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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?

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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.

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  • $\begingroup$ thank you! I took this phrase from a video lecture, also "a classical LC circuit is harmonic oscillator, but if we replace the inductor with a josephson junction the LC circuit becomes anharmonic oscillator (a non linear circuit) , which is basically a superconducting charge qubit." am I right? Please guide me $\endgroup$ – maryam malik Sep 16 at 13:24
  • $\begingroup$ I think in this case the Hamiltonian is something like: $H = \frac{p^2}{2m}+V_0\sin(x)$, i.e. it is literally a non-linear oscillator, where oscillations occur near the minima of the potential, where whereas qubit has only two states. So I think there is more to superconducting qubit than just being a Josephson junction. But I don't know superconductivity well, so I may be mistaken. $\endgroup$ – Vadim Sep 16 at 13:30
  • $\begingroup$ okay. Thank you somuch. $\endgroup$ – maryam malik Sep 16 at 13:41
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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 :

enter image description here

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  • $\begingroup$ Thank you I've understood the anharmonic oscillator now, but do you know the relation between anharmonic oscillator and a superconducting charge qubit? $\endgroup$ – maryam malik Sep 16 at 13:25

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