Schrödinger Equation in Spherical Coordinates
I am trying to learn how to solve three dimensional Schrödinger Equation in Spherical Coordinates. I was reading a text book and I found that there is a missed step in the solution, seen below:
The $theta$ equation,$$sinthetafrac{mathrm d}{mathrm dtheta}left(sinthetafrac{mathrm dTheta}{mathrm dtheta}right)+left[ellleft(ell+1right)sin^2theta-m^2right]Theta=0.tag{4.25}$$is not so simple. The solution is $$Theta(theta)=AP_ell^mleft(costhetaright).tag{4.26}$$where $P^m_ell$ is the associated Legendre function, defined by $$P_ell^mleft(xright)equivleft(1-x^2right)^{vert mvert/2}left(frac{mathrm d}{mathrm dx}right)^{vert mvert}P_ell(x).tag{4.27}$$
It says that the solution of equation (4.25) is not simple and gives directly as equation (4.26). Can you help me to learn how to solve such differential equations?
mathematical-physics coordinate-systems differential-equations
asked Nov 24 '18 at 12:27
Diamond
143
migrated from physics.stackexchange.com Nov 27 '18 at 18:47
This question came from our site for active researchers, academics and students of physics.
|
show 1 more comment
I am trying to learn how to solve three dimensional Schrödinger Equation in Spherical Coordinates. I was reading a text book and I found that there is a missed step in the solution, seen below:
The $theta$ equation,$$sinthetafrac{mathrm d}{mathrm dtheta}left(sinthetafrac{mathrm dTheta}{mathrm dtheta}right)+left[ellleft(ell+1right)sin^2theta-m^2right]Theta=0.tag{4.25}$$is not so simple. The solution is $$Theta(theta)=AP_ell^mleft(costhetaright).tag{4.26}$$where $P^m_ell$ is the associated Legendre function, defined by $$P_ell^mleft(xright)equivleft(1-x^2right)^{vert mvert/2}left(frac{mathrm d}{mathrm dx}right)^{vert mvert}P_ell(x).tag{4.27}$$
It says that the solution of equation (4.25) is not simple and gives directly as equation (4.26). Can you help me to learn how to solve such differential equations?
mathematical-physics coordinate-systems differential-equations
asked Nov 24 '18 at 12:27
Diamond
143
migrated from physics.stackexchange.com Nov 27 '18 at 18:47
This question came from our site for active researchers, academics and students of physics.
1
Would this be better on Mathematics SE?
– Aaron Stevens
Nov 24 '18 at 12:40
@Diamond May you please provide the name of the textbook that you are using?
– N. Steinle
Nov 24 '18 at 13:36
@Aaron Stevens Sorry I did not understand what do you mean.
– Diamond
Nov 24 '18 at 14:53
1
@Diamond Because your question has a mathematical nature (even if it's related to Schrodinger's equation) Aaron believes that it should be posted on Mathematics StackExchange.
– IchVerloren
Nov 24 '18 at 15:34
1
This ODE can be transformed into a standard hypergeometric form. Probably the most useful one as a companion to QM is James Seaborn, "Hypergometric Functions and their Applications".
– ZeroTheHero
Nov 24 '18 at 16:35
|
show 1 more comment
I am trying to learn how to solve three dimensional Schrödinger Equation in Spherical Coordinates. I was reading a text book and I found that there is a missed step in the solution, seen below:
The $theta$ equation,$$sinthetafrac{mathrm d}{mathrm dtheta}left(sinthetafrac{mathrm dTheta}{mathrm dtheta}right)+left[ellleft(ell+1right)sin^2theta-m^2right]Theta=0.tag{4.25}$$is not so simple. The solution is $$Theta(theta)=AP_ell^mleft(costhetaright).tag{4.26}$$where $P^m_ell$ is the associated Legendre function, defined by $$P_ell^mleft(xright)equivleft(1-x^2right)^{vert mvert/2}left(frac{mathrm d}{mathrm dx}right)^{vert mvert}P_ell(x).tag{4.27}$$
It says that the solution of equation (4.25) is not simple and gives directly as equation (4.26). Can you help me to learn how to solve such differential equations?
mathematical-physics coordinate-systems differential-equations
asked Nov 24 '18 at 12:27
Diamond
143
I am trying to learn how to solve three dimensional Schrödinger Equation in Spherical Coordinates. I was reading a text book and I found that there is a missed step in the solution, seen below:
The $theta$ equation,$$sinthetafrac{mathrm d}{mathrm dtheta}left(sinthetafrac{mathrm dTheta}{mathrm dtheta}right)+left[ellleft(ell+1right)sin^2theta-m^2right]Theta=0.tag{4.25}$$is not so simple. The solution is $$Theta(theta)=AP_ell^mleft(costhetaright).tag{4.26}$$where $P^m_ell$ is the associated Legendre function, defined by $$P_ell^mleft(xright)equivleft(1-x^2right)^{vert mvert/2}left(frac{mathrm d}{mathrm dx}right)^{vert mvert}P_ell(x).tag{4.27}$$
It says that the solution of equation (4.25) is not simple and gives directly as equation (4.26). Can you help me to learn how to solve such differential equations?
mathematical-physics coordinate-systems differential-equations
mathematical-physics coordinate-systems differential-equations
asked Nov 24 '18 at 12:27
Diamond
143
asked Nov 24 '18 at 12:27
Diamond
143
asked Nov 24 '18 at 12:27
Diamond
143
asked Nov 24 '18 at 12:27
Diamond
143
asked Nov 24 '18 at 12:27
Diamond
143
143
migrated from physics.stackexchange.com Nov 27 '18 at 18:47
This question came from our site for active researchers, academics and students of physics.
migrated from physics.stackexchange.com Nov 27 '18 at 18:47
This question came from our site for active researchers, academics and students of physics.
1
Would this be better on Mathematics SE?
– Aaron Stevens
Nov 24 '18 at 12:40
@Diamond May you please provide the name of the textbook that you are using?
– N. Steinle
Nov 24 '18 at 13:36
@Aaron Stevens Sorry I did not understand what do you mean.
– Diamond
Nov 24 '18 at 14:53
1
@Diamond Because your question has a mathematical nature (even if it's related to Schrodinger's equation) Aaron believes that it should be posted on Mathematics StackExchange.
– IchVerloren
Nov 24 '18 at 15:34
1
This ODE can be transformed into a standard hypergeometric form. Probably the most useful one as a companion to QM is James Seaborn, "Hypergometric Functions and their Applications".
– ZeroTheHero
Nov 24 '18 at 16:35
|
show 1 more comment
1
Would this be better on Mathematics SE?
– Aaron Stevens
Nov 24 '18 at 12:40
@Diamond May you please provide the name of the textbook that you are using?
– N. Steinle
Nov 24 '18 at 13:36
@Aaron Stevens Sorry I did not understand what do you mean.
– Diamond
Nov 24 '18 at 14:53
1
@Diamond Because your question has a mathematical nature (even if it's related to Schrodinger's equation) Aaron believes that it should be posted on Mathematics StackExchange.
– IchVerloren
Nov 24 '18 at 15:34
1
This ODE can be transformed into a standard hypergeometric form. Probably the most useful one as a companion to QM is James Seaborn, "Hypergometric Functions and their Applications".
– ZeroTheHero
Nov 24 '18 at 16:35
1
1
Would this be better on Mathematics SE?
– Aaron Stevens
Nov 24 '18 at 12:40
Would this be better on Mathematics SE?
– Aaron Stevens
Nov 24 '18 at 12:40
@Diamond May you please provide the name of the textbook that you are using?
– N. Steinle
Nov 24 '18 at 13:36
@Diamond May you please provide the name of the textbook that you are using?
– N. Steinle
Nov 24 '18 at 13:36
@Aaron Stevens Sorry I did not understand what do you mean.
– Diamond
Nov 24 '18 at 14:53
@Aaron Stevens Sorry I did not understand what do you mean.
– Diamond
Nov 24 '18 at 14:53
1
1
@Diamond Because your question has a mathematical nature (even if it's related to Schrodinger's equation) Aaron believes that it should be posted on Mathematics StackExchange.
– IchVerloren
Nov 24 '18 at 15:34
@Diamond Because your question has a mathematical nature (even if it's related to Schrodinger's equation) Aaron believes that it should be posted on Mathematics StackExchange.
– IchVerloren
Nov 24 '18 at 15:34
1
1
This ODE can be transformed into a standard hypergeometric form. Probably the most useful one as a companion to QM is James Seaborn, "Hypergometric Functions and their Applications".
– ZeroTheHero
Nov 24 '18 at 16:35
This ODE can be transformed into a standard hypergeometric form. Probably the most useful one as a companion to QM is James Seaborn, "Hypergometric Functions and their Applications".
– ZeroTheHero
Nov 24 '18 at 16:35
|
show 1 more comment
1 Answer
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The $theta$ equation,$$sinthetafrac{mathrm d}{mathrm dtheta}left(sinthetafrac{mathrm dTheta}{mathrm dtheta}right)+left[ellleft(ell+1right)sin^2theta-m^2right]Theta=0.tag{4.25}$$
You'll have to apply a variable change: let $x=cos(theta)$. That will lead you to the associated Legendre Differential Equation
begin{equation}
(1-x^{2})frac{mathrm{d}^{2}Theta}{mathrm{d}x^{2}}-left(2xfrac{mathrm{d}Theta}{mathrm{d}x}+ellleft(ell+1right)-frac{m^{2}}{1-x^{2}}right)Theta=0
end{equation}
This is satisfied for values $xin [-1,1]$ using Legendre Polynomials given by Rodrigues' formula:
begin{equation}
P_{ell m}(x)=frac{(-1)^{m}}{2^{ell}ell!}(1-x^{2})^{m/2}frac{mathrm{d}^{m+ell}}{mathrm{d}x^{m+ell}}(x^{2}-1)
end{equation}
where $-ellleq m leq ell$
Here you can see a detailed solution.
I may recommend Arfken & Weber's Mathematical Methods for Physicists text.
answered Nov 24 '18 at 13:34
IchVerloren
969
add a comment |
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The $theta$ equation,$$sinthetafrac{mathrm d}{mathrm dtheta}left(sinthetafrac{mathrm dTheta}{mathrm dtheta}right)+left[ellleft(ell+1right)sin^2theta-m^2right]Theta=0.tag{4.25}$$
You'll have to apply a variable change: let $x=cos(theta)$. That will lead you to the associated Legendre Differential Equation
begin{equation}
(1-x^{2})frac{mathrm{d}^{2}Theta}{mathrm{d}x^{2}}-left(2xfrac{mathrm{d}Theta}{mathrm{d}x}+ellleft(ell+1right)-frac{m^{2}}{1-x^{2}}right)Theta=0
end{equation}
This is satisfied for values $xin [-1,1]$ using Legendre Polynomials given by Rodrigues' formula:
begin{equation}
P_{ell m}(x)=frac{(-1)^{m}}{2^{ell}ell!}(1-x^{2})^{m/2}frac{mathrm{d}^{m+ell}}{mathrm{d}x^{m+ell}}(x^{2}-1)
end{equation}
where $-ellleq m leq ell$
Here you can see a detailed solution.
I may recommend Arfken & Weber's Mathematical Methods for Physicists text.
answered Nov 24 '18 at 13:34
IchVerloren
969
add a comment |
The $theta$ equation,$$sinthetafrac{mathrm d}{mathrm dtheta}left(sinthetafrac{mathrm dTheta}{mathrm dtheta}right)+left[ellleft(ell+1right)sin^2theta-m^2right]Theta=0.tag{4.25}$$
You'll have to apply a variable change: let $x=cos(theta)$. That will lead you to the associated Legendre Differential Equation
begin{equation}
(1-x^{2})frac{mathrm{d}^{2}Theta}{mathrm{d}x^{2}}-left(2xfrac{mathrm{d}Theta}{mathrm{d}x}+ellleft(ell+1right)-frac{m^{2}}{1-x^{2}}right)Theta=0
end{equation}
This is satisfied for values $xin [-1,1]$ using Legendre Polynomials given by Rodrigues' formula:
begin{equation}
P_{ell m}(x)=frac{(-1)^{m}}{2^{ell}ell!}(1-x^{2})^{m/2}frac{mathrm{d}^{m+ell}}{mathrm{d}x^{m+ell}}(x^{2}-1)
end{equation}
where $-ellleq m leq ell$
Here you can see a detailed solution.
I may recommend Arfken & Weber's Mathematical Methods for Physicists text.
answered Nov 24 '18 at 13:34
IchVerloren
969
add a comment |
The $theta$ equation,$$sinthetafrac{mathrm d}{mathrm dtheta}left(sinthetafrac{mathrm dTheta}{mathrm dtheta}right)+left[ellleft(ell+1right)sin^2theta-m^2right]Theta=0.tag{4.25}$$
You'll have to apply a variable change: let $x=cos(theta)$. That will lead you to the associated Legendre Differential Equation
begin{equation}
(1-x^{2})frac{mathrm{d}^{2}Theta}{mathrm{d}x^{2}}-left(2xfrac{mathrm{d}Theta}{mathrm{d}x}+ellleft(ell+1right)-frac{m^{2}}{1-x^{2}}right)Theta=0
end{equation}
This is satisfied for values $xin [-1,1]$ using Legendre Polynomials given by Rodrigues' formula:
begin{equation}
P_{ell m}(x)=frac{(-1)^{m}}{2^{ell}ell!}(1-x^{2})^{m/2}frac{mathrm{d}^{m+ell}}{mathrm{d}x^{m+ell}}(x^{2}-1)
end{equation}
where $-ellleq m leq ell$
Here you can see a detailed solution.
I may recommend Arfken & Weber's Mathematical Methods for Physicists text.
answered Nov 24 '18 at 13:34
IchVerloren
969
The $theta$ equation,$$sinthetafrac{mathrm d}{mathrm dtheta}left(sinthetafrac{mathrm dTheta}{mathrm dtheta}right)+left[ellleft(ell+1right)sin^2theta-m^2right]Theta=0.tag{4.25}$$
You'll have to apply a variable change: let $x=cos(theta)$. That will lead you to the associated Legendre Differential Equation
begin{equation}
(1-x^{2})frac{mathrm{d}^{2}Theta}{mathrm{d}x^{2}}-left(2xfrac{mathrm{d}Theta}{mathrm{d}x}+ellleft(ell+1right)-frac{m^{2}}{1-x^{2}}right)Theta=0
end{equation}
This is satisfied for values $xin [-1,1]$ using Legendre Polynomials given by Rodrigues' formula:
begin{equation}
P_{ell m}(x)=frac{(-1)^{m}}{2^{ell}ell!}(1-x^{2})^{m/2}frac{mathrm{d}^{m+ell}}{mathrm{d}x^{m+ell}}(x^{2}-1)
end{equation}
where $-ellleq m leq ell$
Here you can see a detailed solution.
I may recommend Arfken & Weber's Mathematical Methods for Physicists text.
answered Nov 24 '18 at 13:34
IchVerloren
969
answered Nov 24 '18 at 13:34
IchVerloren
969
answered Nov 24 '18 at 13:34
IchVerloren
969
answered Nov 24 '18 at 13:34
IchVerloren
969
969
add a comment |
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1
Would this be better on Mathematics SE?
– Aaron Stevens
Nov 24 '18 at 12:40
@Diamond May you please provide the name of the textbook that you are using?
– N. Steinle
Nov 24 '18 at 13:36
@Aaron Stevens Sorry I did not understand what do you mean.
– Diamond
Nov 24 '18 at 14:53
1
@Diamond Because your question has a mathematical nature (even if it's related to Schrodinger's equation) Aaron believes that it should be posted on Mathematics StackExchange.
– IchVerloren
Nov 24 '18 at 15:34
1
This ODE can be transformed into a standard hypergeometric form. Probably the most useful one as a companion to QM is James Seaborn, "Hypergometric Functions and their Applications".
– ZeroTheHero
Nov 24 '18 at 16:35