Ammonia molecule, from Chapter 9 of the Feynman Lectures, Vol 3. Example of a 2-state system, with a small energy difference between the symmetric

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Ammonia molecule, from Chapter 9 of the Feynman Lectures, Vol 3. Eample of a -state system, with a small energy difference between the symmetric and antisymmetric combinations of states and. This energy difference is eploited by the Ammonia Maser.

Transmission and Reflection Coefficients for Barrier Penetration 6.6 8 m 3 3 π a. V C m V a.75 The graph depends only on this combination of parameters. T( ) 4 sinh C ( ) ( ) MathCad will evaluate the hyperbolic functions properly when the argument becomes imaginary (when the energy is above the barrier height). R( ) T( ) In these epressions, is the ratio E/V Transmission and Reflection Coefficients TE ( ) RE ( ).5 3 4 5 6 7 8 E Energy/Potential Note that the transmission is % when the width of the barrier is an integer multiple / wavelength, such that the wave reflected from the second step interferes destructively with that reflected from the first, completely cancelling the reflected wave. There is a small gap at E/V= in the plot, due to the fact that MathCad cannot calculate numerically the limit there correctly, but you can see by eye that the curve appears to be continuous and smooth there. In fact, the limit is well defined.

Barrier Penetration (tunneling) 6.6 8 m 3 π Units: time in ps, distance in nm, mass in kg a. V E.5V Barrier width and potential, and the energy k me 3.48 m V E κ κ a.6 Wave number outside of the barrier region Decay coefficient within the barrier region V( ) if( if ( a )) Potential shape, for plotting Wave eigenfunction for a plane wave incident from the left with unit amplitude: e ika κ k sinh( κa) F( kκ ) κ k Bkκ ( ) k κ sinh( κa) i κkcosh( κa) cosh( κa) i sinh( κa) κk C( kκ ) ( κ ik ) ( κ ik ) B( kκ ) κ Dk ( κ ) ( κ ik ) ( κ ik ) B( kκ ) κ ψ( kκ ) if e ik Bkκ ( ) e ik if afk ( κ ) e ik Ckκ ( ) e κ P ( ) ψ k κ ψ k κ.9 R Bk κ.7 T Fk κ R T Probability density Check that they add up to unity. Reflection coefficient Transmission coefficient ( ) e κ Dkκ 3 (not normalized) P ( ) V ( ) E V....

Now, use the eigenfunctions calculated above to simulate a wave packet incident on the barrier. The behavior is rather different in this case, because there is a significant range of wave numbers included. In fact, the upper end of the range included below etends above the well, so the transmission is not entirely from tunneling. On the other hand, the lower end of the range has an eponentially lower tunneling probability. a σ a Initial position and gaussian width of the wave packet mv 45.746 Wave number for an energy right at the top of the potential barrier σ k 5 k σ 3.48 k v t m 6.35 8 v Width and central value in k space (momentum space) Packet speed, and elapsed time when it hits the barrier. ϕ( k) πσ k 4 e kk 4σ k i kk e This makes a gaussian distribution of wave numbers designed to make a gaussian-shaped wave packet centered around at time and moving to the right with speed v. Now make a wave function by summing (integrating) eigenfunctions for all values of k within 4-sigma of the peak. The integral is done numerically, which is rather time consuming but within the capability of a modern PC. κ( k) mv k ω( k) k ω k m 4.76 7 Ψ( t) k 4σ k ϕ( k) ψ( k κ ( k) ) e iω( k) t dk π k 4σ k ρ( t) Ψ( t)

t FRAMEt 4 ρ( t ) V ( ) 6a E 6aV 5 t..5.5. ρ t V ( ) 6a E 6aV 5..5.5.

Passing over a Potential Barrier (E>V) 6.6 8 m 3 π Units: time in ps, distance in nm, mass in kg a.8 E V.7E Barrier width and potential, and the energy me k k 6.96 m E V k k a 6.383 Wave number outside of the barrier region V( ) if( if ( a )) Potential shape, for plotting Wave number in the barrier region Wave eigenfunction for a plane wave incident from the left with unit amplitude: F kk C kk if ψ k k e ika k k Bkk cosk a sink i k k a k k k kbkk k Dkk k k k k sin k a sin k a k k k kbkk k i kk cos k a e ik Bkk e ik if afkk e ik ik Ckk ik e Dkk e P ( ) ψ k k ψ k k.69 R Bk k.73 T Fk k Reflection coefficient Transmission coefficient R T Check that they add up to unity.

6 (not normalized) P ( ) V ( ) 4.. Now, use the eigenfunctions calculated above to simulate a wave packet incident on the barrier. The behavior is rather different in this case, because there is a significant range of wave numbers included, including some corresponding to energies below the top of the barrier. a a σ 5 mv 53.75 σ k 3.5 k σ 6.96 Initial position and gaussian width of the wave packet. In this case, the packet is smaller than the barrier thickness, so we epect to see two distinct reflections, rather than a superposition of two reflections. Wave number for an energy right at the top of the potential barrier Width and central value in k space (momentum space) k v t m.53 7 v Packet speed, and elapsed time when it hits the barrier. ϕ( k) πσ k 4 e kk 4σ k i kk e This makes a gaussian distribution of wave numbers designed to make a gaussian-shaped wave packet centered around at time and moving to the right with speed v. ω( k) k m κ( k) k mv κ k 39.78.94 8 ω k Frequency and the wave number in the barrier region. Ψ( t) k 4σ k ϕ( k) ψ( k κ ( k) ) e iω( k) t dk π k 4σ k The wave function. The limits are not set to infinity, in order to save some calculation time.

N i N ( 8a ) 4a i i N Evaluate the wave function at a fied grid of N values for plotting. If MathCad is told to plot the function itself, instead of the array of function values, then it tries to evaluate the function too many times and takes forever. t FRAMEt 4 Time at which to evaluate the wave function, set up for making an animation. ρ Ψ t v 5 V 5E e i i i i i V Arrays of values to plot 3 ρ v e t..

t t.53 7 A later time at which to evaluate the wave function ρ Ψ t v 5 V 5E e i i i i i V Arrays of values to plot 3 ρ v e t.53 7..