$S^+$ and $S^{-}$ operators formula

In summary: Now let's go back to##S^- |m\rangle = \alpha_m |m-1\rangle##We can also write:##S^- S^+ |m\rangle = S^- \beta_m |m+1\rangle = \alpha_m \beta_m |m\rangle = |\alpha_m|^2 |m-1\rangle##But by equation 7,##|\alpha_m|^2 = (S^2 - m(m-1))/|\beta_m|^2 = s(s+1) - m(m-1)/(s(s+1) - m(m+1)) = (s(s+1) - m(m-1))(s(s+1) - m(m+
  • #1
LagrangeEuler
717
20
For ##\hat{S}^+## and ##\hat{S}^{-}## operators for any given spin ##S## relation
[tex]\hat{S}^+|S,m \rangle=\sqrt{S(S+1)-m(m+1)}\hbar|S,m+1 \rangle [/tex]
[tex]\hat{S}^-|S,m \rangle=\sqrt{S(S+1)-m(m-1)}\hbar|S,m-1 \rangle [/tex]
Can someone please explain how we get those factors ##\sqrt{S(S+1)-m(m+1)}\hbar## and ##\sqrt{S(S+1)-m(m-1)}\hbar##?
In ##|S,m \rangle## ##S## denotes spin, and ##m## spin projection.
 
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  • #2
I thought someone would already have answered this. For people who like algebra, it's one of those fun things to derive in quantum mechanics.

Let me list some facts about these operators
  1. ##S^{+} = S_x + i S_y##
  2. ##S^{-} = S_x - i S_y##
  3. ##S^2 = S_x^2 + S_y^2 + S_z^2##
  4. ##S_x S_y - S_y S_x = i S_z##
  5. ##S_z S^{+} = S^{+} (S_z + 1)##
  6. ##S_z S^{-} = S^{-} (S_z - 1)##
  7. ##S^{-} S^{+} = S^2 - S_z(S_z + 1)##
  8. ##(S^{+})^\dagger = S^{-}##
(The last 4 are provable from the first 4).

So if we let ##|m\rangle## be the state with ##S_z |m \rangle = m | m \rangle##,
then ##S_z S^- |m\rangle = S^- (S_z - 1) |m\rangle = S^- (m - 1) |m\rangle = (m-1) S^- |m\rangle##. So ##S^-|m\rangle## is an eigenstate of ##S_z## with eigenvalue ##m-1##. That means (assuming nondegeneracy of eigenvalues---I'm going to skip the argument for why this is the case, because I'm not sure why) that ##S^- |m\rangle## must be a multiple of ##|m-1\rangle##. So let's let ##\alpha_m## be the multiplier:

##S^- |m\rangle = \alpha_m |m-1\rangle##

Analogously, we can show that ##S^+ |m\rangle## has to be an eigenstate of ##S_z## with eigenvalue ##m+1##. So

##S^+ |m\rangle = \beta_m |m+1\rangle##

where ##\beta_m## is some unknown multiplier. We can relate ##\alpha_m## and ##\beta_m## by considering

##\langle m| S^- S^+ |m \rangle##

By our assumptions about ##S^+## and ##S^{-}##, we get:

##\langle m| S^- S^+ |m \rangle = \langle m| (S^+ \beta_m |m+1\rangle) = \langle m |\alpha_{m+1} \beta_m |m\rangle = \alpha_{m+1} \beta_m##. But we also know that ##\langle m| S^- S^+ |m \rangle = (S^+ |m\rangle)^\dagger (S^+ |m\rangle) = \langle m | \beta_m^* \beta_m |m \rangle##. So this proves that

##\beta_m^* = \alpha_{m+1}##

Now on the one hand:

##S^- S^+ |m\rangle = S^- \beta_m |m+1\rangle = \alpha_{m+1} \beta_m |m\rangle = |\beta_m|^2 |m \rangle##

On the other hand, by equation 7 above,

##S^- S^+ |m\rangle = (S^2 - S_z(S_z+1)) |m\rangle = (S^2 - m(m+1)) |m\rangle##

So we conclude that ##\beta_m = \sqrt{S^2 - m(m+1)}##.

But what are the eigenvalues of ##S^2##? Well, since ##S^2## must be greater than or equal to ##S_z^2##, we know that we can't keep raising the value of ##S_z## forever. But since ##S^+ |m\rangle = \beta_m |m+1\rangle##, the only way to prevent raising ##m## indefinitely is if for some maximal value of ##m##,

##S^+ |m_{\text{max}}\rangle = 0##

That implies that ##\beta_{m_{\text{max}}} = 0##.

So since we know an expression for ##\beta_m##,

##\sqrt{S^2 - m_{\text{max}}(m_{\text{max}}+1)} = 0##

That implies that ##S^2 = m_{\text{max}}(m_{\text{max}}+1)##

Rewriting ##m_{\text{max}} \equiv s## gives:
##S^2 =s (s+1)##

and ##\beta_m = \sqrt{s(s+1) - m(m+1)}##
 
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1. What is the purpose of the $S^+$ and $S^{-}$ operators formula?

The $S^+$ and $S^{-}$ operators formula is used in quantum mechanics to describe the spin of a particle. These operators represent the spin-up and spin-down states of a particle, respectively.

2. How are the $S^+$ and $S^{-}$ operators related to each other?

The $S^+$ and $S^{-}$ operators are related as complex conjugates of each other. This means that if the $S^+$ operator is applied to a state, the $S^{-}$ operator will produce the same result when applied to the complex conjugate of that state.

3. Can the $S^+$ and $S^{-}$ operators be used to measure the spin of a particle?

No, the $S^+$ and $S^{-}$ operators cannot be used to directly measure the spin of a particle. They are mathematical representations of the spin states and are used to calculate the probabilities of obtaining a certain spin measurement.

4. How are the $S^+$ and $S^{-}$ operators used in the Schrödinger equation?

The $S^+$ and $S^{-}$ operators are used in the Schrödinger equation to describe the time evolution of a quantum system. They are included in the Hamiltonian operator, which represents the total energy of the system.

5. Are the $S^+$ and $S^{-}$ operators the only operators used to describe spin states?

No, there are other operators used to describe spin states, such as the $S^z$ operator which represents the spin in the z-direction. The $S^+$ and $S^{-}$ operators are just one part of the larger set of spin operators used in quantum mechanics.

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