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Region II. Region III. Region I. V(x)=0. V(x)= ∞. V(x)= ∞. L. 0. x. Particle in a 1-Dimensional Box. Time Dependent Schrödinger Equation. PE. KE. TE. Wave function is dependent on time and position function:. 1. Time Independent Schrödinger Equation. V(x)=0 for L>x>0
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Region II Region III Region I V(x)=0 V(x)=∞ V(x)=∞ L 0 x Particle in a 1-Dimensional Box Time Dependent Schrödinger Equation PE KE TE Wave function is dependent on time and position function: 1 Time Independent Schrödinger Equation V(x)=0 for L>x>0 V(x)=∞ for x≥L, x≤0 Applying boundary conditions: 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. Region I and III: Region II:
Finding the Wave Function Our new wave function: But what is ‘A’? This is similar to the general differential equation: Normalizing wave function: So we can start applying boundary conditions: x=0 ψ=0 x=L ψ=0 where n= * Calculating Energy Levels: Since n= * Our normalized wave function is:
Particle in a 1-Dimensional Box Applying the Born Interpretation n=4 n=4 n=3 E n=3 E n=2 n=2 n=1 n=1 x/L x/L
Particle in a 2-Dimensional Box Doing the same thing do these differential equations that we did in one dimension we get: A similar argument can be made: In one dimension we needed only one ‘n’ But in two dimensions we need an ‘n’ for the x and y component. Lots of Boring Math Our Wave Equations: Since For energy levels:
v E u w u w 0 a C3 u w v σ2 σ1 v u v w σ2 w u v Particle in a 2-Dimensional Equilateral Triangle Let’s apply some Boundary Conditions: Types of Symmetry: A C23 Defining some more variables: w v u So our new coordinate system: Our 2-Dimensional Schrödinger Equation: Solution: Substituting in our definitions of x and y in terms of u and v gives: Where p and q are our nx and ny variables from the 2-D box!
Finding the Wave Function Substituting gives: So what is the wave equation? It can be generated from a super position of all of the symmetry operations! So if: And we recall our original definitions: But what plugs into these? If you recall: Substituting and simplifying gives: A1 Continuing with the others: A2 Energy Levels:
Plotting in Mathematica p=1 q=0