- #1
FatPhysicsBoy
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Homework Statement
Given two arbitrary vectors |\phi_{1}\rangle and |\phi_{2}\rangle belonging to the inner product space \mathcal{H}, the Cauchy-Schwartz inequality states that:
|\langle\phi_{1}|\phi_{2}\rangle|^{2} \leq \langle\phi_{1}|\phi_{1}\rangle \langle\phi_{2}|\phi_{2}\rangle.
Consider |\Psi\rangle = |\phi_{1}\rangle + \lambda|\phi_{2}\rangle
where \lambda is a complex number that can be written as \lambda = a + ib.
a) Write an expression for \langle\Psi|\Psi\rangle \geq 0 as a function of \lambda then rewrite as a function of a and b (f(a,b)).
b) Show that the value of \lambda that minimises \langle\Psi|\Psi\rangle is:
\lambda_{min} = -\frac{\langle\phi_{2}|\phi_{1}\rangle}{\langle\phi_{2}|\phi_{2}\rangle}.
Hint: Compute the derivatives of f(a,b) wrt a and b, solve these to get a_{min} and b_{min} and then compute \lambda_{min}.
Homework Equations
N/A
The Attempt at a Solution
I get \langle\Psi|\Psi\rangle = \langle\phi_{1}|\phi_{1}\rangle + a(\langle\phi_{1}|\phi_{2}\rangle + \langle\phi_{2}|\phi_{1}\rangle) + ib(\langle\phi_{1}|\phi_{2}\rangle - \langle\phi_{2}\phi_{1}\rangle) + (a^{2} + b^{2})\langle\phi_{2}|\phi_{2}\rangle = f(a,b). However I can only show:
\lambda_{min} = -\frac{\langle\phi_{1}|\phi_{2}\rangle}{\langle\phi_{2}|\phi_{2}\rangle},
by combining Re and I am parts in f(a,b) as follows (and finding the relevant derivatives etc.):
\langle\Psi|\Psi\rangle = \langle\phi_{1}|\phi_{1}\rangle + 2a\textrm{Re}(\langle\phi_{1}|\phi_{2}\rangle) + 2b\textrm{Im}(\langle\phi_{1}|\phi_{2}\rangle) + (a^{2} + b^{2})\langle\phi_{2}|\phi_{2}\rangle.
This is the only way I understand how to do it, however, in the solutions for this problem the collection of Re and I am parts is done as follows which I don't understand (in particular the imaginary part):
\langle\Psi|\Psi\rangle = \langle\phi_{1}|\phi_{1}\rangle + 2a\textrm{Re}(\langle\phi_{2}|\phi_{1}\rangle) + 2b\textrm{Im}(\langle\phi_{2}|\phi_{1}\rangle) + (a^{2} + b^{2})\langle\phi_{2}|\phi_{2}\rangle
Thank you
