# Why pesistent current in a normal metal ring is a surprise?

1. Sep 29, 2010

### wdlang

i cannot understand why persistent current in a normal metal ring threaded by a magnetic field is a surprise.

the hamiltonian is

$$H=\frac{1}{2I}\left(-i \frac{\partial}{\partial \theta}-A\right)^2$$

and the eigenstates are

$$\phi_m(\theta)=\frac{1}{\sqrt{2\pi}} e^{i m \theta}$$

with eigenvalues

$$E_m=\frac{1}{2I}(m-A)^2$$.

It is ready to see that generally every eigenstate carries a current, a persistent one.

so why people think it is a surprise?

2. Sep 29, 2010

### ZapperZ

Staff Emeritus
Try doing it, and see if what you think you understand matches reality!

Zz.

3. Sep 29, 2010

### wdlang

of course, in reality, the situation is more complicated, e.g., the potential is not uniform, there is decoherence.

but i think the basic idea is just too simple.

4. Sep 29, 2010

### ZapperZ

Staff Emeritus
We shouldn't be surprised with quantum tunneling and quantum entanglement either, because the basic ideas are also very simple. Yet, we still do!

Zz.

5. Sep 30, 2010

### DrDu

In a many particle state, the contributions of the states with different m nearly cancel and there is only a very tiny fraction of that effect that survives. It forms the basis of diamagnetism.