Problem with angular momentum operator math

[Pchemaaah]

Homework Statement

I am basically trying to show that L_xL_y-L_yL_x=i(hbar)L_z

Homework Equations

L_x=yP_z-zP_y
L_y=zP_x-xP_z

The Attempt at a Solution

I get to the end where I have i(hbar)L_z-z*y*P_xP_z+z*xP_yP_z. How do I get these last two terms to cancel out? I am not too strong in operator math (it hasnt been taught to me, just assumed I can do it), so I don't know if the d/dz part of the P_z operator can still apply to a z that is to the left of it, but if it does then I get L_xL_y-L_yL_x=2*i(hbar)P_z which is wrong according to my book... Can anyone help?

 

[borgwal]

Do you know how to calculate z*P_z-P_z*z?

The equation you're trying to prove is wrong anyway, so that will be quite hard...

 

[Pchemaaah]

Is the solution to your calculation 0 or zP_z+i(hbar)? This is where my operator math gets fuzzy.

And I extracted that equation from a larger problem that asks me to prove that the angular momentum operator (L) squared commutes with each component of the angular momentum. The book goes about solving this by showing that:

L_y²L_x-L_xL_y²=L_y²L_x-(L_yL_x+i(hbar)L_z)L_y

It does not explain how or why this is equivalent, so seeing how the left has to equal the right, I extracted my question. Can you explain why this is not correct or what I am doing wrong?

AH! I apologize, I had a P instead of an L in my first post, sorry if that caused confusion.

 

[borgwal]

Is the solution to your calculation 0 or zP_z+i(hbar)? This is where my operator math gets fuzzy.

Neither: you should apply the operator z*P_z -P_z *z to a dummy wavefunction:

(z*P_z -P_z *z) psi = S psi

where S turns out to be a very simple operator (so, you'll have to find what that simple operator S is). That gives you

z*P_z -P_z *z=S

Then you can use that to figure out what L_x L_y-L_yL_x is.

 

[Pchemaaah]

Neither: you should apply the operator z*P_z -P_z *z to a dummy wavefunction:

(z*P_z -P_z *z) psi = S psi

where S turns out to be a very simple operator (so, you'll have to find what that simple operator S is). That gives you

z*P_z -P_z *z=S

Then you can use that to figure out what L_x L_y-L_yL_x is.

Im sorry but I don't understand what you did. It seems like you threw up an arbitrary operator S and used that to define the solution to (z*P_z -P_z *z) as itself. If it is not 0 or i(hbar), then what is its exact solution?

 

[borgwal]

It is i\hbar (but that's not what you wrote in your question). That is, I wanted you to find the simple operator S: the answer is indeed S=i\hbar I (with I the identity, which is usually left out)

 

[borgwal]

Now substitute zP_z-P_z z=i\hbar in:

L_xL_y-L_yL_x,

what do you get?

 

[Pchemaaah]

Before I do that, can you show me how it ends up being i/hbar? I cannot for the life of me get to that answer which is probably the root of my difficulties haha.

Ill show my steps I guess so you can point out where I am wrong:
zP_z-P_z z=(z*-i*hbar*d/dz)-(-i*hbar*d/dz*z)=-i*z*hbar*d/dz+ihbar=zP_z+ihbar (if the d/dz operator cannot apply to z since z is to its left)

OR -ihbar+ihbar=0 (if it can apply to z)

 

[borgwal]

I tried to tell you how to calculate it: apply the operators to a (dummy) function. The operator d/dz then applies to everything to its right: for one term, the z*P_z term, it only applies to the function, but for the term P_z*z it will apply to the product of z and the function.

 

[Pchemaaah]

Ok, so even if I do put a function in I don't see how you get your result:

(zP_z-P_z z)psi(z)=(z*(-i*hbar*d/dz*psi(z))-(-i*hbar*d/dz*z*psi(z))=-i*z*hbar*Psi(dz)+ihbar*Psi(dz)=(-i*z*hbar+i*hbar)Psi(dz)

Where is i/hbar coming from?

 

[borgwal]

First, it's not i/hbar, it's i\hbar, which is LaTeX for i*hbar.

Second, from your manipulations I get the impression you don't know how to apply d/dz to a function.

 

[Pchemaaah]

Well I have a hard time typing it out, but it is just Psi(dz) instead of Psi(z) when d/dz is applied to it correct?

And can you show me where I am going wrong or how much further I have to go to get to that result because that will essentially enable me to solve this problem since I keep running into the same issues that I am having with your simple sample.

Ill edit the above post to reflect this.

 

[borgwal]

Psi(dz)?? I have no idea where you would get this from.

Do you know what d/dz stands for?

 

[Pchemaaah]

Yes it is the derivative with respect to z. So if I have a dummy function psi and I take the derivative of it with respect to z, then I get Psi'(z) (prime of z) or I was calling it Psi(dz) in the event that it had other variables (d/dz psi(x,y,z)=psi (x,y,dz)). If this is incorrect notation I apologize and I do not know the correct way to express it.

 

[borgwal]

Okay, good: let's go with Psi' then. What is d/dz (z*Psi)? Then subtract z*Psi', and you have your answer.

 

[Pchemaaah]

Okay, good: let's go with Psi' then. What is d/dz (z*Psi)? Then subtract z*Psi', and you have your answer.

d/dz (z*Psi)=Psi(z)+zPsi'(z)

Ahhhh ok I totally forgot about product rule and was stopping the derivative operator before psi. Thank you so much!

 

[borgwal]

okidoki!