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?

  • $\begingroup$ For finite dimensional spaces I'd recon a proof by induction might work. Or a base transform to the standard unit vectors. $\endgroup$ – Floris Claassens Feb 14 '19 at 10:30
  • $\begingroup$ @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. $\endgroup$ – Zixuan Liu Feb 14 '19 at 11:57
  • $\begingroup$ @Dap, I am wondering how to use the possible duplicate problem to solve my problem? $\endgroup$ – Zixuan Liu Feb 14 '19 at 14:38
  • $\begingroup$ @ZixuanLiu: yes I didn't read that question carefully enough and have retracted my comment/duplicate vote. $\endgroup$ – Dap Feb 14 '19 at 14:42

(Not an answer.)

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.

  • $\begingroup$ And I would like to know why ''this imposes fewer than n linear conditions on p''. $\endgroup$ – Zixuan Liu Feb 14 '19 at 16:32
  • $\begingroup$ 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} $. $\endgroup$ – Zixuan Liu Feb 14 '19 at 17:11
  • $\begingroup$ $\frac {\Vert v \Vert + \epsilon^2}{1+\epsilon^2}$ is between 1 and $\Vert v \Vert$. $\endgroup$ – Zixuan Liu Feb 14 '19 at 17:25
  • $\begingroup$ @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. $\endgroup$ – Dap Mar 2 '19 at 20:16
  • $\begingroup$ thanks for keeping attention to this problem. This is a cool idea for the case n=3. $\endgroup$ – Zixuan Liu Mar 8 '19 at 11:35

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