How do I compute the eigenfunctions of an operator that contains another operator?
$begingroup$
Given the operator $A = (Xfrac{d}{dx}+2)$, where $X$ is a linear operator, how can I find the eigenfunction of $A$ corresponding to a zero eigenvalue?
In general, this is just a matter of solving the differential equation for $AF(x) = 0$, however, in this case, that leaves me with the differential equation $X frac{d}{dx} F(x) + 2 F(x) = 0$, and I'm just not really sure what to do with that $X$ operator.
Anyone out there that can get me past this step?
ordinary-differential-equations quantum-mechanics eigenfunctions
edited Sep 24 '18 at 0:03
platty
3,370320
asked Sep 23 '18 at 23:36
Will RussellWill Russell
112
$endgroup$
add a comment |
$begingroup$
Given the operator $A = (Xfrac{d}{dx}+2)$, where $X$ is a linear operator, how can I find the eigenfunction of $A$ corresponding to a zero eigenvalue?
In general, this is just a matter of solving the differential equation for $AF(x) = 0$, however, in this case, that leaves me with the differential equation $X frac{d}{dx} F(x) + 2 F(x) = 0$, and I'm just not really sure what to do with that $X$ operator.
Anyone out there that can get me past this step?
ordinary-differential-equations quantum-mechanics eigenfunctions
edited Sep 24 '18 at 0:03
platty
3,370320
asked Sep 23 '18 at 23:36
Will RussellWill Russell
112
$endgroup$
$begingroup$
$X$ isn't specified in the problem, though we could probably assume that it is the position operator $ihbarfrac{d}{dp_x}$. Obviously a general solution without this assumption is preferred if it is possible.
$endgroup$
– Will Russell
Sep 24 '18 at 1:30
$begingroup$
Yep, that's all the details. This is an exercise from chapter 2 of Quantum Mechanics Concepts and Applications by Nouredine Zettili. As I look at it more, I'm wondering if I can just treat the X operator as a constant for the purpose of solving the differential.
$endgroup$
– Will Russell
Sep 24 '18 at 12:16
$begingroup$
I found the problem in that textbook. $X$ is the position operator! If you included the rest it would be clear as you are asked to compute commutators like $[A,X]$ and $[A,P]$ in the next subproblem. When acting on a real space wavefunction $Xpsi = xpsi$ (you are quoting the momentum space representation of it above) so you are left with the ODE $xpsi' + 2psi = 0$.
$endgroup$
– Winther
Sep 24 '18 at 13:57
$begingroup$
I apologize, I thought I had provided enough context for the problem without complicating it with unnecessary minutia, and clearly didn't provide enough. This is actually much simpler than I was trying to make it, thank you for that. If you don't mind a followup "why?" type question, can you explain why you are allowed to do this when $X$ is acting on the derivative of $psi$ rather than on $psi$ itself? I'm assuming that there is an operator rule that I somehow missed.
$endgroup$
– Will Russell
Sep 24 '18 at 23:05
$begingroup$
If it's not in the book take a look here: en.wikipedia.org/wiki/Position_operator
$endgroup$
– Winther
Sep 24 '18 at 23:08
add a comment |
$begingroup$
Given the operator $A = (Xfrac{d}{dx}+2)$, where $X$ is a linear operator, how can I find the eigenfunction of $A$ corresponding to a zero eigenvalue?
In general, this is just a matter of solving the differential equation for $AF(x) = 0$, however, in this case, that leaves me with the differential equation $X frac{d}{dx} F(x) + 2 F(x) = 0$, and I'm just not really sure what to do with that $X$ operator.
Anyone out there that can get me past this step?
ordinary-differential-equations quantum-mechanics eigenfunctions
edited Sep 24 '18 at 0:03
platty
3,370320
asked Sep 23 '18 at 23:36
Will RussellWill Russell
112
$endgroup$
Given the operator $A = (Xfrac{d}{dx}+2)$, where $X$ is a linear operator, how can I find the eigenfunction of $A$ corresponding to a zero eigenvalue?
In general, this is just a matter of solving the differential equation for $AF(x) = 0$, however, in this case, that leaves me with the differential equation $X frac{d}{dx} F(x) + 2 F(x) = 0$, and I'm just not really sure what to do with that $X$ operator.
Anyone out there that can get me past this step?
ordinary-differential-equations quantum-mechanics eigenfunctions
ordinary-differential-equations quantum-mechanics eigenfunctions
edited Sep 24 '18 at 0:03
platty
3,370320
asked Sep 23 '18 at 23:36
Will RussellWill Russell
112
edited Sep 24 '18 at 0:03
platty
3,370320
asked Sep 23 '18 at 23:36
Will RussellWill Russell
112
edited Sep 24 '18 at 0:03
platty
3,370320
edited Sep 24 '18 at 0:03
platty
3,370320
edited Sep 24 '18 at 0:03
platty
3,370320
3,370320
asked Sep 23 '18 at 23:36
Will RussellWill Russell
112
asked Sep 23 '18 at 23:36
Will RussellWill Russell
112
asked Sep 23 '18 at 23:36
Will RussellWill Russell
112
112
$begingroup$
$X$ isn't specified in the problem, though we could probably assume that it is the position operator $ihbarfrac{d}{dp_x}$. Obviously a general solution without this assumption is preferred if it is possible.
$endgroup$
– Will Russell
Sep 24 '18 at 1:30
$begingroup$
Yep, that's all the details. This is an exercise from chapter 2 of Quantum Mechanics Concepts and Applications by Nouredine Zettili. As I look at it more, I'm wondering if I can just treat the X operator as a constant for the purpose of solving the differential.
$endgroup$
– Will Russell
Sep 24 '18 at 12:16
$begingroup$
I found the problem in that textbook. $X$ is the position operator! If you included the rest it would be clear as you are asked to compute commutators like $[A,X]$ and $[A,P]$ in the next subproblem. When acting on a real space wavefunction $Xpsi = xpsi$ (you are quoting the momentum space representation of it above) so you are left with the ODE $xpsi' + 2psi = 0$.
$endgroup$
– Winther
Sep 24 '18 at 13:57
$begingroup$
I apologize, I thought I had provided enough context for the problem without complicating it with unnecessary minutia, and clearly didn't provide enough. This is actually much simpler than I was trying to make it, thank you for that. If you don't mind a followup "why?" type question, can you explain why you are allowed to do this when $X$ is acting on the derivative of $psi$ rather than on $psi$ itself? I'm assuming that there is an operator rule that I somehow missed.
$endgroup$
– Will Russell
Sep 24 '18 at 23:05
$begingroup$
If it's not in the book take a look here: en.wikipedia.org/wiki/Position_operator
$endgroup$
– Winther
Sep 24 '18 at 23:08
add a comment |
$begingroup$
$X$ isn't specified in the problem, though we could probably assume that it is the position operator $ihbarfrac{d}{dp_x}$. Obviously a general solution without this assumption is preferred if it is possible.
$endgroup$
– Will Russell
Sep 24 '18 at 1:30
$begingroup$
Yep, that's all the details. This is an exercise from chapter 2 of Quantum Mechanics Concepts and Applications by Nouredine Zettili. As I look at it more, I'm wondering if I can just treat the X operator as a constant for the purpose of solving the differential.
$endgroup$
– Will Russell
Sep 24 '18 at 12:16
$begingroup$
I found the problem in that textbook. $X$ is the position operator! If you included the rest it would be clear as you are asked to compute commutators like $[A,X]$ and $[A,P]$ in the next subproblem. When acting on a real space wavefunction $Xpsi = xpsi$ (you are quoting the momentum space representation of it above) so you are left with the ODE $xpsi' + 2psi = 0$.
$endgroup$
– Winther
Sep 24 '18 at 13:57
$begingroup$
I apologize, I thought I had provided enough context for the problem without complicating it with unnecessary minutia, and clearly didn't provide enough. This is actually much simpler than I was trying to make it, thank you for that. If you don't mind a followup "why?" type question, can you explain why you are allowed to do this when $X$ is acting on the derivative of $psi$ rather than on $psi$ itself? I'm assuming that there is an operator rule that I somehow missed.
$endgroup$
– Will Russell
Sep 24 '18 at 23:05
$begingroup$
If it's not in the book take a look here: en.wikipedia.org/wiki/Position_operator
$endgroup$
– Winther
Sep 24 '18 at 23:08
$begingroup$
$X$ isn't specified in the problem, though we could probably assume that it is the position operator $ihbarfrac{d}{dp_x}$. Obviously a general solution without this assumption is preferred if it is possible.
$endgroup$
– Will Russell
Sep 24 '18 at 1:30
$begingroup$
$X$ isn't specified in the problem, though we could probably assume that it is the position operator $ihbarfrac{d}{dp_x}$. Obviously a general solution without this assumption is preferred if it is possible.
$endgroup$
– Will Russell
Sep 24 '18 at 1:30
$begingroup$
Yep, that's all the details. This is an exercise from chapter 2 of Quantum Mechanics Concepts and Applications by Nouredine Zettili. As I look at it more, I'm wondering if I can just treat the X operator as a constant for the purpose of solving the differential.
$endgroup$
– Will Russell
Sep 24 '18 at 12:16
$begingroup$
Yep, that's all the details. This is an exercise from chapter 2 of Quantum Mechanics Concepts and Applications by Nouredine Zettili. As I look at it more, I'm wondering if I can just treat the X operator as a constant for the purpose of solving the differential.
$endgroup$
– Will Russell
Sep 24 '18 at 12:16
$begingroup$
I found the problem in that textbook. $X$ is the position operator! If you included the rest it would be clear as you are asked to compute commutators like $[A,X]$ and $[A,P]$ in the next subproblem. When acting on a real space wavefunction $Xpsi = xpsi$ (you are quoting the momentum space representation of it above) so you are left with the ODE $xpsi' + 2psi = 0$.
$endgroup$
– Winther
Sep 24 '18 at 13:57
$begingroup$
I found the problem in that textbook. $X$ is the position operator! If you included the rest it would be clear as you are asked to compute commutators like $[A,X]$ and $[A,P]$ in the next subproblem. When acting on a real space wavefunction $Xpsi = xpsi$ (you are quoting the momentum space representation of it above) so you are left with the ODE $xpsi' + 2psi = 0$.
$endgroup$
– Winther
Sep 24 '18 at 13:57
$begingroup$
I apologize, I thought I had provided enough context for the problem without complicating it with unnecessary minutia, and clearly didn't provide enough. This is actually much simpler than I was trying to make it, thank you for that. If you don't mind a followup "why?" type question, can you explain why you are allowed to do this when $X$ is acting on the derivative of $psi$ rather than on $psi$ itself? I'm assuming that there is an operator rule that I somehow missed.
$endgroup$
– Will Russell
Sep 24 '18 at 23:05
$begingroup$
I apologize, I thought I had provided enough context for the problem without complicating it with unnecessary minutia, and clearly didn't provide enough. This is actually much simpler than I was trying to make it, thank you for that. If you don't mind a followup "why?" type question, can you explain why you are allowed to do this when $X$ is acting on the derivative of $psi$ rather than on $psi$ itself? I'm assuming that there is an operator rule that I somehow missed.
$endgroup$
– Will Russell
Sep 24 '18 at 23:05
$begingroup$
If it's not in the book take a look here: en.wikipedia.org/wiki/Position_operator
$endgroup$
– Winther
Sep 24 '18 at 23:08
$begingroup$
If it's not in the book take a look here: en.wikipedia.org/wiki/Position_operator
$endgroup$
– Winther
Sep 24 '18 at 23:08
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Your operator X may be thought of as a constant, since it does nothing on "numbers", which include the parameter x, so it commutes with everything just like a constant.
It is then evident that
$$
Xfrac{d}{dx} F(x,X)= -2 F(x,X)
$$
is solved by
$$
F(x,X)=c e^{{-2x}X^{-1}} ,
$$
where $X^{-1}$ is the inverse operator to X.
answered Dec 23 '18 at 21:19
Cosmas ZachosCosmas Zachos
1,582520
$endgroup$
add a comment |
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1 Answer
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1 Answer
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votes
$begingroup$
Your operator X may be thought of as a constant, since it does nothing on "numbers", which include the parameter x, so it commutes with everything just like a constant.
It is then evident that
$$
Xfrac{d}{dx} F(x,X)= -2 F(x,X)
$$
is solved by
$$
F(x,X)=c e^{{-2x}X^{-1}} ,
$$
where $X^{-1}$ is the inverse operator to X.
answered Dec 23 '18 at 21:19
Cosmas ZachosCosmas Zachos
1,582520
$endgroup$
add a comment |
$begingroup$
Your operator X may be thought of as a constant, since it does nothing on "numbers", which include the parameter x, so it commutes with everything just like a constant.
It is then evident that
$$
Xfrac{d}{dx} F(x,X)= -2 F(x,X)
$$
is solved by
$$
F(x,X)=c e^{{-2x}X^{-1}} ,
$$
where $X^{-1}$ is the inverse operator to X.
answered Dec 23 '18 at 21:19
Cosmas ZachosCosmas Zachos
1,582520
$endgroup$
add a comment |
$begingroup$
Your operator X may be thought of as a constant, since it does nothing on "numbers", which include the parameter x, so it commutes with everything just like a constant.
It is then evident that
$$
Xfrac{d}{dx} F(x,X)= -2 F(x,X)
$$
is solved by
$$
F(x,X)=c e^{{-2x}X^{-1}} ,
$$
where $X^{-1}$ is the inverse operator to X.
answered Dec 23 '18 at 21:19
Cosmas ZachosCosmas Zachos
1,582520
$endgroup$
Your operator X may be thought of as a constant, since it does nothing on "numbers", which include the parameter x, so it commutes with everything just like a constant.
It is then evident that
$$
Xfrac{d}{dx} F(x,X)= -2 F(x,X)
$$
is solved by
$$
F(x,X)=c e^{{-2x}X^{-1}} ,
$$
where $X^{-1}$ is the inverse operator to X.
answered Dec 23 '18 at 21:19
Cosmas ZachosCosmas Zachos
1,582520
answered Dec 23 '18 at 21:19
Cosmas ZachosCosmas Zachos
1,582520
answered Dec 23 '18 at 21:19
Cosmas ZachosCosmas Zachos
1,582520
answered Dec 23 '18 at 21:19
Cosmas ZachosCosmas Zachos
1,582520
1,582520
add a comment |
add a comment |
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$begingroup$
$X$ isn't specified in the problem, though we could probably assume that it is the position operator $ihbarfrac{d}{dp_x}$. Obviously a general solution without this assumption is preferred if it is possible.
$endgroup$
– Will Russell
Sep 24 '18 at 1:30
$begingroup$
Yep, that's all the details. This is an exercise from chapter 2 of Quantum Mechanics Concepts and Applications by Nouredine Zettili. As I look at it more, I'm wondering if I can just treat the X operator as a constant for the purpose of solving the differential.
$endgroup$
– Will Russell
Sep 24 '18 at 12:16
$begingroup$
I found the problem in that textbook. $X$ is the position operator! If you included the rest it would be clear as you are asked to compute commutators like $[A,X]$ and $[A,P]$ in the next subproblem. When acting on a real space wavefunction $Xpsi = xpsi$ (you are quoting the momentum space representation of it above) so you are left with the ODE $xpsi' + 2psi = 0$.
$endgroup$
– Winther
Sep 24 '18 at 13:57
$begingroup$
I apologize, I thought I had provided enough context for the problem without complicating it with unnecessary minutia, and clearly didn't provide enough. This is actually much simpler than I was trying to make it, thank you for that. If you don't mind a followup "why?" type question, can you explain why you are allowed to do this when $X$ is acting on the derivative of $psi$ rather than on $psi$ itself? I'm assuming that there is an operator rule that I somehow missed.
$endgroup$
– Will Russell
Sep 24 '18 at 23:05
$begingroup$
If it's not in the book take a look here: en.wikipedia.org/wiki/Position_operator
$endgroup$
– Winther
Sep 24 '18 at 23:08