Technical question about Nikolic' Quantum Myths

[birulami]

I am referring to the http://arxiv.org/abs/quant-ph/0609163" discussed at length in other threads with a purely technical question to help me understand more of the paper.

On page 12, formula 26 defines the norm of a vector as

$$\langle\psi|\psi\rangle = \psi_1^*\psi_1 + \psi_2^*\psi_2 .$$

My question is: isn't there a square root missing to get a norm. When I look up my lecture notes on functional analysis, a norm on a Hilbert space is defined by

$$||f|| = \sqrt{(f,f)}$$

where $(\cdot,\cdot)$ is the scalar product. If the above formula 26 would nevertheless be correct, I would end up with a scalar product with a square in, which is not linear and therefore not a scalar product.

What am I confusing here?

Thanks,
Harald.

 

[lbrits]

Square root is indeed missing. Or he meant to write norm squared.

 

[Demystifier]

Or he meant to write norm squared.

Yes he did.

Even the the mass squared is sometimes called mass by physicists, especially relativists.

 

[birulami]

technical question about Nikolic' Quantum Myths (2nd round)

Thanks for the answers so far. Reading on in http://arxiv.org/abs/quant-ph/0609163" , I next stumble over formula 28. There I get

$$p_1=|\langle \phi_1|\psi\rangle|^2 = |\sqrt{1^*\psi_1 + 0\psi_2}|^2 = |\sqrt{\psi_1}|^2 = <br /> |\psi_1| = \sqrt{\psi_1^*\psi_1} .$$

This, however, contradicts the equation in the paragraph before formula 24, where it reads $p_1 = \psi_1^*\psi_1$.

My question is, whether formula 28 should rather start with $\sqrt{p_1}$ or whether actually it should read $p_1 = \sqrt{\psi_1^*\psi1}$ in the paragraph before 24?

Thanks,
Harald

 

[Demystifier]

$$|\langle \phi_1|\psi\rangle|^2 = |\sqrt{1^*\psi_1 + 0\psi_2}|^2$$

This is wrong. The correct statement is
$$|\langle \phi_1|\psi\rangle|^2 = |1^*\psi_1 + 0\psi_2|^2$$
To repeat, <a|a> denotes the norm SQUARED.

 

[birulami]

This is wrong. The correct statement is
$$|\langle \phi_1|\psi\rangle|^2 = |1^*\psi_1 + 0\psi_2|^2$$
To repeat, <a|a> denotes the norm SQUARED.

Arrrgh, this missing square of the norm got me all messed up. I should have read my own initial post.

Thanks for getting me back on the right track.
Harald.