# Schr - Physical Chemistry Course Number C362 5 The...

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Unformatted text preview: Physical Chemistry Course Number: C362 5 The time-dependent Schr¨odinger Equation ı ¯ h ∂ ∂t ψ ( x, t ) = Hψ ( x, t ) (5.1) where, H is called the Hamiltonian operator. It is the operator in quantum me- chanics that corresponds to the energy of the system. In quantum mechanics, every physically measurable quantity, has a corresponding operator. The operator H can be written as a sum of the kinetic and potential energy: H = K + V = p 2 2 m + V (5.2) Now if we substitute the momentum operator in Eq. ( 4.4 ), i.e. ˆ p = − ı ¯ h ∂ ∂x , we can write down the Hamiltonian operator as H = p 2 2 m + V = 1 2 m bracketleftBigg − ı ¯ h ∂ ∂x bracketrightBigg 2 + V = − ¯ h 2 2 m ∂ 2 ∂x 2 + V (5.3) where − ¯ h 2 2 m ∂ 2 ∂x 2 = 1 2 m bracketleftBig − ı ¯ h ∂ ∂x bracketrightBig 2 is the kinetic energy operator. The time-dependent Schr¨odinger Equation ( 5.1 ) can be rationalized from the wave-particle duality. We will proceed to show this below. 1. As a result of the wave-particle duality, let us write ψ ( x, t ) as a collection of waves. h ( x ) = integraldisplay dpf ( p ) exp braceleftBigg ı p ¯ h x bracerightBigg (5.4) Such a collection of waves, or packet of waves, is called a wavepacket. 2. Note further that waves exp { ıkx } are eigenstates of the momentum operator. 3. h ( x ) in Eq. ( 5.4 ) does not have a time-dependence. So, we go ahead and mul- tiply Eq. ( 5.4 ) by the quantity exp {− ıωt } = exp braceleftBig − ı E ¯ h t bracerightBig so as to maintain the same wave-form as in Eq. ( ?? ), ψ ( x, t ) = integraldisplay integraldisplay dpdEf ( p ) g ( E ) exp braceleftBigg ı bracketleftBigg p ¯ h x − E ¯ h t bracketrightBiggbracerightBigg (5.5) Why is this okay? 4. Go ahead substitute the right hand side of Eq. ( 5.5 ) into Eq. ( 5.1 ) to see what happens. (Forget about V for now.) Chemistry, Indiana University 23 c circlecopyrt 2011, Srinivasan S. Iyengar (instructor) Physical Chemistry Course Number: C362 5. Differentiating Eq. ( 5.5 ) twice with respect to x we obtain: − ¯ h 2 2 m ∂ 2 ∂x 2 ψ ( x, t ) = − ¯ h 2 2 m ∂ 2 ∂x 2 integraldisplay dpdEf ( p ) g ( E ) exp braceleftBigg ı bracketleftBigg p ¯ h x − E ¯ h t bracketrightBiggbracerightBigg = − ¯ h 2 2 m integraldisplay integraldisplay dpdEf ( p ) g ( E ) bracketleftBigg ı p ¯...
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Schr - Physical Chemistry Course Number C362 5 The...

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