- #1
Nezva
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Calculate [Lz,L+]
By defintion ladder operators are:
L+=Lx+iLy
L-=Lx-iLy
Important Relations:
LxLy = i[tex]\hbar[/tex]Lz, LyLz = i[tex]\hbar[/tex]Lx, LzLx = i[tex]\hbar[/tex]Ly
Lx = ypz - zpy, Ly = xpz - zpx, Lz = xpy - ypx
To start solving;
[Lz,L+]
Lz - (Lx + iLy) = 0
Multiply through by [tex]\hbar[/tex]:
[tex]\hbar[/tex]Lz - [tex]\hbar[/tex]Lx + i[tex]\hbar[/tex]Ly
The i[tex]\hbar[/tex]Ly is equal to LzLx. From this point
I've tried varying approaches in attempt to cancel variable out, but have failed. I have a feeling this problem can be solved easier. Should I try to use spherical coordinates instead of Cartesian? From trying to figure this out I have stumbled upon the answer but I would like to know how to produce the answer.
By defintion ladder operators are:
L+=Lx+iLy
L-=Lx-iLy
Important Relations:
LxLy = i[tex]\hbar[/tex]Lz, LyLz = i[tex]\hbar[/tex]Lx, LzLx = i[tex]\hbar[/tex]Ly
Lx = ypz - zpy, Ly = xpz - zpx, Lz = xpy - ypx
To start solving;
[Lz,L+]
Lz - (Lx + iLy) = 0
Multiply through by [tex]\hbar[/tex]:
[tex]\hbar[/tex]Lz - [tex]\hbar[/tex]Lx + i[tex]\hbar[/tex]Ly
The i[tex]\hbar[/tex]Ly is equal to LzLx. From this point
I've tried varying approaches in attempt to cancel variable out, but have failed. I have a feeling this problem can be solved easier. Should I try to use spherical coordinates instead of Cartesian? From trying to figure this out I have stumbled upon the answer but I would like to know how to produce the answer.
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