# Correlation amplitude and continuous spectra

The correlation amplitude is defined in Modern Quantum Mechanics by JJ Sakurai, chapter 2 as

$$C(t)=\langle\alpha|\alpha,t_0=0;t\rangle=\langle\alpha|\mathcal{U}(t,0)|\alpha\rangle \, .$$

In Eq. (2.1.65), for an initial state $|\alpha\rangle$ represented by a superposition of $|a'\rangle$, i.e, $\sum_{a'} c_{a'}|a'\rangle$ the correlation amplitude is $$C(t)=\sum_{a'}|c_{a'}|^2 \exp\left(\frac{-iE_{a'}t}{\hbar}\right) \, .$$

Consider a state ket which is a superposition of many energy eigenkets with similar energies, i.e, a quasi-continuous spectrum. Here we replace $$\sum_{a'}\rightarrow\int dE\rho(E) \\ c_{a'} \rightarrow g(E) \Bigg|_{E\approx E_0}$$ where $\rho(E)$ characterizes the density of energy eigenstates. The correlation amplitude becomes $$C(t)=\int dE|g(E)|^2\rho(E) \exp\left(\frac{-iEt}{\hbar}\right)$$ and subjected to the normalization condition $$\int dE|g(E)|^{2}\rho(E)=1 \, .$$

My understanding:

For a continuous spectrum $$\sum_{a'}|a'\rangle\langle{a'}|\alpha\rangle=\sum_{a'}\mathcal{C}_{a'}|a'\rangle\rightarrow \int d\mathcal{E'}|\mathcal{E'}\rangle\langle\mathcal{E'}|\alpha\rangle=\int d\mathcal{E'}\mathcal{C}_{\mathcal{E'}}|\mathcal{E'}\rangle$$

$$\sum_{a'}|\langle{a'}|\alpha\rangle|^{2}=1 \rightarrow \int d\mathcal{E'}|\langle{\mathcal{E'}}|\alpha\rangle|^2=1 \, .$$

Doubt:

Why do we define the correlation amplitude $C(t)$ using $\rho(E)$ and $g(E)$ instead of $\mathcal{C}(E)$?

What is the physical meaning of this expression and of $\rho(E)$ and $g(E)$?

• i'd appreciate if someone could comment onto it. – ss1729 Feb 22 '16 at 7:05
• $g(E)$ seems to be just your $C(E)$. As for $\rho(E)$, it accounts for the degeneracy of the continuous spectrum. For instance, if the states would be indexed by the 3-d momentum ${\bf p}$, the integral would be simply $\int{d^3\{bf p}$ since each vector ${\bf p}$ labels a single state and the "spectrum" is non-degenerate. But if the states are indexed by the energy $E={\bf p}^2/2m$, then for each $E$ there is a continuum of states corresponding to the momentum-space sphere $|{\bf p}| = \sqrt{2mE}$. – udrv Feb 23 '16 at 5:20

The correlation amplitude tells you how much this latter system resembles the former. If it's value is one it means that the system has no change. You will get the exact state ket from which you started. In general since we expect some change (i.e, an increase in the degree of mismatching between the two states of a physical system), it's value will be less than one. I.e, it's value changes from 1 (corresponding to the same initial state) to some value less than 1. If the amplitude value is closer to one, then the degree of resemblence between the two states is too good. Now let's see what this $g(E)$ means? If you refer to the text book (Modern Quantum Mechanics by J.J.Sakurai) there the author states that the state ket is assumed to be a linear combination of so many eigen kets with "similar" (which means they are close but not equal) energy eigen values. So it could represent a quasi-continuos spectrum. Hence we replace the summation by integral (changing from discrete to continuous). Next, since we are dealing with the summation of linear combination of eigen vectors with similar energy values we place inside the integral the energy density $ρ(E)$ at different points represented by the eigen vectors with the variable $dE$. Now the complex number Ca′ should be replaced by $g(E)$. $g(E)$ represents the degeneracy (or degenerative multiplicity). Degeneracy of an eigenvalue correspond to how many times the eigenvalues repeated. For example when I solve the equation $X|a\rangle=a|a\rangle$, if I get the eigen values as $a=-1,2,-1$, it means the value $a=-1$ has a degeneracy of 2. The eigen vectors corresponding to them will be identical. The reason for using degeneracy here is because we postulated that the energy eigenvalues are similar. I hope you understand.