The Hydrogen Atom Redux The first three wavefunctions (n=1,2,3) of a - PowerPoint PPT Presentation

The Hydrogen Atom Redux

The first three wavefunctions (n=1,2,3) of a 1D particle in a box of length L. n=1 1.0 0.5 ψ (x/L) x/L 0.2 0.4 0.6 0.8 1.0 - 0.5 n=3 - 1.0 n=2 Physics 273 – The Hydrogen Atom Redux 2

Example: Let’s consider a particle-in-a-box in the n=2 state. What are the probabilities that the particle is located between: r a) 0 < x < L/3 2 ⇣ n π x ⌘ Ψ ( x ) = L sin b) L/3 < x < 2L/3 L c) 2L/3 < x < L Z L/ 3 ✓ 2 π x ◆ 2 L sin 2 P 1 = dx ≈ 0 . 402 L 0 Z 2 L/ 3 ✓ 2 π x ◆ 2 L sin 2 P 2 = dx ≈ 0 . 196 L L/ 3 Z L ✓ 2 π x ◆ 2 L sin 2 P 2 = dx ≈ 0 . 402 L 2 L/ 3 Notice that 0.402+.196+0.402=1! Good: we’ve normalized the wavefunction properly. Physics 273 – The Hydrogen Atom Redux 3

The n=2 wavefunction of a particle in a box 1.0 0.5 ψ (x/L) x/L 0.2 0.4 0.6 0.8 1.0 - 0.5 - 1.0 The n=2 wavefunction squared of a particle in a box 2.0 1.5 [ ψ (x/L)] 2 Total area=1 1.0 0.5 x/L 0.2 0.4 0.6 0.8 1.0 Physics 273 – The Hydrogen Atom Redux 4

Ok, let’s start tackling the hydrogen atom. The first problem we encounter is the fact that it is a 3- dimensional system. Since the potential has a radial symmetry , it will be best to move into spherical coordinates. Coulomb Potential: e 2 1 V ( r ) = − 4 ⇡✏ 0 r As before, we can move into an inertial frame by substituting the reduced mass for the electron mass. Physics 273 – The Hydrogen Atom Redux 5

The potential is independent of time, so we use the 3D time-independent Schrodinger equation: � ~ 2 2 m r 2 Ψ ( ~ x ) + V ( ~ x ) Ψ ( ~ x ) = E Ψ ( ~ x ) We need to do a coordinate transformation. It turns out that this yields the following:  1 − ~ 2 ✓ ◆ ✓ ◆ @ r 2 @ Ψ @ sin ✓@ Ψ 1 + + r 2 sin ✓ r 2 @ r @ r @✓ @✓ 2 µ @ 2 Ψ e 2 � 1 1 r Ψ = E Ψ r 2 sin 2 ✓ − @� 2 4 ⇡✏ 0 Yuck. Physics 273 – The Hydrogen Atom Redux 6

What we want to do is get this into the form [...] + [...]=0 depends only on θ , Φ depends only on r This essentially turns the problem into solve two separate di ff erential equations. So, let’s rewrite ψ by separating variables: Ψ ( ~ x ) = Ψ ( r, ✓ , � ) = R ( r ) Y ( ✓ , � ) and plug it into Schrodinger’s equation... Physics 273 – The Hydrogen Atom Redux 7

We get...  Y − ~ 2 ✓ ◆ ✓ ◆ ∂ r 2 ∂ R ∂ sin θ∂ Y R + + r 2 sin θ 2 µ r 2 ∂ r ∂ r ∂θ ∂θ ∂ 2 Y � R + V ( r ) RY = ERY r 2 sin 2 θ ∂φ 2 r 2 − 2 µ Now, multiply (on the left) by to get ~ 2 Y R  1 − 2 µr 2 ✓ ◆ � ∂ r 2 ∂ R ~ 2 ( V − E ) + ∂ r ∂ r R ∂ 2 Y  ✓ ◆ � 1 1 ∂ sin θ∂ Y + = 0 Y sin 2 θ Y sin θ ∂φ 2 ∂θ ∂θ This is exactly the form we want: [...] + [...] = 0 Physics 273 – The Hydrogen Atom Redux 8

So we have separated the “r” part from the “ θ , Φ ” part. Since varying one “part” cannot a ff ect varying the other “part,” both parts must be independent and constant :  1 − 2 µr 2 ✓ ◆ � @ r 2 @ R ~ 2 ( V − E ) = ` ( ` + 1) @ r @ r R constants @ 2 Y  1 ✓ ◆ 1 � @ sin ✓@ Y + = − ` ( ` + 1) Y sin 2 ✓ Y sin ✓ @� 2 @✓ @✓ Notice that the angular part is also independent of V. It turns out that the solution to this di ff erential equation is known as the spherical harmonics . We can ignore it for now. It’s only important that it’s independent of r. Physics 273 – The Hydrogen Atom Redux 9

Let’s first consider the 𝓂 =0 case. This is the easiest to solve. + 2 µr 2 e 2 ✓ ◆ ✓ ◆ @ r 2 @ R 1 1 r + E = 0 ~ 2 @ r @ r 4 ⇡✏ 0 R The solution will be of the form: R ( r ) = Ce − r/a 0 Let’s plug it in. From the chain rule, we have ∂ r + r 2 ∂ 2 R ✓ ◆  � ∂ r 2 ∂ R 2 r ∂ R 1 = 1 ∂ r 2 ∂ r ∂ r R R So we have: R , ∂ 2 R + r 2 ✓ ◆ ∂ R ∂ r 2 ∂ R = − 2 r ∂ r = − 1 ∂ r 2 = 1 1 R → a 2 a 2 ∂ r ∂ r a 0 R a 0 0 0 Physics 273 – The Hydrogen Atom Redux 10

Rearranging terms, we get ✓ ~ 2 ◆ 1 e 2 ~ 2 E = − r − 2 µa 2 µa 0 4 ⇡✏ 0 0 For this to hold for all values of r, the expression (...) must be equal to 0! Thus we can solve for a 0 and E. ✓ ~ 2 e 2 = 0 → a 0 = 4 ⇡✏ 0 ~ 2 ◆ − µe 2 4 ⇡✏ 0 µa 0 This is the Bohr radius! Similarly, if (...)=0, then the energy is E = − ~ 2 µe 4 = − 0 ~ 2 ≈ − 13 . 6 eV 2 µa 2 32 ⇡ 2 ✏ 2 0 Physics 273 – The Hydrogen Atom Redux 11

Finally, the last bit is to get the normalization. Ψ ( r, θ , φ ) = Ce − r/a 0 For the ground-state equation, the angular solution to Schrodinger’s equation is a constant. Therefore: Z ∞ Z Ψ ? Ψ d 3 r = Ψ ( r ) ? Ψ ( r )4 π r 2 dr = 1 0 s Z ∞ 1 4 π C 2 r 2 e − 2 r/a 0 dr = 1 → C = π a 3 0 0 Therefore, the ground state wavefunction of the hydrogen atom is s 1 e − r/a 0 Ψ ( r, θ , φ ) = π a 3 0 Physics 273 – The Hydrogen Atom Redux 12

Example: Calculate the most probable value of the radius in the ground state. The probability for an electron between r and r+dr is: 1 e − 2 r/a 0 · 4 π r 2 dr P ( r ) dr = Ψ ? Ψ · 4 π r 2 dr = π a 3 0 Find the maximum value of P(r):  �  � dP ( r ) = 0 = 4 2 re − 2 r/a 0 − 2 = 8 1 − r r 2 e − 2 r/a 0 re − 2 r/a 0 a 3 a 3 dr a 0 a 0 0 0 The only local maxima occurs when r=a 0 , that is, the Bohr radius is the most probable location of the electron. Physics 273 – The Hydrogen Atom Redux 13