Physics Asked by mevis on January 10, 2021
I have just started studying about quantum computers (hardware side) and I am really confused about what is a quantum anharmonic oscillator. I have read somewhere that a qubit is the physical realization of a quantum anharmonic oscillator. How is it so?
Harmonic quantum oscillator has same displacement between each consecutive energy levels, i.e. : $$ E_{n+1} - E_n = hbar,omega $$ In anharmonic quantum oscillator energy difference between next levels is not a constant and usual follows some non-linear form. Like in for example Morse_potential which helps to define molecule vibrational energy levels. Energy difference between consecutive levels in that case is : $$ E_{n+1}-E_{n}=hbar,omega -alpha(n+1)~hbar^{2} ,omega^{2} $$ So it's not constant, i.e. depends on exact energy level where you are starting from and is non-linear too,- follows a polynomial form of $a,omega-b,omega^2$. That's why it is anharmonic quantum oscillator.
Sometimes picture is worth a thousand words, so here it is - a graph with harmonic and Morse anharmonic oscillators depicted :
Answered by Agnius Vasiliauskas on January 10, 2021
Anharmonic oscillator is an oscillator (i.e. a physical system that exhibits a periodic motion), which is not described by a linear differential equation (i.e. not harmonic). For example, a system (classical or quantum) with Hamiltonian $$ H = frac{p^2}{2m} + lambda x^4 $$ is clearly an oscilator, but clearly not a harmonic one.
Since in quantum mechanics everything can be considered as period oscillations, one can extend the term anharmonic oscillator to pretty much everything. Still, qubits (i.e. spin-1/2 systems) are special in quantum mechanics in that they are elementary systems alongside the harmonic oscillator - the two can be consider as giving rise respectively to Bose and Fermi statistics, and their equation of motion are very similar when expressed in terms of creation/annihilation operators. So calling qubit an anharmonic oscillator is loaded with meaning... but likely not very essential for whatever is discussed. I think the phrase cited in the question is a colorful abuse of language.
Answered by Vadim on January 10, 2021
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