# Hamiltonian of a quantum circuit including a diode?

• The LC circuit has a Hamiltonian: $$\hat{H}={E_L\over2} \hat{\varphi}^2 + 4E_C \hat{n}^2$$
where $$\hat{\varphi}$$ is the magnetic flux and $$\hat{n}$$ is the number of charge. • What is the Hamiltonian for the following circuit? I have never seen a Hamiltonian for a circuit with a diode, and I doubt that it exists or used - for the reasons I describe below. However, from purely academic perspective it is an interesting question to ponder.

• Unlike inductance and capacitance, which can be characterized by linear response and therefore described by quadratic Hamiltonians, a diode is a non-linear element. Non-linearity is alien to the structure of quantum mechanics, although one could think about including it via a potential term.
• The Hamiltonian for LC circuit is actually not written for a real LC circuit with a macroscopic capacitance and inductance, but for a microscopic object. Any conducting object, however small, has some capacitance and inductance, which at microscales have to be treated quantum mechanically. On the other hand, a diode is an essentially macrosopic, human-engineered, device - it never happens at micro-/nano-scale.

One can reason from the point of view of the physical content of variables $$\hat{n}$$ and $$\hat{\varphi}$$ - the former is the operator of charge on the capacitor, whereas the latter is its conjugate (typically bias on the inductance), related by a commutation relation $$[\hat{\varphi},\hat{n}]=i$$ (the coefficients $$\pm1$$ or $$\hbar$$ in this commutation relation are a matter of personal choice). Thus, the equation for motion for the charge, which we can identify with the current, is $$\hat{I}=\dot{n} = [\hat{n},\hat{H}] = -\frac{E_L}{\hbar}\hat{\varphi}.$$ (Writing likewise the EOM for $$\varphi$$ we obtain the oscillator equation of the LC circuit.)

If we now replace the inductance term by a potential $$V(\varphi)$$, we have $$\hat{H} = 4E_C\hat{n}^2 + V(\hat{\varphi}),\\ \hat{I}=\dot{n} = [\hat{n},\hat{H}] =-\frac{1}{\hbar}\frac{\partial V(\hat{\varphi})}{\partial \hat{\varphi}}$$ We can the further identify $$\hat{I}$$ with the current given by the Shokley diode equation and $$\hat{\varphi}$$ with the bias in this equation: $$\hat{I} = I_S\left(e^{\frac{\hat{\varphi}}{nV_T}} -1\right)$$ and integrating this equation $$\hat{\varphi}$$ would give us the potential $$V(\hat{\varphi})$$.

• - Are you sure $\hat{\varphi}$ is the voltage across the diode ? - Since $E_C={e^2\over 2C}$, it should be $U={Q\over C}={ 4E_C \over e } \hat{n}$ instead of $\hat{\varphi}$ – ruiqi ding Sep 18 at 14:47
• I might be mistaken, but I do not agree with your reasoning either. The coherent approach would be to write a lagrangian for the classical circuit, identify momentum operator, pass to the Hamiltonian formalism, and then quantize by imposing canonical commutation relations. – Vadim Sep 18 at 15:16
• As far as I remember, there is some arbitrariness in how one defines lagrangian and hence what one chooses as position and momentum operators, since what we really know for sure is the equation of motion. – Vadim Sep 18 at 15:19

According to one of my professors:

I am not 100% sure, but I guess for the ideal diode it should be something like $$4 E_C \hat{n}^2 H(\hat{n})$$ where $$H(\hat{n})$$ is the Heaviside step function.

If the charge is negative, the capacitor is immediately discharged and the energy of the system is $$0$$. If the charge is positive, the energy is just the energy of the capacitor.

I think he is absolutely right.

Now I am wondering what the eigenvalues and eigenstates are for this Hamiltonian.

Oh silly me, I just realized that what I have here is like a free particle ($$V(\hat{\varphi})=0$$), except for the Heaviside part. And I quote Griffith (under eqn 2.82): there is no such thing as a free particle with a definite energy

• Thinking of it, this could make some sense, if the diode is in place of capacitor rather than inductance. – Vadim Sep 18 at 16:58