Orbital angular momentum problem

In summary: Begin by calculating the commutator [L^2,L_x]...use the "go advanced" and "preview post" buttons to make sure whatever you write is actually readable, before submitting your post.[L^2,L_x]\varphi=(L_y*i\hbar*L_z)+(i*\hbar*L_z)(L_y)+L_z(i*\hbar*L_y)+(i*\hbar*L_y)L_z[L^2,L_y]\varphi= L_x(i*\hbar*L_z)+(i*\
  • #1
noblegas
268
0

Homework Statement



Consider a particle that moves in three dimensions with wave function [tex] \varphi[/tex] . Use operator methods to show that if [tex]\varphi[/tex] has total angular momentm quantum number l=0 , then [tex]\varphi[/tex] satifies

[tex] L\varphi=0[/tex]

for all three components [tex]L_\alpha[/tex] of the total angular momentum L

Homework Equations



[tex] [L^2,L_\alpha][/tex]?

[tex] L^2=L_x^2+L_y^2+L_z^2 [/tex]?

[tex] L^2=\hbar*l(l+1)[/tex] , l=0,1/2,1,3/2,... ?

[tex] L_z=m\hbar[/tex] , m=-l, -l+1,...,l-1,l. ?

The Attempt at a Solution



[tex] [L^2,L_\alpha]=[L_x^2+L_y^2+L_z^2,L_\alpha]=[L_x^2,L_\alpha]+[L_y^2,L_\alpha]+[L_z^2,L_\alpha][/tex]. [tex][AB,C]=A[B,C]+[A,C]B[/tex]; Therefore,[tex] [L_x^2,L_\alpha]+[L_y^2,L_\alpha]+[L_z^2,L_\alpha]=L_x[L_x,L_\alpha]+[L_x,L_\alpha]L_x+L_y[L_y,L_\alpha]+[L_y,L_\alpha]L_y+L_z[L_z,L_\alpha]+[L_z,L_\alpha]L_z[/tex]. Now what?
 
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  • #2
People having a hard time reading the problem?
 
  • #3
Do you not already know [itex][L^2,L_x][/itex], [itex][L^2,L_y][/itex] and [itex][L^2,L_z][/itex]?

If so, just calculate [itex][L^2,L_x]\varphi[/itex], [itex][L^2,L_y]\varphi[/itex] and [itex][L^2,L_z]\varphi[/itex] and use the fact that [itex]L^2\varphi=l(l+1)\varphi=0[/itex] for [itex]l=0[/itex].
 
  • #4
I don't understand gabbagabbahey's method. I would (using bra-ket notation) define three states [itex]|\psi_i\rangle = L_i|\phi\rangle[/itex], [itex]i=x,y,z[/itex], and then compute [itex]\sum_{i}\langle\psi_i|\psi_i\rangle[/itex].
 
  • #5
gabbagabbahey said:
Do you not already know [itex][L^2,L_x][/itex], [itex][L^2,L_y][/itex] and [itex][L^2,L_z][/itex]?

If so, just calculate [itex][L^2,L_x]\varphi[/itex], [itex][L^2,L_y]\varphi[/itex] and [itex][L^2,L_z]\varphi[/itex] and use the fact that [itex]L^2\varphi=l(l+1)\varphi=0[/itex] for [itex]l=0[/itex].

Yes I calculated those commutators [tex][L^2,L_x]\varphi=(L_y*i\hbar*L_z)+(i*\hbar*L_z)(L_y)+L_z(i*\hbar*L_y)+(i*\hbar*L_y)L_z,[L^2,L_y]= L_x(i*\hbar*L_z)+(i*\hbar*L_z)L_y+L_z(i*\hbar*L_x)+(i*\hbar*L_x) +(i*\hbar*L_x)L_z, [L^2,L_z]=L_x(i*\hbar*L_y)+(i*\hbar*L_y)(L_x)+L_y(i*\hbar*L_x)+(i*\hbar*L_x)L_y [/tex], The comutators simply expand more and do not simplify . at [tex] l=0, l(l+1)\varphi=0[/tex]

Please look at my latex for the expression [tex][L^2,L_z][/tex] PF is not completely showing the solution to this expression for some reason. Now is it readable
 
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  • #6
What you've just written is completely unreadable...How do you expect others to be able to help you, if you don't take the time to make sure they can read what you are posting?

Begin by calculating the commutator [itex][L^2,L_x][/itex]...use the "go advanced" and "preview post" buttons to make sure whatever you write is actually readable, before submitting your post.
 
  • #7
[tex]
[L^2,L_x]\varphi=(L_y*i\hbar*L_z)+(i*\hbar*L_z)(L_y)+L_z(i* \hbar*L_y)+(i*\hbar*L_y)L_z,[L^2,L_y]= L_x(i*\hbar*L_z)+(i*\hbar*L_z)L_y+L_z(i*\hbar*L_x) +(i*\hbar*L_x) +(i*\hbar*L_x)L_z, [L^2,L_z]=L_x(i*\hbar*L_y)+(i*\hbar*L_y)(L_x)+L_y(i*\hbar*L _x)+(i*\hbar*L_x)L_y
[/tex] Now is it readable? [tex][L^2,L_z]=L_x(i*\hbar*L_y)+(i*\hbar*L_y)(L_x)+L_y(i*\hbar*L _x)+(i*\hbar*L_x)L_y[/tex]
 
  • #8
I still can't make heads or tails of what you've written...When you read a textbook or article, or even someone else's posts here...do they lay things out in a jumbled mess like that?
 
  • #9
gabbagabbahey said:
I still can't make heads or tails of what you've written...When you read a textbook or article, or even someone else's posts here...do they lay things out in a jumbled mess like that?

I will separate my commutators into separate latex code and aligned the commutators vertically. Maybe it should then be easier for you to read:

[tex][L^2,L_x]\varphi=(L_y*i\hbar*L_z)+(i*\hbar*L_z)(L_y)+L_z(i* \hbar*L_y)+(i*\hbar*L_y)L_z[/tex]
[tex][L^2,L_y]\varphi= L_x(i*\hbar*L_z)+(i*\hbar*L_z)L_y+L_z(i*\hbar*L_x) +(i*\hbar*L_x) +(i*\hbar*L_x)L_z[/tex]
[tex][L^2,L_z]\varphi=L_x(i*\hbar*L_y)+(i*\hbar*L_y)(L_x)+L_y(i*\hbar*L _x)+(i*\hbar*L_x)L_y[/tex],
at[tex]l=0, l(l+1)\varphi=0[/tex]
 
  • #10
That's a little easier to read (Although I really wish you'd stop using that ugly [itex]*[/itex] symbol for multiplication). Why do you have a [itex]\varphi[/itex] on the LHS of each equation? And, you are missing some negative signs on the RHS...why don't you show me your steps for calculating [itex][L^2,L_x][/itex]?
 
  • #11
gabbagabbahey said:
That's a little easier to read (Although I really wish you'd stop using that ugly [itex]*[/itex] symbol for multiplication). Why do you have a [itex]\varphi[/itex] on the LHS of each equation? And, you are missing some negative signs on the RHS...why don't you show me your steps for calculating [itex][L^2,L_x][/itex]?

okay.[tex] [L^2,L_x]=[L_x^2+L_y^2+L_z^2,L_x]=[L_x^2,L_x]+[L_y^2,L_x]+[L_z^2,L_z][/tex]

[tex][L_x^2,L_x]+[L_y^2,L_x]+[L_z^2,L_x]=[L_y^2,L_x]+[L_z^2,L_x][/tex]

[tex][L_y^2,L_x]+[L_z^2,L_z]=L_y[L_y,L_x]+[L_y,L_x]L_y+L_z[L_z,L_x]+[L_z,L_x][L_z][/tex]

[tex] [L_y,L_x]=i*\hbar*L_z, [L_z,L_x]=i*\hbar*L_y[/tex], therefore

[tex]L_y[L_y,L_x]+[L_y,L_x]L_y+L_z[L_z,L_x]+[L_z,L_x][L_z]=i*L_y(\hbar*L_z),+i*\hbar*L_z*L_y+i*L_z*(\hbar*L_y)+i(\hbar*L_y)*L_z[/tex] .How did you get a negative term? (Sorry, if I don't used the [tex] * [/tex] symbol, latex will read the i's and L's as a subscript rather than a coeffcient
 
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  • #12
[itex][L_y,L_x]=-[L_x,L_y]=-i\hbar L_z[/itex]:wink:
 
  • #13
My calculations showed that [tex] [L^2,L_z]=0,[L^2,L_y]=0,[L^2,L_x]=0[/tex], now what?
 
  • #14
Now, by definition what is [itex][L^2,L_x]\varphi[/itex]?...compare that to what you've just calculated...
 
  • #15
gabbagabbahey said:
Now, by definition what is [itex][L^2,L_x]\varphi[/itex]?...compare that to what you've just calculated...

since[tex][L^2,L_x]=0[/tex],then[tex][L^2,L_x]\varphi=0\varphi=0[/tex]?
 
  • #16
If I asked you the definition of [itex][A,B]\varphi[/itex], you would say [itex](AB-BA)\varphi[/itex], right?

So, when I say "by definition what is [itex][L^2,L_x]\varphi[/itex]?", you say ____?
 
  • #17
gabbagabbahey said:
If I asked you the definition of [itex][A,B]\varphi[/itex], you would say [itex](AB-BA)\varphi[/itex], right?



So, when I say "by definition what is [itex][L^2,L_x]\varphi[/itex]?", you say ____?

yes that's right. [tex][L^2,L_x]=(L^2L_x-L_xL^2)\varphi[/tex], should I let [tex]L_x=(\hbar/i)*d/dx[/tex] and [tex] L=(\hbar/i)*d/dx+(\hbar/i)*d/dy+(\hbar/i)*d/dz[/tex] and let [tex] (-\hbar/i)*d/dx+(-\hbar/i)*d/dy+(-\hbar/i)*d/dz[/tex]
 
  • #18
No need for all that, by definition you have [itex][L^2,L_x]\varphi=(L^2L_x-L_xL^2)\varphi[/itex], and you also just calculated that this was equal to zero, so

[tex][L^2,L_x]\varphi=(L^2L_x-L_xL^2)\varphi=L^2L_x\varphi-L_xL^2\varphi=0[/tex]

Now, express [itex]\varphi[/itex] as a linear combination of the eigenfunctions [itex]\psi_l[/itex] of [itex]L^2[/itex], and use the fact that [itex]L^2\psi_l=l(l+1)\psi_l=0[/itex] for [itex]l=0[/itex] to show that [itex]L_xL^2\varphi=0[/itex]
 
  • #19
I still don't think this works. We now have [itex]L^2L_x\varphi=0[/itex], but how do we go from there to prove that [itex]L_x\varphi=0[/itex]?
 
  • #20
Avodyne said:
I still don't think this works. We now have [itex]L^2L_x\varphi=0[/itex], but how do we go from there to prove that [itex]L_x\varphi=0[/itex]?

[tex]L^2L_x\varphi=0[/tex] why not this expression? Can't I assume [tex]L^2=0[/tex] or [tex]L_x\varphi=0[/tex]
 

1. What is orbital angular momentum?

Orbital angular momentum is a measure of the rotational motion of an object around a fixed point or axis. It is a vector quantity that takes into account the mass, velocity, and distance of an object from the axis of rotation.

2. How is orbital angular momentum calculated?

The formula for calculating orbital angular momentum is L = mvr, where L is the angular momentum, m is the mass of the object, v is its velocity, and r is the distance from the axis of rotation.

3. What is the significance of orbital angular momentum in physics?

Orbital angular momentum is an important concept in physics as it is conserved in isolated systems and plays a crucial role in understanding the behavior of objects in motion, such as planets orbiting around a star or electrons orbiting around an atom.

4. Can orbital angular momentum be negative?

Yes, orbital angular momentum can have a negative value. This indicates that the direction of rotation is opposite to the direction of the angular velocity vector, resulting in a negative magnitude of angular momentum.

5. How is orbital angular momentum related to spin angular momentum?

Orbital angular momentum and spin angular momentum are both forms of angular momentum, but they differ in the way they are generated. Orbital angular momentum is associated with the motion of an object around an axis, while spin angular momentum is a property of particles and is inherent to their rotation on their own axis.

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