Certainly! Let's tackle the segments of your question step-by-step.
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### Part (i) Deriving the One-Dimensional Time Dependent Schrödinger Equation
To derive the one-dimensional time-dependent Schrödinger equation, we start by considering the general principles of quantum mechanics. The wave function [tex]\(\psi(x, t)\)[/tex] describes the quantum state of a particle in terms of its position [tex]\(x\)[/tex] and time [tex]\(t\)[/tex].
For a particle in a potential [tex]\(V(x,t)\)[/tex], the time-dependent Schrödinger equation is given by:
[tex]\[
i \hbar \frac{\partial \psi(x, t)}{\partial t} = \hat{H} \psi(x, t)
\][/tex]
where [tex]\(\hat{H}\)[/tex] is the Hamiltonian operator, which represents the total energy of the system. For a particle of mass [tex]\(m\)[/tex] moving in a potential [tex]\(V(x, t)\)[/tex], the Hamiltonian operator in one dimension is:
[tex]\[
\hat{H} = \frac{\hat{p}^2}{2m} + V(x, t)
\][/tex]
Here, [tex]\(\hat{p}\)[/tex] is the momentum operator, which in the position representation is given by [tex]\(\hat{p} = -i\hbar \frac{\partial}{\partial x}\)[/tex]. Thus,
[tex]\[
\hat{H} = \frac{-\hbar^2}{2m} \frac{\partial^2}{\partial x^2} + V(x, t)
\][/tex]
Substituting this Hamiltonian into the time-dependent Schrödinger equation gives:
[tex]\[
i \hbar \frac{\partial \psi(x, t)}{\partial t} = \left( \frac{-\hbar^2}{2 m} \frac{\partial^2}{\partial x^2} + V(x, t) \right) \psi(x, t)
\][/tex]
Rearranging terms, we get the desired form of the Schrödinger equation:
[tex]\[
\frac{-\hbar^2}{2 m} \frac{\partial^2 \psi(x, t)}{\partial x^2} + V(x, t) \psi(x, t) = i \hbar \frac{\partial \psi(x, t)}{\partial t}
\][/tex]
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### Part (ii) Time-Independent Schrödinger Equation
If the potential [tex]\(V\)[/tex] is a function of position only, i.e., [tex]\(V = V(x)\)[/tex], the wave function [tex]\(\psi(x, t)\)[/tex] can be expressed as a product of a spatial part and a temporal part:
[tex]\[
\psi(x, t) = u(x) T(t)
\][/tex]
Substituting this into the time-dependent Schrödinger equation:
[tex]\[
\frac{-\hbar^2}{2 m} \frac{\partial^2 (u(x)T(t))}{\partial x^2} + V(x) (u(x)T(t)) = i \hbar \frac{\partial (u(x)T(t))}{\partial t}
\][/tex]
Separating variables, we get:
[tex]\[
\frac{-\hbar^2}{2 m} T(t) \frac{\partial^2 u(x)}{\partial x^2} + V(x) u(x) T(t) = i\hbar u(x) \frac{dT(t)}{\partial t}
\][/tex]
Dividing through by [tex]\(u(x)T(t)\)[/tex]:
[tex]\[
\frac{-\hbar^2}{2 m} \frac{1}{u(x)} \frac{\partial^2 u(x)}{\partial x^2} + V(x) = \frac{i \hbar}{T(t)} \frac{dT(t)}{\partial t}
\][/tex]
The left side depends only on [tex]\(x\)[/tex], and the right side depends only on [tex]\(t\)[/tex]. Therefore, each side must equal a constant, which we denote as [tex]\(E\)[/tex]:
[tex]\[
\frac{-\hbar^2}{2 m} \frac{1}{u(x)} \frac{\partial^2 u(x)}{\partial x^2} + V(x) = E
\][/tex]
[tex]\[
i \hbar \frac{1}{T(t)} \frac{dT(t)}{\partial t} = E
\][/tex]
Now, focusing on the spatial part, we get the time-independent Schrödinger equation:
[tex]\[
\frac{-\hbar^2}{2 m} \frac{\partial^2 u(x)}{\partial x^2} + V(x) u(x) = E u(x)
\][/tex]
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### Part (iii) One-Dimensional Hamiltonian Operator
The one-dimensional Hamiltonian operator [tex]\(\hat{H}\)[/tex] is comprised of the kinetic energy operator and the potential energy operator. For a particle of mass [tex]\(m\)[/tex] in a potential [tex]\(V(x)\)[/tex], the Hamiltonian operator in one dimension is given by:
[tex]\[
\hat{H} = \frac{\hat{p}^2}{2m} + V(x)
\][/tex]
where [tex]\(\hat{p}\)[/tex] is the momentum operator. In the position representation, the momentum operator is [tex]\(\hat{p} = -i\hbar \frac{\partial}{\partial x}\)[/tex]. Therefore, the kinetic energy operator is:
[tex]\[
\frac{\hat{p}^2}{2m} = \frac{(-i\hbar \frac{\partial}{\partial x})^2}{2m} = \frac{-\hbar^2}{2m} \frac{\partial^2}{\partial x^2}
\][/tex]
Putting it all together, the Hamiltonian operator is:
[tex]\[
\hat{H} = \frac{-\hbar^2}{2m} \frac{\partial^2}{\partial x^2} + V(x)
\][/tex]
This concludes the derivations and expressions for parts (i), (ii), and (iii).
