Inequality in overlap of quantum states

Question

For quantum states $vertpsi_1rangle, vertpsi_2rangle, vertphirangle$, is it true that:
$$tag{1}langle phivertpsi_1ranglelanglepsi_1vertphiranglelangle phivertpsi_2ranglelanglepsi_2vertphirangle + langle phivertpsi_2ranglelanglepsi_2vertphiranglelangle phivertpsi_1ranglelanglepsi_1vertphirangle leq langle phivertpsi_1ranglelanglepsi_2vertphirangle + langle phivertpsi_2ranglelanglepsi_1vertphirangle.$$
My argument is that each number $c_i = langlephivertpsi_irangle$ is a complex number with modulus smaller than 1 since it is the square root of a probability. So we have to show:
$$2|c_1|^2|c_2|^2 leq c_1c_2^* + c_1^*c_2tag{2}.$$

user1271772 · Accepted Answer

$langle a | b rangle$ 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 phivertpsi_1ranglelanglepsi_1vertphiranglelangle phivertpsi_2ranglelanglepsi_2vertphirangle + langle phivertpsi_2ranglelanglepsi_2vertphiranglelangle phivertpsi_1ranglelanglepsi_1vertphirangle  &= 2langle phivertpsi_2ranglelanglepsi_2vertphiranglelangle phivertpsi_1ranglelanglepsi_1vertphirangle 
&=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_2tag{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:
$$
c_1c_2^* + c_1^*c_2 = c_1(c_2 - c_2) = 0tag{5},
$$
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.

Norbert Schuch · Answer

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

Inequality in overlap of quantum states

2 Answers

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

$c_1$ and $c_2$ have different signs:

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