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V(x)=0for L>x>0 V(x)=∞for x≥L, x≤0 Particle in a 1-Dimensional Box Classical Physics: The particle can exist anywhere in the box and follow a path in accordance.

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Presentation on theme: "V(x)=0for L>x>0 V(x)=∞for x≥L, x≤0 Particle in a 1-Dimensional Box Classical Physics: The particle can exist anywhere in the box and follow a path in accordance."— Presentation transcript:

1 V(x)=0for L>x>0 V(x)=∞for x≥L, x≤0 Particle in a 1-Dimensional Box Classical Physics: The particle can exist anywhere in the box and follow a path in accordance to Newton’s Laws. Quantum Physics: The particle is expressed by a wave function and there are certain areas more likely to contain the particle within the box. KE PE TE Time Dependent Schrödinger Equation Wave function is dependent on time and position function: 1 Time Independent Schrödinger Equation Applying boundary conditions: Region I and III: Region II: V(x)=0V(x)=∞ 0 L x Region I Region IIRegion III

2 Finding the Wave Function This is similar to the general differential equation: So we can start applying boundary conditions: x=0 ψ=0 x=L ψ=0 where n= * Our new wave function: But what is ‘A’? Calculating Energy Levels: Normalizing wave function: Since n= * Our normalized wave function is:

3 E x/L E Particle in a 1-Dimensional Box n=1 n=2 n=3 n=4 n=1 n=2 n=3 n=4 Applying the Born Interpretation

4 Particle in a 2-Dimensional Box A similar argument can be made: Lots of Boring Math Our Wave Equations: Doing the same thing do these differential equations that we did in one dimension we get: In one dimension we needed only one ‘n’ But in two dimensions we need an ‘n’ for the x and y component. Since For energy levels:

5 Particle in a 2-Dimensional Equilateral Triangle Types of Symmetry: C3C3 v u w v w u σ1σ1 C23C23 vu w u v w σ2σ2 w u v σ2σ2 Let’s apply some Boundary Conditions: a0 Defining some more variables: So our new coordinate system: Our 2-Dimensional Schrödinger Equation: Substituting in our definitions of x and y in terms of u and v gives: Where p and q are our n x and n y variables from the 2-D box! Solution: A v u w E

6 But what plugs into these? So what is the wave equation? It can be generated from a super position of all of the symmetry operations! Finding the Wave Function If you recall: Continuing with the others: Substituting gives: And we recall our original definitions: Substituting and simplifying gives: A1A1 A2A2 So if: Energy Levels:

7 Plotting in Mathematica p=1 q=0

8 A1A1

9 A2A2


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