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welcome back to the lectures on CY 1001
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or introductory chemistry in this group
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of lectures consisting of several parts
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I shall describe the quantum mechanics
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associated with the hydrogen atom the
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solution of the Schrodinger equation i
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will give you the results the solution
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of the Schrodinger equation has its
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first major achievement in arriving at
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the spectra of the hydrogen atom which
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were known for many decades before that
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and it's a spectrum of the hydrogen atom
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which prompted Niels Bohr to came up
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with his first model of quantizing the
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energy and quantizing the angular
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momentum of an atom Schrodinger equation
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of course does this using his
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prescription and the wave function and
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we shall see some of the details in the
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calculation of the energies and in the
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calculation of probabilities of the
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electrons and so on so this is part one
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of the quantum mechanics of hydrogen
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atom and in the solution of this
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equation H sy is equal to e sy we're now
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sigh is three-dimensional Cartesian
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coordinate we do use a classical
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starting point of the nucleus with an
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electron somewhere and the nucleus
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having a positive charge plus z DS and
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it's one and the electron with a minus
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charge and a distance of or I shall not
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describe this as a two-body problem it
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will even though that's the right way of
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doing it the two-body problem and then
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remove the center of mass
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from the two body problem and study only
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the relative motion of the two particle
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system which in this case the relative
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mass for the or the reduced mass for the
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two particle system is the mass of the
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electron times the mass of the proton
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divided by the mass of the electron plus
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that of the proton which is
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approximately the mass of the electron /
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the times mass of the proton divided by
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the mass of the proton since MP is much
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much greater than Emmy and therefore mu
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turns out to be approximately m e when
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you cancel the MP so we shall worry
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about making that approximation and
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write the mass as nothing but the mass
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of the electron therefore we needed the
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Hamiltonian for the electron so let's
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assume that the nucleus is stationary
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does not contribute to the overall
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kinetic energy of the atom that's
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already there in the center of mass
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which is not considered here therefore
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if you write to the kinetic energy and
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the potential energy operator for the
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hydrogen atom it will be in terms of the
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operators it will be P squared 1 by 2 m
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e minus z e square by 4 pi epsilon
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naught r which is the classical columbic
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energy of interaction between the
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positive and the negative charge and
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this p square which is an operator is
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given by P X square plus py square plus
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preset square the three components of
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the momentum in a coordinate system
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which is probably fixed in the nucleus
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itself for argument's and then of course
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px is replaced by the derivative
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operators so that the Hamiltonian
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becomes minus H bar square by 2 m.e dou
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square by dou X square
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plus dou square by dou y square plus dou
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square by dou Z a square where these are
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the coordinates of the electron with
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respect to that origin and then you have
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the potential energy minus the d square
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by 4 pi epsilon naught r and r is in
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principle greater than 0 and less than
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or equal to infinity at infinity of
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course the coulombic interaction is zero
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therefore here the boundary includes the
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entire three-dimensional world the whole
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universe so the boundaries are
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explicitly accessed from minus infinity
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to plus infinity Y is from minus
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infinity to plus infinity and also Z is
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from minus infinity to plus infinity so
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this is the three-dimensional region and
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the volume element that we talked about
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for the particle but the electron
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probability is the psy is the volume
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element is DX dy DZ and then the
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probability is sy XE ye is AD absolute
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square d x dy DZ n as the probability of
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finding the electron in the region or in
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the cube between X a and X c plus d XE
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ye and ye plus d y e + ze and ze + the
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ze this is the three-dimensional
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Cartesian coordinate representation for
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the hydrogen electron problem the
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nucleus electron problem instead of
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Cartesian coordinates in the case of
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hydrogen atom one uses spherical polar
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coordinates
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and the reason for that is that if you
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look at the hydrogen atom the potential
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energy is spherically symmetric
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therefore the important contribution to
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the stability of the hydrogen atom which
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is the binding energy between the
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coulombic charges being spherically
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symmetrical the system is better
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described using the spherical polar
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coordinates which if you recall have
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three variables the radius of the sphere
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and then the polar angles theta and Phi
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on the sphere the standard relations for
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these are X is equal to R sine theta cos
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T and Y is or sine theta sine fein and z
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is or cos theta these are the equations
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for the transformation between polar and
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Cartesian coordinates and the inverse
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transformation is of course or is square
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root of x square plus y square plus Z
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square and if you take the ratio of X by
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Y or sine theta cancels off and you have
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caught fire or Phi is tan inverse Y by X
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and the last relation is theta which is
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given in terms of tan inverse square
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root of x square plus y square divided
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by Z so the coordinate transformation
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allows you to either use the spherical
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or the Cartesian coordinate by using the
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relationship between them and this one
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animation gives you the relation or the
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visualization of the spherical polar
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coordinate system and the values let me
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play the animation here the relation
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between the Cartesian and the spherical
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system is given for a given
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one value of R the radius of a sphere
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and you can see that if you fix your
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polar axis called as an axis then the
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polar angle theta is the angle theta
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varying from zero to PI as shown by
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these different radii so that's a
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variation of theta and theta varies from
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0 to PI only and the other angle is of
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course the azimuthal angle phi which is
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perpendicular in a plane perpendicular
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to this and if you rotate this arc semi
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arc by 2 pi you generate the surface of
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the sphere so that's the the azimuthal
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angle phi with respect to your chosen
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x-axis
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so that's a spherical coordinate system
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in which you can see the variation in
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theta given by these very different arcs
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and the value of Phi corresponding to
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each one of these arcs starting from the
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x-axis here at some arbitrary point and
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then going around the x-axis to the plus
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y to the minus X to the minus y and back
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you have two pi therefore if we recall
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our lecture component or varies from 0
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to infinity being the radius of the
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sphere the the sphere is from zero
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radius to all over the universe and
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theta varies from 0 to PI as the the
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polar angle varying from 0 to PI as you
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have seen with respect to the z axis and
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the Phi which goes around the circle in
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2 pi pi is 0 to 2 pi and these are
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relations in parallel to the XY Z all
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going from minus infinity to plus
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infinity in the Cartesian axis in taking
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care of the the whole universal space
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therefore this these are the limits and
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DX dy DZ which is a volume element in
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Cartesian coordinate space will have to
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be expressed in terms of the volume
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elements in spherical polar coordinate
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system and that's given by r square d or
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sine theta D theta D Phi those of you
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who are not familiar with this
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transformation must go back and look at
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the coordinate transformation and the
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simple differentials expressed from one
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coordinate to the other and the
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relations are given by what is known as
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the Jacobian the magnitude of the
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Jacobian the Jacobian being the partial
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derivative of X with respect to R X with
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respect to theta and X with respect to
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Phi and the
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derivative of Y with respect to R with
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respect to theta and with respect to Phi
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E and likewise the partial derivative
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with respect to Z with a Z with respect
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to R and with respect to theta with
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respect to Phi the determinant of this
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multiplied by D or D theta D Phi is this
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is called the Jacobian and this is in
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elementary transformation matrix that
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transforms volume elements from one
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coordinate system another coordinate
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system and this Jacobian with the
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magnitude has the r square sine theta
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with the DRD theta D Phi therefore when
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you calculate the volume elements and
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when you calculate the probability is
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using spherical polar coordinate system
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if you are using Cartesian coordinates
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transform the relation from Cartesian to
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the polar and these are the mathematical
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formulas already well-known and derive
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from elementary differential algebraic
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al culous keep this in mind therefore
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now we have this Hamiltonian expressed
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in terms of minus H bar square by 2 mu
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dou square by dou X a square plus dou
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square by dou y squared plus dou square
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my dose at a square minus h ze square by
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4 pie epsilon not or this needs to be
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changed to polar coordinates spherical
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polar coordinates
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that's not a trivial exercise but it's
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not a very hard exercise the derivatives
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for example no by dou X of any function
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of XYZ or expressed in another
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coordinate system like our theta fee if
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we have 2x plus I in terms of our theta
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P the derivatives are expressed using
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the partial derivatives of the
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coordinates with respect to the new
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coordinates so for example domain 0 X of
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sigh if you wanted to write the
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appropriate wave function in the polar
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coordinate nameless i XYZ is replaced by
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the corresponding substitution of the x
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and y MZ using r theta P using this
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function then there is a very simple
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partial derivative chain rule which
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tells you how to calculate those I by
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dou X as nothing other than though by
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dou R with dough or by dou X derivative
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plus know theta by dou X dou by dou
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theta plus dou Phi by dou X dou by dou
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Phi acting on the wave function sigh or
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theta fee so this is the transformation
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of the derivative form of the Cartesian
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coordinate into the corresponding polar
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coordinates and of course you can
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calculate though or by dou X dou theta
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by dou X and 0 Phi by dou X from the
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inverse relations that you already have
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already have that okay from this you can
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calculate the derivative of R with
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respect to x y&z the derivative of Phi
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with respect to XY and Z and the
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derivative of theta with respect to X 0
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NZ therefore the partial derivatives
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that you need to calculate for
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expressing the kinetic energy in
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spherical polar coordinate system
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involves three such quantities
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namely those square by dou X square
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which is operating this once more but
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being careful that the terms contain
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already or theta and fee and therefore
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the partial derivatives have to be taken
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carefully and you have to do the same
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thing for those square by dou Y square
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and dou square by dou Z square therefore
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let me summarize this particular part of
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the lecture with the corresponding
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expressions namely those I by dou Y as
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the bar by dou Y dou by dou R plus dou
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theta by dou Y dou by dou theta plus dou
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Phi by dou Y dou by dou Phi acting on
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the wave function sine of theta and fee
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and similarly dose I by dou Z as though
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or by dou zhi dou by dou r plus dou
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theta by dou zhi dou by dou theta plus
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dou phi my nose at dou by dou phi acting
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on the corresponding wave functions i or
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theta fee these are the derivative
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equivalents and you calculate likewise
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the those square terms that those square
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by dou y square terms and the dou square
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by does that square term so the summary
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of doing that calculation and if you are
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doing it for the first time about two to
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three hours is what the time that you
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have to give in order to add all these
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terms and cancel and arrive at the final
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form but I will write the final magic
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form that everybody uses for solving the
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hydrogen atom Hamiltonian in polar
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spherical coordinates
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the hamiltonian is minus h bar square by
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2 m IE 1 by r square dou by dou r of r
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square dou by dou r plus 1 by r square
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sine theta dou by dou theta sine theta
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dou by dou theta plus 1 by r square sine
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square theta rho square by 2 pi square
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all of which is the transformation of
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the derivatives to the spherical polar
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form and for this is nothing but the
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kinetic energy term in terms of the
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circle polar coordinates with the
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potential energy minus the d square by 4
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pi epsilon naught or and the equation
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that you are looking for solving is the
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h sigh or theta and v is equal to e sigh
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or theta and fee instead of the h sigh
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XYZ in terms of e sy XYZ okay the wave
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functions are different in the different
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polar different coordinate systems but
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please remember the energy which is
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independent of the coordinate
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representation will not be different
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between different coordinate system how
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you represent your coordinates should
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not lead to any changes in the eigen
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value for the hydrogen electron and
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therefore we the traditional method is
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to use the spherical polar coordinates
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and that allows the wave function to be
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separated into an or dependent wave
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function only yet theta dependent they
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function only in the fire dependent wave
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function only if you recall the particle
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in the two dimensional box where we had
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an XY dependent wave function being
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separated into an X only wave function
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and then why one big wave function term
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and we were able to get the energies and
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the solutions etc therefore separation
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of variables is far more detailed here
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in the case of hydrogen atom and let me
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stop with this as the focal point for
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the next part of the lecture on what is
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called the substitution of the wave
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function in terms of the three radial
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only polar theta angle dependent only
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and azimuthal angle Phi dependent only
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functions and how we separate this into
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three different equations we will not
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solve them but in the second part we
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will look at the solution and the third
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part we will see some physical
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representations of the wave functions
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themselves the real and imaginary parts
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so let me stop with part one year will
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continue exactly from this in the next
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part until then thank you
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you