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Orbital angular momentum problem

by noblegas
Tags: angular, momentum, orbital
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noblegas
#1
Oct18-09, 11:31 PM
P: 386
1. The problem statement, all variables and given/known data

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
2. Relevant 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. ?
3. 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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noblegas
#2
Oct19-09, 04:34 PM
P: 386
People having a hard time reading the problem?
gabbagabbahey
#3
Oct19-09, 04:40 PM
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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].

Avodyne
#4
Oct19-09, 08:50 PM
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Orbital angular momentum problem

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].
noblegas
#5
Oct19-09, 09:25 PM
P: 386
Quote Quote by gabbagabbahey View Post
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
gabbagabbahey
#6
Oct19-09, 09:34 PM
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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.
noblegas
#7
Oct19-09, 09:44 PM
P: 386
[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]
gabbagabbahey
#8
Oct19-09, 09:49 PM
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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?
noblegas
#9
Oct19-09, 10:38 PM
P: 386
Quote Quote by gabbagabbahey View Post
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 seperate my commutators into seperate 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]
gabbagabbahey
#10
Oct19-09, 10:44 PM
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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]?
noblegas
#11
Oct19-09, 11:07 PM
P: 386
Quote Quote by gabbagabbahey View Post
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
gabbagabbahey
#12
Oct20-09, 01:26 AM
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[itex][L_y,L_x]=-[L_x,L_y]=-i\hbar L_z[/itex]
noblegas
#13
Oct20-09, 06:18 AM
P: 386
My calculations showed that [tex] [L^2,L_z]=0,[L^2,L_y]=0,[L^2,L_x]=0[/tex], now what?
gabbagabbahey
#14
Oct20-09, 07:00 AM
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Now, by definition what is [itex][L^2,L_x]\varphi[/itex]?...compare that to what you've just calculated....
noblegas
#15
Oct20-09, 07:09 AM
P: 386
Quote Quote by gabbagabbahey View Post
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]?
gabbagabbahey
#16
Oct20-09, 07:24 AM
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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 ____?
noblegas
#17
Oct20-09, 07:51 AM
P: 386
Quote Quote by gabbagabbahey View Post
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 thats 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]
gabbagabbahey
#18
Oct20-09, 08:05 AM
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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]


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