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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:

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

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  • $\begingroup$ Welcome to Stack Exchange, and +1 for the question! $\endgroup$ Nov 29 '20 at 3:00
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$\langle a | b \rangle$ 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.

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  • $\begingroup$ 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} $. $\endgroup$
    – KAJ226
    Nov 29 '20 at 4:49
  • $\begingroup$ @KAJ226 good point. $\endgroup$ Nov 29 '20 at 5:02
  • $\begingroup$ This is completely wrong. $\endgroup$ Dec 12 '20 at 15:09
  • $\begingroup$ @NorbertSchuch Which part. $\endgroup$ Dec 12 '20 at 15:11
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    $\begingroup$ 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".) $\endgroup$ Dec 12 '20 at 15:17
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This is wrong, just take $c_1>0$ and $c_2<0$.

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