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I understand how a transmon qubit is analogous to an LC circuit, but has an anharmonic energy spectrum due to the nonlinearity of the Josephson junction. I also understand how to write out the Hamiltonian and the significance of the contributing terms and find the energy eigenstates.

However, I am wondering if there is a good way to conceptualize what is happening in the circuit in a more physical sense, i.e., in the ground ($|0\rangle$) state, what are the cooper pairs doing versus what are the cooper pairs doing in the excited ($|1\rangle$) state? I am basically trying to understand what physically differentiates the ground and excited states.

epelaez
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taryn3194
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1 Answers1

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Basically, the current and voltage oscillate - they have uncertain values that would correspond to oscillation in the classical limit. In the ground state, the uncertainty is at or near the minimum allowed by the uncertainty principle. In the first excited state, the uncertainty is greater. That is, if you were to perform a measurement you would get a larger range of results.

Think of a pendulum swinging back and forth. Because of the uncertainty principle, it cannot be at rest but it can be as close as possible to a resting state. If energy is added, it starts to swing more and there is a wider range in the position of the pendulum.

Anna Naden
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