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


Basically I got mad with conventions.

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


Here is another link for one of his review articles:


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


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)


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...


  • $\begingroup$ 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. $\endgroup$ – Yunlong Lian Nov 23 '12 at 11:37

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 $l_B^2$.

  • $\begingroup$ 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 $\endgroup$ – Yunlong Lian Nov 24 '12 at 7:37

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