## Quantum Transport: Atom to Transistor - Questions & Answers

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**Lecture 4: Charging/Coulomb Blockade**, slide 6

**Q:**
Why is the broadening in G vs Vg plot equal to "gamma + kT + 2*Uo ". My question is about the 2*Uo part.
With your reasoning that the electronic level cannot raise its own energy by coulombic means it should end up being just Uo.

**A:**
This is a somewhat subtle point - slide 6 refers to the limit when
the broadening is large (far from the Coulomb blockade limit) - in
this limit since gamma is large it is probably hard to tell
experimentally whether the remainder is U0 or 2*U0 - so I am not sure
if the answer is known experimentally.

Conceptually I am inclined to vote for 2*U0 for the following reason: A large broadening means that one could view the entire broadened level, not as two levels, but as a collection of 2N closely spaced levels with only (1/N) of each wavefunction residing on the dot (N is a very large number, say a thousand or a million). Filling up the entire broadened level adds 2N electrons to the entire system (dot + contact), but only two to the dot.

You are arguing that since there are only two electrons on the dot, each should feel the charging due to 2-1 = 1 electron. I would argue that we are adding 2N electrons to the system and each electron should feel the charging due to (1/N) of (2*N - 1) electrons which is = 2 - (1/N) ~ 2.

As I said, this is somewhat subtle, hard to convey by email...

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- molecular electronics
- nanoelectronics