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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:

How/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 $g(E)$?

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 $g(E)$?

The correlation amplitude is defined in Modern Quantum Mechanics by JJ Sakurai, chapter 2 Eq. (2.1.65) as

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

In Eq. (2.1.66), 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:

How 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 $g(E)$?

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:

How/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 $g(E)$?

The correlation amplitude , $$ C(t)=\langle\alpha|\alpha,t_{o}=0;t\rangle=\langle\alpha|\mathcal{U}(t,0)|\alpha\rangle $$

In Modern Quantum Mechanics: Sakurai, Chapter:2, eqn(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: $$ C(t)=\sum_{a'}|c_{a'}|^{2}exp\Bigg(\frac{-iE_{a'}t}{\hbar}\Bigg) $$ In Modern Quantum Mechanics: Sakurai, Chapter:2, eqn(2.1.66):

Consider a state ket which is a superposition of many energy eigenkets with similar energies, i.e, a quasi-continuous spectrum. Here we replace:

$$ \sum_{a'}\rightarrow\int dE\rho(E) \\ c_{a'} \rightarrow g(E) \Bigg|_{E\approx E_{o}} $$ where $\rho(E)$ characterizes the density of energy eigenstates. The correlation amplitude becomes

$$ C(t)=\int dE|g(E)|^{2}\rho(E)exp\Bigg(\frac{-iEt}{\hbar}\Bigg) $$ and subjected to the normalization condition $$ \int dE|g(E)|^{2}\rho(E)=1 $$ My understanding:

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

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

Doubt:

How 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 $g(E)$  ?

The correlation amplitude is defined in Modern Quantum Mechanics by JJ Sakurai, chapter 2 Eq. (2.1.65) as

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

In Eq. (2.1.66), 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:

How 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 $g(E)$?