(Algebra) Quantum Theory - Cauchy-Schwartz inequality proof

Join the discussion
Ask a follow-up here, or get your own question answered by working scientists, mathematicians and engineers — people, not an autocomplete.
Real named experts · corrections over time · the nuance an AI answer skips
1 reply · 2K views
FatPhysicsBoy
Messages
62
Reaction score
0

Homework Statement


Given two arbitrary vectors [itex]|\phi_{1}\rangle[/itex] and [itex]|\phi_{2}\rangle[/itex] belonging to the inner product space [itex]\mathcal{H}[/itex], the Cauchy-Schwartz inequality states that:

[itex]|\langle\phi_{1}|\phi_{2}\rangle|^{2} \leq \langle\phi_{1}|\phi_{1}\rangle \langle\phi_{2}|\phi_{2}\rangle[/itex].

Consider [itex]|\Psi\rangle = |\phi_{1}\rangle + \lambda|\phi_{2}\rangle[/itex]

where [itex]\lambda[/itex] is a complex number that can be written as [itex]\lambda = a + ib[/itex].

a) Write an expression for [itex]\langle\Psi|\Psi\rangle \geq 0[/itex] as a function of [itex]\lambda[/itex] then rewrite as a function of a and b ([itex]f(a,b)[/itex]).

b) Show that the value of [itex]\lambda[/itex] that minimises [itex]\langle\Psi|\Psi\rangle[/itex] is:

[itex]\lambda_{min} = -\frac{\langle\phi_{2}|\phi_{1}\rangle}{\langle\phi_{2}|\phi_{2}\rangle}[/itex].

Hint: Compute the derivatives of [itex]f(a,b)[/itex] wrt a and b, solve these to get [itex]a_{min}[/itex] and [itex]b_{min}[/itex] and then compute [itex]\lambda_{min}[/itex].

Homework Equations



N/A

The Attempt at a Solution



I get [itex]\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)[/itex]. However I can only show:

[itex]\lambda_{min} = -\frac{\langle\phi_{1}|\phi_{2}\rangle}{\langle\phi_{2}|\phi_{2}\rangle}[/itex],

by combining Re and I am parts in [itex]f(a,b)[/itex] as follows (and finding the relevant derivatives etc.):

[itex]\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[/itex].

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):

[itex]\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[/itex]

Thank you
 
Physics news on Phys.org
Anyone? Parts 1) and 2) can pretty much be ignored they just provide context for the problem.. I think ultimately it's just a complex number/conjugation question which I haven't understood properly.