Hamiltonian of a quantum harmonic oscillator On page 286-287 of Nielsen Chuang's Quantum Information and Quantum Computation (10th edition) book, the Hamiltonian for a quantum harmonic oscillator is approximated as $H=a^\dagger a.$ What are the assumptions involved in such an approximation and why is this approximation needed?
 A: I have not looked at the book , however the sense in which it is an approximation is that it is neglecting the constant term
The Hamiltonian of a SHO is ,
$H= (a^{\dagger}a + 1/2)\hbar\omega$
This means that the ground state energy of the SHO is $1/2\hbar\omega$. This is what is being neglected since it is only a constant.
A: I have a lot of experience with this particular book, but unfortunately it is in my office right now so I can't reference the exact page you're on.
In QIQC, and this book in particular, you're going to be doing a lot of manipulating the Hamiltonian of the QSHO using commutators.  Since constants always commute, the constant term of the QSHO falls out of every one of these operations and would be extremely burdensome to include.  You'll see what I mean when you start doing problems with:
$$[H,a]$$
$$[H,a^2]$$
$$[H,\left(a^\dagger\right)^2\left(a\right)^2]$$
$$...$$
To clarify one thing: this is not an approximation but more of a shorthand for these types of algebraic problems.  It is very important to keep the $1/2\hbar\omega$ for some problems in the book, like those involving expectation values!
