Quantum Computing Asked by Oli on January 6, 2021

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:

$$2|c_1|^2|c_2|^2 leq c_1c_2^* + c_1^*c_2tag{2}.$$

$langle a | b rangle$ 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:

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 $|c_2|^2>0.$

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

Correct answer by user1271772 on January 6, 2021

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

Answered by Norbert Schuch on January 6, 2021

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