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Hydrogen atom • Next weeks homework should be available by 5pm today and is due next week, 4/22. • The last homework set will be out on 4/22 and will be due on 4/30 (one day later than normal). It is a normal assignment, not an extra credit assignment. http://www.colorado.edu/physics/phys2170/ Physics 2170 – Spring 2009 1 Rest of semester • Investigate hydrogen atom (Wednesday 4/15 and Friday 4/17) • Learn about intrinsic angular momentum (spin) of particles like electrons (Monday 4/20) • Take a peak at multielectron atoms including the Pauli Exclusion Principle (Wednesday 4/22) • Describe some of the fundamentals of quantum mechanics (expectation values, eigenstates, superpositions of states, measurements, wave function collapse, etc.) (Friday 4/24 and Monday 4/27) • Review of semester (Wednesday 4/29 and Friday 5/1) • Final exam: Saturday 5/2 from 1:30pm-4:00pm in G125 (this room) http://www.colorado.edu/physics/phys2170/ Physics 2170 – Spring 2009 2 Reading quiz 1 Set frequency to DA Please answer this question on your own. No discussion until after. Q. The potential seen by the electron in a hydrogen atom is… V (r ) A. Independent of distance r B. Spherically symmetric C. An example of a central force potential D. Constant E. More than one of the above 2 ke V (r) r 2 ke The potential seen by the electron is V (r) r Spherically symmetric (doesn’t depend on direction). It depends only on distance from proton so it is a central force potential. http://www.colorado.edu/physics/phys2170/ Physics 2170 – Spring 2009 3 3-D central force problems The hydrogen atom is an example x r sin cos of a 3D central force problem. y r sin sin z r cos The potential energy depends z r only on the distance from a point (spherically symmetric) Spherical coordinates is the natural coordinate system for this problem. General potential: V(r,,) Central force potential: V(r) y x Engineering & math types sometimes swap and . The Time Independent Schrödinger Equation (TISE) becomes: 2 2 1 1 1 2 r 2 sin 2 2 V (r ) E 2 2 2me r r r r sin r sin We can use separation of variables so (r, , ) R(r )( )( ) http://www.colorado.edu/physics/phys2170/ Physics 2170 – Spring 2009 4 The () part 2 The variable only E 2 appears in the TISE as So we should not be surprised that the solution is ( ) eim Note that m is a separation variable and not the electron mass. We use me for the electron mass. x r sin cos y r sin sin z r cos z r y x Are there any constraints on m? ( 2 ) ? They have to be the same! ( ) ( 2 ) What can we say about ( ) and () eim cos m i sin m Since cosine and sine have periods of 2, as long as m is an integer (positive, negative, or 0) the constraint is satisfied. http://www.colorado.edu/physics/phys2170/ Physics 2170 – Spring 2009 5 Angular momentum quantization about z-axis Note ( ) eim is similar to eikx which is the solution to the free particle with p k x r sin cos y r sin sin z r cos As k gives the momentum in the x direction, m gives the momentum in the direction (angular momentum). z r y x Angular momentum about the z-axis is quantized: Lz m There is nothing truly special about the z-axis. We can point the z-axis anywhere we want to. It is just the nature of the coordinate system that treats the z-axis differently than the x and y axes. http://www.colorado.edu/physics/phys2170/ Physics 2170 – Spring 2009 6 The () part The solution to the () part is more complicated so we skip it. The end result is that there is another quantum variable ℓ which must be a non-negative integer and ℓ ≥ |m|. The ℓ variable quantizes the total angular momentum: x r sin cos y r sin sin z r cos L ( 1) z r x y Note, for large ℓ, L so ℓ is basically the total angular momentum and m is the z-component of the angular momentum. Since the z-component cannot be larger than the total, |m| ≤ ℓ. http://www.colorado.edu/physics/phys2170/ Physics 2170 – Spring 2009 7 Clicker question 1 Set frequency to DA Total angular momentum is L ( 1) ℓ can be 0, 1, 2, 3, … The z-component of the angular momentum is Lz m where m can be 0, ±1, ±2, … ±ℓ Just like with any vector, the total angular momentum can be written L L2x L2y L2z Q. Given the rules above, can Lx=Ly=0? That is, can L=Lz? A. Yes, in more than one case It is possible for Lx=Ly=Lz=0 in B. Yes, but only in one case which case L = 0 so ℓ=0 and m=0 C. Never In general, if Lx=Ly=0 then L L2x L2y L2z simplifies to L Lz which means m ( 1) or m2 ( 1). But 2 ( 1) ( 1)2 and there is no integer between ℓ and ℓ+1 so m2 ( 1) (except for m=ℓ=0) http://www.colorado.edu/physics/phys2170/ Physics 2170 – Spring 2009 8 Spherical harmonics We have determined the angular part of the wave function so (r, , ) R(r)( )( ) has become (r, , ) R(r)m ( )eim with the quantum numbers ℓ and m specifying the total angular momentum and the z-component of angular momentum. This angular solution works for any central force problem. The combination m ( )eim are the spherical harmonics Ym ( , ) Spherical harmonics are 3-D and complex (real and imaginary terms), making it very difficult to display. http://www.colorado.edu/physics/phys2170/ Physics 2170 – Spring 2009 9 A few of the spherical harmonics Real part only 0 Colors give phase Red = +1 Y0 1 Cyan = -1 1 Purple = +i 0 Green = -i 1 1 Y cos( ) Y sin( )ei 0 1 1 Y2 3 cos ( ) 1 Y2 sin( ) cos( )e 0 2 i 2 Y2 sin 2 ( )e2i 2 3 m3 m2 m 1 m 0 m 1 m2 m3 http://oak.ucc.nau.edu/jws8/dpgraph/Yellm.html http://www.colorado.edu/physics/phys2170/ Physics 2170 – Spring 2009 10 Clicker question 2 For any central force potential we can write the wave function as Set frequency to DA (r, , ) R(r)m ( )eim (r, , ) R(r)Ym ( , ) Q. What are the boundary conditions on the radial part R(r)? A. R(r) must go to zero as r goes to 0 B. R(r) must go to zero as r goes to infinity C. R(∞) must equal R(0) D. R(r) must equal R(r+2). E. More than one of the above. In order for (r,,) to be normalizable, it must go to zero as r goes to infinity. Therefore, R(r)→0 as r→∞. Physically makes sense as well. Probability of finding the electron very far away from the proton is very small. http://www.colorado.edu/physics/phys2170/ Physics 2170 – Spring 2009 11