Introduction to Quantum Mechanics

Classical Mechanics: Quantum Mechanics The actors: Cause and effect Future history of particle is determined by its initial position and momentum together with the force acts upon it Quantum Mechanics Uncertainty principle Exploring the probability The actors: Erwin Schrodinger, Werner Heisenberg, Max Born, Paul Dirac, and others

Wave function  𝜓 2 is proportional to the probability to find the particle at certain time and place Example 𝜓=𝐴+𝑖𝐵 𝜓 ∗ =𝐴−𝑖𝐵 𝜓 2 =𝜓𝜓 ∗ = 𝐴 2 + 𝐵 2 Normalization −∞ ∞ 𝜓 2 𝑑𝑉=1 −∞ ∞ 𝜓 2 𝑑𝑉=1

Summary of the rules 𝜓 must be continous and single valued everywhere 𝑑𝜓 𝑑𝑥 , 𝑑𝜓 𝑑y , 𝑑𝜓 𝑑z , must be continous and single valued everywhere 𝜓 must normalizable, meaning 𝜓 must be 0 as x, y,z → ±∞, so that 𝜓 2 𝑑𝑉 over all space be finite

Wave Equation 𝑑 2 𝑦 𝑑 𝑥 2 = 1 𝑣 2 𝑑 2 𝑦 𝑑 𝑡 2 𝑑 2 𝑦 𝑑 𝑥 2 = 1 𝑣 2 𝑑 2 𝑦 𝑑 𝑡 2 Try to apply above formula to the case of 𝑦 𝑥 =𝐴 exp−iω(𝑡− 𝑥 𝑣 )

Formulating the wave function  𝑥 =𝐴 exp−iω(𝑡− 𝑥 𝑣 ) 𝜔=2𝜋𝑓 , 𝑣=𝜆𝑓, 𝐸=ℎ𝑓=2𝜋ℏ𝑓, 𝜆= ℎ 𝑝 = 2𝜋ℏ 𝑝 Remember:  𝑥 =𝐴 exp−i/ℏ(𝐸𝑡−𝑝𝑥) 𝜕 2 𝜓 𝜕 𝑥 2 =− 𝑝 2 ℏ 2 𝜓 𝑝 2 𝜓=− ℏ 2 𝜕 2 𝜓 𝜕 𝑥 2 𝜕𝜓 𝑑𝑡 =− 𝑖𝐸 ℏ 𝜓 E𝜓= - ℏ 𝑖 𝜕𝜓 𝑑𝑡 𝐸= 𝑝 2 2𝑚 +𝑈(𝑥,𝑡)

Schrödinger’s equation: Time Dependent From 𝐸𝜓= 𝑝 2 2𝑚 𝜓+𝑈(𝑥,𝑡)𝜓 𝑖ℏ 𝜕𝜓 𝑑𝑡 =− ℏ 2 2𝑚 𝜕 2 𝜓 𝑑 𝑥 2 +𝑈(𝑥,𝑡)𝜓 𝑖ℏ 𝜕𝜓 𝑑𝑡 =− ℏ 2 2𝑚 𝜕 2 𝜓 𝑑 𝑥 2 + 𝜕 2 𝜓 𝑑 𝑦 2 + 𝜕 2 𝜓 𝑑 𝑧 2 +𝑈𝜓

Wave Function Superposition 𝜓= 𝑎 1 𝜓 1 + 𝑎 2 𝜓 2 Probability: (take a simple case of a1=a2=1 𝑃= 𝜓 2 = 𝜓 1 + 𝜓 2 2 = 𝜓 1 ∗ + 𝜓 2 ∗ 𝜓 1 + 𝜓 2 = 𝜓 1 ∗ 𝜓 1 + 𝜓 2 ∗ 𝜓 2 + 𝜓 1 ∗ 𝜓 2 + 𝜓 2 ∗ 𝜓 1 = 𝑃 1 + 𝑃 2 + 𝜓 1 ∗ 𝜓 2 + 𝜓 2 ∗ 𝜓 1 Expectation value 𝑥 = −∞ ∞ 𝑥 𝜓 2 𝑑𝑥

Operators  𝑥 =𝐴 exp−i/ℏ(𝐸𝑡−𝑝𝑥) 𝜕𝜓 𝜕𝑥 = 𝑖 ℏ 𝑝𝜓 𝑝𝜓= ℏ 𝑖 𝜕𝜓 𝜕𝑥 𝑝𝜓= ℏ 𝑖 𝜕𝜓 𝜕𝑥 𝐸𝜓=𝑖ℏ 𝜕𝜓 𝜕𝑡 𝜕𝜓 𝜕𝑡 =− 𝑖 ℏ 𝐸𝜓 Operator momentum 𝑝 = ℏ 𝑖 𝜕 𝜕𝑥 Operator Energi 𝐸 =𝑖ℏ 𝜕 𝜕𝑡

Another way to obtain Schrodinger’s Equation 𝐸 = 𝐾𝐸 + 𝑈 Kinetic energy operator 𝐸𝐾 = 𝑝 2 2𝑚 = 1 2𝑚 ℏ 𝑖 𝜕 𝑑𝑥 2 =− ℏ 2 2𝑚 𝜕 2 𝜕 𝑥 2 Energy operator 𝐸 =𝑖ℏ 𝜕 𝜕𝑡 Total energy operator in Hamiltonian form 𝑖ℏ 𝜕 𝑑𝑡 =− ℏ 2 2𝑚 𝜕 2 𝑑 𝑥 2 +𝑈(𝑥,𝑡)

Operator and Expectation Values 𝑝 = −∞ ∞ 𝜓 ∗ 𝑝 𝜓𝑑𝑥= −∞ ∞ 𝜓 ∗ ℏ 𝑖 𝜕 𝜕𝑥 𝜓𝑑𝑥 = ℏ 𝑖 −∞ ∞ 𝜓 ∗ 𝜕 𝜕𝑥 𝜓𝑑𝑥 𝐸 = −∞ ∞ 𝜓 ∗ 𝐸 𝜓𝑑𝑥= −∞ ∞ 𝜓 ∗ 𝑖ℏ 𝜕 𝜕𝑡 𝜓𝑑𝑥 =𝑖ℏ −∞ ∞ 𝜓 ∗ 𝜕𝜓 𝜕𝑡 𝑑𝑥

Steady State Condition  𝑥 =𝐴 exp− i ℏ 𝐸𝑡−𝑝𝑥 =𝐴 𝑒 − i ℏ 𝐸𝑡 𝑒 i ℏ 𝑝𝑥 = 𝜓 𝑠 𝑒 − i ℏ 𝐸𝑡 Subtitute in to Hamiltonian form 𝐸=− ℏ 2 2𝑚 𝜕 2 𝑑 𝑥 2 +𝑈(𝑥,𝑡) Time dependent Schrodinger ‘s equation 𝐸 𝜓 𝑠 𝑒 −𝑖𝐸𝑡/ℏ =− ℏ 2 2𝑚 𝜕 2 𝑑 𝑥 2 𝜓 𝑠 𝑒 −𝑖𝐸𝑡/ℏ +𝑈(𝑥,𝑡) 𝜓 𝑠 𝑒 −𝑖𝐸𝑡/ℏ Stready State for one dimension 𝜕 2 𝑑 𝑥 2 𝜓 𝑠 + 2𝑚 ℏ 2 𝐸−𝑈 𝜓 𝑠 =0 Stready State for three dimension 𝜕 2 𝑑 𝑥 2 𝜓 𝑠 + 𝜕 2 𝑑 𝑦 2 𝜓 𝑠 + 𝜕 2 𝑑 𝑧 2 𝜓 𝑠 + 2𝑚 ℏ 2 𝐸−𝑈 𝜓 𝑠 =0

Particle in a box with infinite potential wall U=∞ U=∞ Time independet, the particle is always in the box for any time. 𝜕 2 𝑑 𝑥 2 𝜓 𝑠 + 2𝑚 ℏ 2 𝐸−𝑈 𝜓 𝑠 =0 𝜓=0 𝜓≠0 𝜓=0 U=0 𝜕 2 𝑑 𝑥 2 𝜓+ 2𝑚 ℏ 2 𝐸𝜓=0 X=0 X=L Solution: 𝜓 𝑥 =𝐴𝑒𝑥𝑝(𝑖𝛼𝑥) Question: A?,  ?

Determine  𝜕 2 𝑑 𝑥 2 𝜓=− 2𝑚 ℏ 2 𝐸𝜓 Subtitute the solution 𝜓 𝑥 =𝐴𝑒𝑥𝑝 𝑖𝛼𝑥 to the left side of equation 𝛼= 2𝑚𝐸 ℏ Remeber: 𝑒 𝑖𝛼𝑥 =𝑠𝑖𝑛𝛼𝑥+𝑖𝑐𝑜𝑠𝛼𝑥 Boundary condition: 𝜓=0 at x=0 and x =L, the imaginary part should be zero 2𝑚𝐸 ℏ 𝐿=𝑛𝜋 n=1,2,3,.... 𝐸 𝑛 = 𝑛 2 𝜋 2 ℏ 2 2𝑚 𝐿 2 Wave function: 𝜓 𝑛 𝑥 =𝐴𝑠𝑖𝑛 2𝑚 𝐸 𝑛 ℏ 𝑥 𝜓 𝑛 𝑥 =𝐴𝑠𝑖𝑛 𝑛𝜋𝑥 𝐿

Probability −∞ ∞ 𝜓 𝑛 2 𝑑𝑥= 0 𝐿 𝜓 𝑛 2 𝑑𝑥= 𝐴 2 0 𝐿 sin 2 𝑛𝜋𝑥 𝐿 𝑑𝑥 = 𝐴 2 2 0 𝐿 𝑑𝑥 − 0 𝐿 𝑐𝑜𝑠 2 𝑛𝜋𝑥 𝐿 𝑑𝑥 = 𝐴 2 2 𝑥− 𝐿 2𝑛𝜋 𝑠𝑖𝑛 2 𝑛𝜋𝑥 𝐿 0 𝐿 = 𝐴 2 𝐿 2 𝐴= 2 𝐿 Normalization −∞ ∞ 𝜓 𝑛 2 𝑑𝑥= 1 Solution of wave function of particle in a box with infinite potential energy in the boundary 𝜓 𝑛 𝑥 = 2 𝐿 𝑠𝑖𝑛 𝑛𝜋𝑥 𝐿

Box with finite potential energy higher than electron energy III I L E U +x Regions I and III (x<0 and x>L) 𝜕 2 𝑑 𝑥 2 𝜓+ 2𝑚 ℏ 2 (𝐸−𝑈)𝜓=0 II Remember: U>E -x Or 𝑎= 2𝑚 ℏ 2 (𝑈−𝐸) 𝜕 2 𝑑 𝑥 2 𝜓− 𝑎 2 𝜓=0 With Solution: 𝜓 𝐼 =𝐶 𝑒 𝑎𝑥 +𝐷 𝑒 −𝑎𝑥 𝜓 𝐼𝐼𝐼 =𝐹 𝑒 𝑎𝑥 +𝐺 𝑒 −𝑎𝑥 Boundary condition: 𝜓 should be finite everywhere Solution: 𝜓 𝐼 =𝐶 𝑒 𝑎𝑥 𝜓 𝐼𝐼𝐼 =𝐺 𝑒 −𝑎𝑥

III I L E U +x Region II (0<x<L) 𝜕 2 𝑑 𝑥 2 𝜓+ 2𝑚 ℏ 2 𝐸𝜓=0 Solution: II 𝜓 𝐼𝐼 𝑥 =𝐴𝑠𝑖𝑛 2𝑚𝐸 ℏ 𝑥+ Bcos 2𝑚𝐸 ℏ 𝑥

Tunneling The Barrier Potential 𝜓_𝐼𝐼 U x=0 x=L 𝜓 𝐼 + 𝜓 𝐼 − 𝜓 𝐼𝐼𝐼 + 𝜕 2 𝑑 𝑥 2 𝜓 𝐼 + 2𝑚 ℏ 2 𝐸 𝜓 𝐼 =0 𝜕 2 𝑑 𝑥 2 𝜓 𝐼𝐼𝐼 + 2𝑚 ℏ 2 𝐸 𝜓 𝐼𝐼𝐼 =0 Solutions 𝜓 𝐼 𝑥 =𝐴 𝑒 𝑖 𝑘 1 𝑥 +𝐵 𝑒 −𝑖 𝑘 1 𝑥 𝑘 1 = 2𝑚𝐸 ℏ = 2𝜋 𝜆 = 𝑝 ℏ 𝜓 𝐼𝐼𝐼 𝑥 =𝐹 𝑒 𝑖 𝑘 1 𝑥 +𝐺 𝑒 −𝑖 𝑘 1 𝑥

The incoming wave 𝜓 𝐼 𝑥 =𝐴 𝑒 𝑖 𝑘 1 𝑥 The reflected wave 𝜓 𝐼 𝑥 =𝐵 𝑒 −𝑖 𝑘 1 𝑥 The transmitted wave 𝜓 𝐼𝐼𝐼 𝑥 =𝐹 𝑒 𝑖 𝑘 1 𝑥 Transmission Probability 𝑇= 𝜓 𝐼𝐼𝐼 + 2 𝑣 𝐼𝐼𝐼 + 𝜓 𝐼 + 2 𝑣 𝐼 + = 𝐹 𝐹 ∗ 𝑣 𝐼𝐼𝐼 + AA ∗ 𝑣 𝐼 +

Inside the barrier 𝜕 2 𝑑 𝑥 2 𝜓+ 2𝑚 ℏ 2 (𝐸−𝑈)𝜓=0 U 𝜓 𝐼𝐼𝐼 + 𝜓 𝐼 + 𝜓_𝐼𝐼 𝜕 2 𝑑 𝑥 2 𝜓+ 2𝑚 ℏ 2 (𝐸−𝑈)𝜓=0 𝜓_𝐼𝐼 U x=0 x=L 𝜓 𝐼 + 𝜓 𝐼 − 𝜓 𝐼𝐼𝐼 + 𝜕 2 𝑑 𝑥 2 𝜓− 2𝑚 ℏ 2 (𝑈−𝐸)𝜓=0 Since U>E Solution 𝜓 𝐼𝐼 =𝐶 𝑒 − 𝑘 2 𝑥 +𝐷 𝑒 𝑘 2 𝑥 Wave number inside the barrier 𝑘 2 = 2𝑚 ℏ 2 (𝑈−𝐸)

Boundary Conditions At x=0 𝜓 𝐼 = 𝜓 𝐼𝐼 𝜕𝜓 𝐼 𝜕𝑥 = 𝜕𝜓 𝐼𝐼 𝜕𝑥 𝜓 𝐼𝐼𝐼 = 𝜓 𝐼𝐼 𝜓 𝐼 = 𝜓 𝐼𝐼 𝜕𝜓 𝐼 𝜕𝑥 = 𝜕𝜓 𝐼𝐼 𝜕𝑥 𝜓 𝐼𝐼𝐼 = 𝜓 𝐼𝐼 𝜕𝜓 𝐼𝐼𝐼 𝜕𝑥 = 𝜕𝜓 𝐼𝐼 𝜕𝑥 At x=L

How tunneling effect can occur in p-n juction? Wikipedia.com

Quantumaniac.tumblr.com

Scanning Tunneling Microscope Nobel Prize 1986

Harmonic Oscilator 𝜕 2 𝑑 𝑥 2 𝜓+ 2𝑚 ℏ 2 (𝐸−𝑈)𝜓=0 𝜕 2 𝑑 𝑥 2 𝜓+ 2𝑚 ℏ 2 (𝐸−𝑈)𝜓=0 𝜕 2 𝑑 𝑥 2 𝜓+ 2𝑚 ℏ 2 (𝐸− 1 2 𝑘 𝑥 2 )𝜓=0 Solusi: 𝜓 𝑛 = 2𝑚𝑓 ℏ 1 4 2 𝑛 𝑛! − 1 2 𝐻 𝑛 𝑦 𝑒 − 𝑦 2 /2

𝐸 𝑛 = 𝑛+ 1 2 ℎ𝑓 www.pci.tu-bs.de