# Quantum-mechanical Schwarz inequality: Proving $\langle \psi \mid \phi \rangle^* \langle \psi \mid \phi \rangle \ge 0$ for 1D case.

I am currently studying the textbook Mathematical methods of quantum optics by Ravinder R. Puri. When discussing the postulates of quantum mechanics, the author introduces the quantum-mechanical version of the Schwarz inequality as follows:

An important consequence of the axioms defining the scalar product is the Schwarz inequality $$\langle \phi \mid \phi \rangle \langle \psi \mid \psi \rangle \ge \langle \phi \mid \psi \rangle \langle \psi \mid \phi \rangle, \tag{1.5}$$ where the equality holds if and only if the two vectors in question are linearly dependent i.e. if $$\mid \psi \rangle = \mu \mid \phi \rangle, \tag{1.6}$$ $$\mu$$ being a complex number. In order to establish this, show that the minimum value of $$\langle \Psi(\mu) \mid \Psi(\mu) \rangle$$, where $$\mid \Psi \rangle \ = \ \mid \psi \rangle - \mu \mid \phi \rangle$$, as a function of $$\mu$$ is $$\langle \psi \mid \psi \rangle - |\langle \psi \mid \phi \rangle |^2 / \langle \phi \mid \phi \rangle$$. The requirement that this value, due to axiom 3 of the scalar product, be positive leads to the Schwarz inequality in (1.5). Also, according to the axiom 4 above, $$\langle \Psi(\mu) \mid \Psi(\mu) \rangle = 0$$ iff $$\mid \Psi(\mu) \rangle - 0$$ i.e. iff (1.6) holds. it may be verified easily that (1.5) then holds with equality. In a similar way we can derive the generalized Schwarz inequality $$\det(\langle \psi_\mu \mid \psi_\nu \rangle ) \ge 0, \tag{1.7}$$ where $$\det(\langle \psi_\mu \mid \psi_\nu \rangle )$$ is the determinant of the matrix constituted by the elements $$\det\langle \psi_\mu \mid \psi_\nu \rangle$$, $$\mu, \nu = 1, \dots, n$$. Invoking the fact that the determinant of a matrix is zero if its rows (or columns) are linearly dependent, it follows that the equality in (1.7) holds iff $$\mid \psi_\mu \rangle$$ are linearly dependent.

Axiom 3 is as follows:

$$\langle \psi \mid \psi \rangle > 0$$

Axiom 4 is as follows:

$$\langle \psi \mid \psi \rangle = 0 \ \text{if and only if \mid \psi \rangle = 0}$$

My goal is to prove this case of the Schwarz inequality for myself.

In researching this problem, I found this webpage. The author claims that

$$| (\psi, \phi) |^2 \le (\psi, \psi)(\phi, \phi).$$

Trying to connect this with (1.5), I get

$$\langle \phi \mid \phi \rangle \langle \psi \mid \psi \rangle \ge \langle \phi \mid \psi \rangle \langle \psi \mid \phi \rangle = \langle \psi \mid \phi \rangle^* \langle \psi \mid \phi \rangle \ge 0,$$

where $$*$$ is the complex conjugate.

So then I wonder: Is it true that $$\langle \psi \mid \phi \rangle^* \langle \psi \mid \phi \rangle \ge 0$$? It isn't clear to me that this is true. So my objective now is to prove that this is true. To simplify things, I will first try to prove that it is true for the 1-dimensional case:

$$\psi$$ and $$\phi$$ are complex numbers, right? So let $$\psi = x_1 + i y_1$$ and $$\phi = x_2 + i y_2$$.

\begin{align} \langle \psi \mid \phi \rangle &= (x_1 - i y_1) \cdot (x_2 + i y_2) \\ &= x_1 x_2 - i x_1 y_2 - i x_2 y_1 + y_1 y_2 \\ &= (x_1 x_2 + y_1 y_2) - i(x_1 y_2 + x_2 y_1) \end{align}

$$\langle \psi \mid \phi \rangle^* = (x_1 x_2 + y_1 y_2) + i (x_1 y_2 + x_2 y_1)$$

\begin{align} \langle \psi \mid \phi \rangle^* \langle \psi \mid \phi \rangle &= [(x_1 x_2 + y_1 y_2) + i (x_1 y_2 + x_2 y_1)] \cdot [(x_1 x_2 + y_1 y_2) - i(x_1 y_2 + x_2 y_1)] \\ &= (x_1 x_2 + y_1 y_2) (x_1 x_2 + y_1 y_2) + i (x_1 x_2 + y_1 y_2)(x_1 y_2 + x_2 y_1) - i (x_1 y_2 + x_2 y_1)(x_1 x_2 + y_1 y_2) - (x_2 y_1 + y_1 y_2)(x_1 y_2 + x_2 y_1) \\ &= (x_1^2 x_2^2 + 2 x_1 x_2 y_1 y_2 + y_1^2 y_2^2) + i (x_1^2 x_2 y_2 + x_1 x_2^2 y_1 + x_1 y_1 y_2^2 + x_2 y_2 y_1^2) - i (x_1^2 x_2 y_2 + x_1 y_1 y_2^2 + x_2^2 x_1 y_1 + x_2 y_1^2 y_2) - (x_1 x_2 y_1 y_2 + x_2^2 y_1^2 + y_2^2 y_1 x_1 + y_1^2 y_2 x_2) \\ &= (x_1^2 x_2^2 + 2 x_1 x_2 y_1 y_2 + y_1^2 y_2^2) - (x_1 x_2 y_1 y_2 + x_2^2 y_1^2 + y_2^2 y_1 x_1 + y_1^2 y_2 x_2) \end{align}

No, $$|\phi\rangle$$ and $$|\psi\rangle$$ are not complex numbers. They're element in a Hilbert space and $$\langle \phi |\psi\rangle$$ is their inner product. So it's $$\langle \phi |\psi\rangle$$ which is a complex number. And then $$\langle \phi |\psi\rangle^\star\langle \phi |\psi\rangle=|\langle \phi |\psi\rangle|^2$$ is the norm of a complex number so it's a real non negative value.