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:

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


2 Answers 2


$\langle a | b \rangle$ 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.

  • $\begingroup$ 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, 2020 at 4:49
  • $\begingroup$ @KAJ226 good point. $\endgroup$ Nov 29, 2020 at 5:02
  • $\begingroup$ This is completely wrong. $\endgroup$ Dec 12, 2020 at 15:09
  • 1
    $\begingroup$ 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, 2020 at 15:17
  • 1
    $\begingroup$ It was a joke. Should I explain it? (See e.g. math.stackexchange.com/questions/3171748/…) $\endgroup$ Dec 12, 2020 at 17:07

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


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.