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He is CORRECT. I use $\mathbf{B}=\left(0,0,B_{\perp}\right)$ and he use $\mathbf{B}=\left(0,0,-B_{\perp}\right)$. $B_{\perp}>0$.

Nov.28.2012

Basically I got mad with conventions.

1.Here is the link of the book (second edition):

http://books.google.fr/books/about/Quantum_Hall_Effects.html?id=p3JpcdbqBPoC

Here is another link for one of his review articles:

http://iopscience.iop.org/0034-4885/72/8/086502

2.I am not happy with the negative sign of the commutator $\left[X,Y\right]=-il_{B}^{2}$. Here is my calculation:

$\left[X,Y\right]=\left(\left[x,p_{x}\right]+\left[-p_{y},y\right]\right)/eB+\left[-P_{y},P_{x}\right]/e^{2}B^{2}=il_{B}^{2}$

3.In my calculation, I used some conventions different from Prof.Ezawa's book and article. Here is my convention:

$X:=x-P_{y}/eB\;;\; Y:=y+P_{x}/eB$

However, Prof.Ezawa use this convention:

$X:=x+P_{y}/eB\;;\; Y:=y-P_{x}/eB$

To be prepared for being driven mad, please compare them carefully.

4.I think he must be wrong somewhere, for example, in his book (2nd.ed), (10.2.5) and in his article (2.15)

$\left[P_{x},P_{y}\right]=i\hbar^{2}/l_{B}^{2}$

but you know we use minimal coupling $\mathbf{p}\rightarrow\mathbf{p}-\frac{q}{c}\mathbf{A}$ in this problem, $q=-e,e>0,{\mathbf{p}+\frac{e}{c}\mathbf{A}}$ for electrons, as Prof.Ezawa suggested in his book (10.2.3) and his article (2.12). Under this convention, I calculated $\left[P_{x},P_{y}\right]$ as following:

$\left[P_{x},P_{y}\right]=-i\hbar e\left(\left[\partial_{x},A_{y}\right]+\left[A_{x},\partial_{y}\right]\right)/c=-i\hbar eB/c=-i\hbar^{2}/l_{B}^{2}$

Oh my god here is a negative sign.

5.To summarize, I think if we take Prof.Ezawa's convention and apply his result for $\left[P_{x},P_{y}\right]=i\hbar^{2}/l_{B}^{2}$ during the calculation of $\left[X,Y\right]$, we will get his result. But his result for $\left[P_{x},P_{y}\right]=i\hbar^{2}/l_{B}^{2}$ seems not correct.

6.Someone save my day...

No!

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I have found another check for the expression (2.14) in Prof.Ezawa's article. If we solve the cyclotron motion problem for a classical particle with negative charge, we will find $\left(R_{x},R_{y}\right)=\left(P_{y},-P_{x}\right)/m\omega_{c}$, which indicates that the sign used in these expressions are inappropriate. –  Yunlong Lian Nov 23 '12 at 11:37
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1 Answer 1

up vote 2 down vote accepted

I have done this calculation some time ago. My convention was:

$$ X = x - \frac{P_y}{m \omega_c}\quad Y = y + \frac{P_x}{m \omega_c} $$ and $$P_i = p_i +\frac{e}{c} A_i$$

And the magnetic field is $B = \nabla \wedge A = B_z \hat{z}$. Note that in particular: $X = x-\frac{1}{m\omega_c}(p_i + \frac{e}{c}A_i)$. My notes say that this gives:

$$ [X, Y] = i l_B^2\qquad \text{and}\qquad [P_x,P_y] = -\frac{i}{l_B^2}$$

If you now take his convention, you essentially flip the magnetic field, $\vec{B} \rightarrow -\vec{B}$. This replaces:

$$ X = x + \frac{P_y}{m \omega_c}\quad Y = y - \frac{P_x}{m \omega_c} $$

but you still have $P_i = p_i +\frac{e}{c} A_i$ -- that stays the same. Therefore we have: $X = x+\frac{1}{m\omega_c}(p_i + \frac{e}{c}A_i)$ (!!!! compare this to the other convention), and to compute the commutator we get:

$$\begin{align}[X,Y] &= \left[x+\frac{1}{m\omega_c}(p_y + \frac{e}{c}A_y),y-\frac{1}{m\omega_c}(p_x + \frac{e}{c}A_x)\right] \\ &= (-[x,p_x] + [p_y,y])/m\omega_c + \frac{e}{c (m\omega_c)^2}(-[p_y,A_x]+[A_y,p_x]) \end{align}$$ Now, $[x,p_x] = i$, as always. The other commutator depends on the orientation of the magnetic field: $$-[p_y,A_x]+[A_y,p_x] = i ([\nabla_y, A_x] - [\nabla_x, A_y]) = -i(\nabla\wedge A)_z = iB_z$$, and so you get

$$[X,Y] = -\frac{2i}{m\omega_c} + \frac{e}{c (m\omega_c)^2} iB_z = -i l_B^2$$

Long story short: your derivation of $[X,Y]$ does not apply to his conventions. Your conve

Final note: If you switch conventions, you essentially replace $B_z \rightarrow -B_z$, so the magnetic length and cyclotron frequency also switch sign, $l_B^2\rightarrow -l_B^2$ and $\omega_c \rightarrow - \omega_c$. So you see that both commutators (involving $X$ and $Y$ and $P_x$ and $P_y$) pick up a minus sign, because they both involve $l_B^2$.

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To further save my N-page calculation based on Ezawa, I came up with another idea - not to flip the B field, but to do a spatial reflection, or change the handness of the xyz-coord frame. Hope you know it :D –  Yunlong Lian Nov 24 '12 at 7:37
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