Deriving Hamiltonian in Landau Gauge Using Symmetric Gauge Transformation

[shinobi20]

Homework Statement

Define n=(x + iy)/(2)^½L and ñ=(x - iy)/(2)^½L.
Also, ∂_n = L(∂_x - i ∂_y)/(2)^½ and ∂_ñ = L(∂_x + i ∂_y)/(2)^½.
with ∂_n=∂/∂_n, ∂_x=∂/∂_x, ∂_y=∂/∂_y, and L being the magnetic length.
a=(1/2)ñ+∂_n and a^†=(1/2)n -∂_ñ
a and a^† are the lowering and raising operators of quantum mechanics.

Show that H=ħω_c(a^†a + ½)

Homework Equations

L=ħc/eB, ω_c=eB/mc (cyclotron frequency), e for the charge of the electron
H = P_x²/2m + ( P_y² + eBx/c )²/2m

The Attempt at a Solution

I have tried to find x,y,∂_x,∂_y in terms of n,ñ,∂_n,∂_ñ. But I ended up getting only some if the right terms to come out but not all, is my first step wrong? Any suggestions?

 

[blue_leaf77]

H = Px2/2m + ( Py2 + eBx/c )2/2m

Should the exponent "2" of $P_y$ be there?

 

[shinobi20]

Should the exponent "2" of $P_y$ be there?

Sorry, it was a typo. Do you have any suggestions?

 

[blue_leaf77]

You should post your initial attempt before we can discuss further. In particular, how the old variables look like in terms of the new ones?

 

[shinobi20]

You should post your initial attempt before we can discuss further. In particular, how the old variables look like in terms of the new ones?

This is what I've done so far. My problem is that everything is there except for the ½. I wrote ∂ for ∂n and ∂(bar) for ∂ñ.

 

[blue_leaf77]

According to this link https://en.wikipedia.org/wiki/Landau_quantization, the Gauge you should be using is the symmetric gauge and hence the original Hamiltonian should be different than that you are using. For instance, in Landau gauge, the operator ${y}$ is not present.

 

[shinobi20]

According to this link https://en.wikipedia.org/wiki/Landau_quantization, the Gauge you should be using is the symmetric gauge and hence the original Hamiltonian should be different than that you are using. For instance, in Landau gauge, the operator ${y}$ is not present.

Why can't I show it using the Landau gauge? The choice is just for simplification of computation right?

 

[blue_leaf77]

Why can't I show it using the Landau gauge? The choice is just for simplification of computation right?

$x$ and $y$ appear symmetrically in the gauge transformation, but they do not in the original Hamiltonian.

 

[shinobi20]

$x$ and $y$ appears symmetrically in the gauge transformation, but they do not in the original Hamiltonian.

Oh I see that, then I'll try it again using the symmetric gauge. Thanks!