# Inequality in overlap of quantum states

For quantum states $$\vert\psi_1\rangle, \vert\psi_2\rangle, \vert\phi\rangle$$, is it true that:

$$\tag{1}\langle \phi\vert\psi_1\rangle\langle\psi_1\vert\phi\rangle\langle \phi\vert\psi_2\rangle\langle\psi_2\vert\phi\rangle + \langle \phi\vert\psi_2\rangle\langle\psi_2\vert\phi\rangle\langle \phi\vert\psi_1\rangle\langle\psi_1\vert\phi\rangle \leq \langle \phi\vert\psi_1\rangle\langle\psi_2\vert\phi\rangle + \langle \phi\vert\psi_2\rangle\langle\psi_1\vert\phi\rangle.$$

My argument is that each number $$c_i = \langle\phi\vert\psi_i\rangle$$ is a complex number with modulus smaller than 1 since it is the square root of a probability. So we have to show:

$$2|c_1|^2|c_2|^2 \leq c_1c_2^* + c_1^*c_2\tag{2}.$$

• Welcome to Stack Exchange, and +1 for the question! Nov 29 '20 at 3:00

$$\langle a | b \rangle$$ is usually a scalar, so you can move such objects around on the left side, and arrive at (assuming all wavefunctions are normalized):

\begin{align} \tag{1} \langle \phi\vert\psi_1\rangle\langle\psi_1\vert\phi\rangle\langle \phi\vert\psi_2\rangle\langle\psi_2\vert\phi\rangle + \langle \phi\vert\psi_2\rangle\langle\psi_2\vert\phi\rangle\langle \phi\vert\psi_1\rangle\langle\psi_1\vert\phi\rangle &= 2\langle \phi\vert\psi_2\rangle\langle\psi_2\vert\phi\rangle\langle \phi\vert\psi_1\rangle\langle\psi_1\vert\phi\rangle \\ &=2|c_1|^2|c_2|^2 \tag{2}\\ &\le2.\tag{3} \end{align}

In the case where $$c_1$$ and $$c_2$$ are positive and real-valued, which does actually happen quite often in quantum mechanics, you'd get:

\begin{align} 2(c_1c_2)^2 ~~~~~&\textrm{vs.} ~~~~~2 c_1c_2\tag{4} \\ \end{align}

Then since $$c_1 c_2 \le 1$$, you would be correct that the left side is $$\le$$ the right side. However there's cases where the inequality is not true, for example:

### $$c_1$$ purely real and $$c_2$$ purely imaginary:

The right-hand side (of your inequality in Eq. 2) becomes:

$$c_1c_2^* + c_1^*c_2 = c_1(c_2 - c_2) = 0\tag{5},$$

but the left-hand side is bigger than 0 because $$|c_2|^2>0.$$

### $$c_1$$ and $$c_2$$ have different signs:

The right-side (of your inequality in Eq. 2) can be negative while the left-side is positive.

• Let $c_1 = a + ib$ and $c_2 = u + iv$ then $c_1 c_2^* + c_1^* c_2 = (a+ib)(u-iv) + (a-ib)(u+iv) = 2(au+bv) \in \mathbb{R}$. Nov 29 '20 at 4:49
• @KAJ226 good point. Nov 29 '20 at 5:02
• This is completely wrong. Dec 12 '20 at 15:09
• @NorbertSchuch Which part. Dec 12 '20 at 15:11
• All except for the one you added after I posted my answer. (And the one you added after my comment above, changing "real-valued" to "positive and real-valued".) Dec 12 '20 at 15:17

This is wrong, just take $$c_1>0$$ and $$c_2<0$$.