About solving time dependent Schrodinger equation (Griffiths – Chapter 2 Time Independent Schrodinger Equation) Given the time dependent Schrodinger Equation:

∂Ψ

2Ψ

Ψ

1. Observe that Schrodinger time dependent equation is variable separable if the potential V

is a function of x only.

2. Suppose Schrodinger time dependent equation is variable separable, then its solution will be variable separable. Let Ψ , and the equation becomes

1

21

3. In the above equation, the left side is a function of t alone, and the right side is a function

of x alone. This is only possible if in fact both sides are constant.

4. Let (a constant). Solving that gives .

5. At the same time, we also get the time independent Schrodinger equation:

2

(i) The energy and stationary states are found via solving the time

independent Schrodinger equations

(ii) The constants of integration are found via normalization of the state function and boundary conditions of the potential V.

(iii) The boundary conditions are:

a. is always continuous

b. is continuous except at points where the potential is infinite.

(Griffiths – Section 2.5 The Delta-Function Potential)

(iv) Discretization of the energy and state function usually comes from the boundary conditions.

(v) The stationary states are orthonormal, i.e.

∗ 1if0if

(vi) Time independent Schrodinger equation (time dependent as well) is a linear differential equation. Thus their solution, the stationary states form a complex vector space.

(vii) Since the solution space of time independent Schrodinger equation is a complex vector space V, we can define a complex inner product:

⟨ , ⟩ ∗

for all , ∈ where f(x) and g(x) are square integrable functions. A square integrable function h(x) satisfy:

| | ∗ ∞

The solution space of the time independent Schrodinger equation together with the complex inner product form a complex inner product space.

6. Finally, the solution to the time dependent Schrodinger equation is a linear combination

of the separable solution:

Ψ ,

7. The coefficients can be evaluated by the Fourier’s trick which make use of the

orthonormality of the stationary states:

⟨ , Ψ⟩ ∗ Ψ , ∗ ∗

Making coefficient the subject:

⟨ , Ψ⟩ ∗Ψ ,

Suppose we know the wave function at some time :

⟨ , Ψ , ⟩ ∗Ψ ,

In particular, if we know the wave function at t = 0:

⟨ , Ψ , 0 ⟩ ∗ Ψ , 0

About solving time independent Schrodinger equation The time independent Schrodinger equation is given by:

……. (1)

In general, the solution to time independent Schrodinger equation (TISE) will depend on the potential (energy) function and its boundary conditions (if any). Solving (1) with a given potential V(x) in general, can be difficult. Suppose we assume that the potential is a constant function, i.e. (constant). The above equation can then be re-written into:

2

where ∈ . Thus the above equation can be simplified into:

------- (2)

Notice that although is a real number (positive when E > V, negative when E < V or zero when V = E), k itself is a complex number when E < V. The general solution to equation (2) for complex function is given by

0 …… (3) where A and B are complex constants of integration. This can be verified quite easily. Using Euler’s equation, we can write the general solution into trigonometric form:

cos sin sin ⟹ cos sin ……. (4) where A and B are complex constants of integration. Case 1

⟹ 0. Let ⟹ where 0. Equation (3) imply that when , or when , But the above two equations are the same since A and B are arbitrary complex constants of integration. We can choose either one to be our general solution. If we need to be a real function, then A and B will be reduced to real constants of integration. Case 2

⟹ 0. Let ⟹ where 0. Equation (4) imply that when , cos sin or when , cos sin Again, the above two equations are the same since A and B are arbitrary complex constants of integration. Suppose is a real function such that:

sin cos with C and D real constants of integration. Comparing with

cos sin we have and . Solve the two equations for A and B we have

and

Case 3

⟹ 0⟹ 0. The differential equation becomes 0 which imply that

(A, B complex integration constants) Similarly, A and B will be real constants of integration if is a real function.

About Normalization A wave function Ψ , that satisfy the Schrodinger equation will have the area under its norm square curve |Ψ , | remain unchanged as time progresses, i.e.

|Ψ , | 0

Griffiths – Section 1.4 Normalization Proof

|Ψ , | |Ψ , |

Total derivative is used on the left as the integral is a function of t only while partial derivative is used on the right as the integrand is a function of both x and t. By product rule,

|Ψ| Ψ∗Ψ Ψ∗ ΨΨ

Ψ∗

Schrodinger equation says that

Ψ2

ΨΨ

Taking conjugate Ψ∗

2Ψ∗

Ψ∗

So

|Ψ| Ψ∗2

ΨΨ Ψ

2Ψ∗

Ψ∗2

Ψ∗ Ψ Ψ∗

Ψ

2Ψ∗ Ψ Ψ∗

Ψ

Taking the integral

|Ψ|2

Ψ∗ ΨΨ

Ψ∗

But Ψ must go to zero as x goes to ± infinity, otherwise the wave function would not be normalizable. Since |Ψ∗| |Ψ| → 0 , this means that Ψ∗ must also go to zero as x goes to ±

infinity. The wave functions ΨandΨ∗ → 0 also imply that ,∗, and

∗⟶ 0 as x

goes to ± infinity. Loosely speaking, when a curve goes to zero, its gradient and thus its curvature all goes to zero. As a result

|Ψ| |Ψ| 0 ⇒ |Ψ| constant

In particular, for normalization, we require the constant to be 1 and the wave function Ψ , will stay normalized as time progresses. Thus |Ψ , | can vary with time but the area under it remain constant.