$|\psi_1\rangle,|\psi_2\rangle,|\psi_3\rangle\in\mathbb C^3,$$ let $$P_i\in\mathbb R^{3\times 3}$$ be the real part of the Hermitian matrix $$|\psi_i\rangle\langle\psi_i|.$$ We want to find $$v\neq 0$$ with $$v^T(P_2-P_1) v=v^T(P_3-P_1)v=0$$ because this would give $$v^TP_1v=v^TP_2v=v^TP_3v.$$

Note that any linear combination $$y_1(P_2-P_1)+y_2(P_3-P_1)$$ has trace zero, so is indefinite. The existence of $$v$$ follows from a technical report of F. Bohnenblust, Joint positiveness of matrices. That report also mentions an argument based on joint diagonalization which I will give here.

Suppose that no such $$v$$ exists. Then we can apply the following theorem of Milnor given in W. H. Greub, Linear Algebra, 3rd Ed, p.256, to the bilinear forms $$\Phi(x)=x^T(P_2-P_1)x$$ and $$\Psi(x)=x^T(P_3-P_1)x.$$

Let $$E$$ be a [real] vector space of dimension $$n\geq 3$$ and let $$\Phi$$ and $$\Psi$$ be two symmetric bilinear functions such that $$\Phi(x)^2+\Phi(x)^2\neq 0$$ if $$x\neq 0.$$ Then $$\Phi$$ and $$\Psi$$ are simultaneously diagonalizable.

So we can write $$M^T(P_2-P_1)M=\operatorname{diag}(a_{11},a_{21},a_{31})$$ and $$M^T(P_3-P_1)M=\operatorname{diag}(a_{12},a_{22},a_{32})$$ for some real matrix $$A\in\mathbb R^{3\times 2}$$ and some non-singular matrix $$M\in\mathbb R^{3\times 3}.$$

As mentioned above, $$M^T(y_1(P_2-P_1)+y_2(P_3-P_1))M$$ cannot be negative definite, which means $$Ay$$ does not lie in the strictly negative orthant $$(-\infty,0)^3.$$ By LP duality ("Gordan's lemma"), there is a non-zero vector $$x\in[0,\infty)^3$$ with $$A^Tx=0.$$ But then any $$w=(\pm\sqrt{x_1},\pm\sqrt{x_2},\pm\sqrt{x_3})$$ would satisfy $$w^T \operatorname{diag}(a_{1i},a_{2i},a_{3i}) w=0$$ for $$i=1,2,$$ which means $$v=Mw$$ would satisfy $$v^T(P_2-P_1)v=v^T(P_3-P_1)v=0.$$

Unfortunately this approach cannot work for $$n=4.$$ The matrices

$$Q_1=\begin{pmatrix} 0&1&0&0\\ 1&0&0&0\\ 0&0&0&0\\ 0&0&0&0 \end{pmatrix}, Q_2=\begin{pmatrix} 1&0&0&0\\ 0&-1&0&0\\ 0&0&0&0\\ 0&0&0&0 \end{pmatrix}, Q_3=\begin{pmatrix} 0&0&0&0\\ 0&-2&0&0\\ 0&0&1&0\\ 0&0&0&1 \end{pmatrix}$$ all have trace zero. If $$v\in\mathbb R^4$$ satisfies $$v^TQ_1v=v^TQ_2v=v^TQ_3v=0$$ then:

• $$v_1v_2=0$$
• $$v_1^2=v_2^2$$
• $$2v_2^2=v_3^2+v_4^2$$

But the first two conditions imply $$v_1=v_2=0,$$ and the third then implies $$v_3=v_4=0.$$

Let $$\epsilon=1/1000,$$ let $$P_i=\tfrac14 I_{4\times 4}+\epsilon Q_i$$ for $$i=1,2,3$$ and $$P_4=I_{4\times 4}-\epsilon(Q_1+Q_2+Q_3).$$ Then $$P_1,P_2,P_3,P_4$$ are symmetric, positive semidefinite, sum to $$I_{4\times 4}$$ and all have trace $$1.$$ This means any proof for $$n=4$$ needs to use properties of the matrices $$\mathrm{Re}(|\psi_i\rangle\langle\psi_i|)$$ other than just these crude properties.

• And I would like to know why ''this imposes fewer than n linear conditions on p''. – Zixuan Liu Feb 14 at 16:32
• there seems to be a problem with your claim that $\Vert v \Vert < \Vert w' \Vert$. We can see that $\Vert w' \Vert^2 > \frac {\Vert w \Vert^2} {1+\epsilon^2} = \frac{\Vert v \Vert^2 + \epsilon^2} {1+\epsilon^2}$. – Zixuan Liu Feb 14 at 17:11
• $\frac {\Vert v \Vert + \epsilon^2}{1+\epsilon^2}$ is between 1 and $\Vert v \Vert$. – Zixuan Liu Feb 14 at 17:25
• @ZixuanLiu: sorry, in fact my statement was wrong for n>3. I've edited my answer to give an argument for the 3x3 case and explain why my argument doesn't work for n>3. – Dap Mar 2 at 20:16
• thanks for keeping attention to this problem. This is a cool idea for the case n=3. – Zixuan Liu Mar 8 at 11:35