# Conceptual question about the transition dipole moment and its components

I'm looking at the following transition: $$1s\to2p$$. For this transition, my textbook says: We have to consider three degenerate $$m$$-components $$m=0,\pm1$$. We put the quantization axis in the $$z$$-direction, so $$m=l_z$$. The matrix element $$M_x+iM_y$$ gives the transitions with $$\Delta m=1$$, $$M_x-iM_y$$ gives the ones with $$\Delta m=-1$$ and $$M_z$$ gives $$m=0$$.

I'm totally puzzled by this last sentence. Why is this case? How did it come up with these relations and why is there the imaginary unit in play? I'm completely new to this subject and would be really thankful for explanations.

– rob
Dec 3, 2020 at 20:44
• Hi Rob! Thanks, I will try to look into it. Is this also explainable by other methods though? I'm just in my third semester of a physics bachelor and I think we will deal with ladder operators at the earliest in fifth semester. The textbook I mentioned is also a book for the lowest basics of QM and the word "ladder operator" doesn't even appear in its index. If there is no other way, I think its weird that this textbook just uses it without showing any derivation of it... Dec 3, 2020 at 20:52
• Do you know some representation for the operators $M_{x,y,z}$?
– rob
Dec 4, 2020 at 17:42
• Do you mean like $(M_{ik})_x=e\int\psi_i^{\ast} \ x\ \psi_k \ d\tau$? When, for example, we perform the integral of $(M_{ik})_x + i(M_{ik})_y$, we see that this is only non zero, when $\Delta m=m_i-m_k=1$. On the other side, when we calculate the integral of $(M_{ik})_z$, we see that this is only non zero when $\Delta m=0$. So I at least know where this comes from. What I'm confused about, is the fact that one time we use a linear combination of $M$ and the other time we use $M$ alone.Does this come from the choice of our quantization axis? (we put it into the $z$-direction) Dec 4, 2020 at 18:50

As mentioned in some of the comments, the operators $$M_+=M_x+iM_y$$ and $$M_-=M_x-iM_y$$ are known as raising and lowering operators, also sometimes called ladder operators or creation/annihilation operators. All of these names can be found for these operators in different contexts, though most often you'll find these names in reference to the quantum harmonic oscillator. In the context of angular momentum, I often see them going without a name. The properties of these operators can be found in any standard source on quantum mechanics to varying levels of detail, but I will try and give you some idea of where they come from here.

The key starting place for angular momentum is the commutation relations of the components. You can think of this as being similar to the commutation relations between the position and momentum operators, $$[x,p]=i\hbar$$. For angular momentum, the commutators are given by $$[M_x,M_y]=i\hbar M_z,\ \ \ \ [M_y,M_z]=i\hbar M_x,\ \ \ \ [M_z,M_x]=i\hbar M_y.$$ These commutators may seem strange, but can be shown on very general grounds using the properties of rotations, though that approach typically can only be found in graduate texts. Somewhat simpler would be to take the definition of angular momentum as $$\boldsymbol{L}=\boldsymbol{x}\times\boldsymbol{p}$$ and then use the commutation relations between the components of position and momentum to work out the expressions above (replace the $$M$$'s by $$L$$'s). So, I will take the above expressions as given. Just in case your class hasn't come to commutation relations yet, I'll note that the definition of the commutator is $$[A,B]=AB-BA$$.

Now, using the commutation relations above, we can show that $$\boldsymbol{M}^2=M_x^2+M_y^2+M_z^2$$ commutes with all the components: $$[M_x,\boldsymbol{M}^2]=[M_y,\boldsymbol{M}^2]=[M_z\boldsymbol{M}^2]=0.$$ It's a theorem that when we have operators that commute with each other, we can find a basis in which they are simultaneously diagonal (this is the statement that we can "choose our quantization axis"). The conventional choice is $$\boldsymbol{M}^2$$ and $$M_z$$. The eigenvalues of $$\boldsymbol{M}^2$$ are $$\hbar^2\ell(\ell+1)$$ and the eigenvalues of $$M_z$$ are $$\hbar m_z$$. I will denote the state with the particular values $$\ell$$ and $$m_z$$ by $$|\ell,m_z\rangle$$. Hence, $$\boldsymbol{M}^2|\ell,m_z\rangle=\hbar^2\ell(\ell+1)|\ell,m_z\rangle,\ \ \ \ M_z|\ell,m_z\rangle=\hbar m_z|\ell,m_z\rangle.$$

Now that we have all the needed setup, let's look at the operators $$M_+$$ and $$M_-$$. First of all, since $$\boldsymbol{M}^2$$ commutes with $$M_x$$ and $$M_y$$, we have $$[\boldsymbol{M}^2,M_+]=[\boldsymbol{M}^2,M_-]=0.$$ This tells us that if we have a state $$|\ell,m_z\rangle$$ and apply, say, $$M_+$$ to it to make the state $$M_+|\ell,m_z\rangle$$, it will have the same $$\ell$$-value: $$\boldsymbol{M}^2(M_+|\ell,m_z\rangle)=M_+\boldsymbol{M}^2|\ell,m_z\rangle=\hbar^2\ell(\ell+1)(M_+|\ell,m_z\rangle).$$ We can work out the same thing for $$M_-$$. This means that $$M_+$$ and $$M_-$$ don't act to change the $$\ell$$-value of the state.

If we instead want to think about $$m_z$$, we need to look at the commutators with $$M_z$$: $$[M_z,M_+]=+\hbar M_+,\ \ \ \ [M_z,M_-]=-\hbar M_-.$$ So if we wanted to investigate the state $$M_+|\ell,m_z\rangle$$, note that the above commutator implies $$M_zM_+=M_+M_z+\hbar M_+$$ so $$M_z(M_+|\ell,m_z\rangle)=(M_+M_z+\hbar M_+)|\ell,m_z\rangle=\hbar(m_z+1)(M_+|\ell,m_z\rangle).$$ So the result of applying $$M_+$$ to the state is a state with one higher $$m_z$$ value (it's the old value plus one). If we did the same for $$M_-$$, we would find that the result of applying $$M_-$$ to the state is a state with $$m_z$$ value decreased by one.

So in the end, the expressions for $$M_+$$ and $$M_-$$ may look a little arbitrary, but there's a good reason why they look that way, and it all comes from the commutation relations between the components of the angular momentum operator that I wrote down at the beginning.

• Thank you for this Richard! This is really well explained and helped me a lot! Now I'm really feeling like to learn more about this! Dec 8, 2020 at 10:49