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Show that psi(-x) is a solution with same eigenvalue as psi(x)
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I have been given the assignment to show that $psi(-x)$ is another solution with the same energy eigenvalue as $psi(x)$, given that the potential is even around $x=0$. The solutions are one-dimensional.
I have tried to get my head around it but I am not sure what to look for. My idea would be to divide through by the wave function in the wave equation, and we would then be left with the energy-eigenvalue $E$ as a function of the potential, which we know is even around the origin:
$$-frachbar^22mfracd^2 psi(x)dx^2 +V(x)psi(x)=psi(x)E$$.
Dividing by $psi(x)$:
$$-frachbar^22mfracd^2dx^2+V(x)=E$$
And the eigenvalue is now a function of the potential only. Is this an illegal move? Or where do I go from here?
quantum-mechanics homework-and-exercises symmetry schroedinger-equation eigenvalue
edited 13 mins ago
Aaron Stevens
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asked 1 hour ago
Jacob Kaldau
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up vote
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I have been given the assignment to show that $psi(-x)$ is another solution with the same energy eigenvalue as $psi(x)$, given that the potential is even around $x=0$. The solutions are one-dimensional.
I have tried to get my head around it but I am not sure what to look for. My idea would be to divide through by the wave function in the wave equation, and we would then be left with the energy-eigenvalue $E$ as a function of the potential, which we know is even around the origin:
$$-frachbar^22mfracd^2 psi(x)dx^2 +V(x)psi(x)=psi(x)E$$.
Dividing by $psi(x)$:
$$-frachbar^22mfracd^2dx^2+V(x)=E$$
And the eigenvalue is now a function of the potential only. Is this an illegal move? Or where do I go from here?
quantum-mechanics homework-and-exercises symmetry schroedinger-equation eigenvalue
edited 13 mins ago
Aaron Stevens
3,224420
asked 1 hour ago
Jacob Kaldau
61
New contributor
Jacob Kaldau is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I have been given the assignment to show that $psi(-x)$ is another solution with the same energy eigenvalue as $psi(x)$, given that the potential is even around $x=0$. The solutions are one-dimensional.
I have tried to get my head around it but I am not sure what to look for. My idea would be to divide through by the wave function in the wave equation, and we would then be left with the energy-eigenvalue $E$ as a function of the potential, which we know is even around the origin:
$$-frachbar^22mfracd^2 psi(x)dx^2 +V(x)psi(x)=psi(x)E$$.
Dividing by $psi(x)$:
$$-frachbar^22mfracd^2dx^2+V(x)=E$$
And the eigenvalue is now a function of the potential only. Is this an illegal move? Or where do I go from here?
quantum-mechanics homework-and-exercises symmetry schroedinger-equation eigenvalue
edited 13 mins ago
Aaron Stevens
3,224420
asked 1 hour ago
Jacob Kaldau
61
New contributor
Jacob Kaldau is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
I have been given the assignment to show that $psi(-x)$ is another solution with the same energy eigenvalue as $psi(x)$, given that the potential is even around $x=0$. The solutions are one-dimensional.
I have tried to get my head around it but I am not sure what to look for. My idea would be to divide through by the wave function in the wave equation, and we would then be left with the energy-eigenvalue $E$ as a function of the potential, which we know is even around the origin:
$$-frachbar^22mfracd^2 psi(x)dx^2 +V(x)psi(x)=psi(x)E$$.
Dividing by $psi(x)$:
$$-frachbar^22mfracd^2dx^2+V(x)=E$$
And the eigenvalue is now a function of the potential only. Is this an illegal move? Or where do I go from here?
quantum-mechanics homework-and-exercises symmetry schroedinger-equation eigenvalue
quantum-mechanics homework-and-exercises symmetry schroedinger-equation eigenvalue
edited 13 mins ago
Aaron Stevens
3,224420
asked 1 hour ago
Jacob Kaldau
61
New contributor
Jacob Kaldau is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 13 mins ago
Aaron Stevens
3,224420
asked 1 hour ago
Jacob Kaldau
61
New contributor
Jacob Kaldau is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 13 mins ago
Aaron Stevens
3,224420
edited 13 mins ago
Aaron Stevens
3,224420
edited 13 mins ago
Aaron Stevens
3,224420
3,224420
asked 1 hour ago
Jacob Kaldau
61
New contributor
Jacob Kaldau is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 1 hour ago
Jacob Kaldau
61
asked 1 hour ago
Jacob Kaldau
61
61
New contributor
Jacob Kaldau is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Jacob Kaldau is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Jacob Kaldau is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
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2 Answers
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up vote
3
down vote
Simply make a change of variable $xrightarrow -x$ in your equation which becomes
$-frachbar^22mfracd^2dx^2psi(-x)+V(-x)psi(-x)=Epsi(-x)$
Since $V(-x)=V(x)$, this reduces to the same equation for $psi(-x)$ as for $psi(x)$ therefore, they have the same energy eigenvalues.
answered 50 mins ago
E. Bellec
311
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E. Bellec is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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First, thank you for taking your time. However, I am completely new to quantum mechanics and I fail to see why the two equation for psi(-x) and psi(x) are the same when the potential is the same. Could you clarify this, or give a hint?
– Jacob Kaldau
41 mins ago
@JacobKaldau This shows that $hat Hpsi(x)=Epsi(x)$ and $hat Hpsi(-x)=Epsi(-x)$, where $hat H$ and $E$ is the same in either case. So $psi(x)$ and $psi(-x)$ have the same eigenvalues $E$ of the same operator $hat H$
– Aaron Stevens
32 mins ago
Ahhh, yes of course. Now I understand it. Thanks a lot! :)
– Jacob Kaldau
24 mins ago
add a comment |Â
up vote
2
down vote
The step you take between your two equations is an illegal operation. This is because dividing $d^2 psi/dx^2$ by $psi(x)$ does not yield $d^2/dx^2$; it yields $psi''(x)/psi(x)$.
An analogous statement would be if you had a scalar $lambda$, a vector $vecv$, and a matrix $A$ such that $A vecv = lambda vecv$. You can't "divide by $vecv$" to yield $A = lambda$.
answered 46 mins ago
Michael Seifert
13.7k12651
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
Simply make a change of variable $xrightarrow -x$ in your equation which becomes
$-frachbar^22mfracd^2dx^2psi(-x)+V(-x)psi(-x)=Epsi(-x)$
Since $V(-x)=V(x)$, this reduces to the same equation for $psi(-x)$ as for $psi(x)$ therefore, they have the same energy eigenvalues.
answered 50 mins ago
E. Bellec
311
New contributor
E. Bellec is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
First, thank you for taking your time. However, I am completely new to quantum mechanics and I fail to see why the two equation for psi(-x) and psi(x) are the same when the potential is the same. Could you clarify this, or give a hint?
– Jacob Kaldau
41 mins ago
@JacobKaldau This shows that $hat Hpsi(x)=Epsi(x)$ and $hat Hpsi(-x)=Epsi(-x)$, where $hat H$ and $E$ is the same in either case. So $psi(x)$ and $psi(-x)$ have the same eigenvalues $E$ of the same operator $hat H$
– Aaron Stevens
32 mins ago
Ahhh, yes of course. Now I understand it. Thanks a lot! :)
– Jacob Kaldau
24 mins ago
add a comment |Â
up vote
3
down vote
Simply make a change of variable $xrightarrow -x$ in your equation which becomes
$-frachbar^22mfracd^2dx^2psi(-x)+V(-x)psi(-x)=Epsi(-x)$
Since $V(-x)=V(x)$, this reduces to the same equation for $psi(-x)$ as for $psi(x)$ therefore, they have the same energy eigenvalues.
answered 50 mins ago
E. Bellec
311
New contributor
E. Bellec is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
First, thank you for taking your time. However, I am completely new to quantum mechanics and I fail to see why the two equation for psi(-x) and psi(x) are the same when the potential is the same. Could you clarify this, or give a hint?
– Jacob Kaldau
41 mins ago
@JacobKaldau This shows that $hat Hpsi(x)=Epsi(x)$ and $hat Hpsi(-x)=Epsi(-x)$, where $hat H$ and $E$ is the same in either case. So $psi(x)$ and $psi(-x)$ have the same eigenvalues $E$ of the same operator $hat H$
– Aaron Stevens
32 mins ago
Ahhh, yes of course. Now I understand it. Thanks a lot! :)
– Jacob Kaldau
24 mins ago
add a comment |Â
up vote
3
down vote
up vote
3
down vote
Simply make a change of variable $xrightarrow -x$ in your equation which becomes
$-frachbar^22mfracd^2dx^2psi(-x)+V(-x)psi(-x)=Epsi(-x)$
Since $V(-x)=V(x)$, this reduces to the same equation for $psi(-x)$ as for $psi(x)$ therefore, they have the same energy eigenvalues.
answered 50 mins ago
E. Bellec
311
New contributor
E. Bellec is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Simply make a change of variable $xrightarrow -x$ in your equation which becomes
$-frachbar^22mfracd^2dx^2psi(-x)+V(-x)psi(-x)=Epsi(-x)$
Since $V(-x)=V(x)$, this reduces to the same equation for $psi(-x)$ as for $psi(x)$ therefore, they have the same energy eigenvalues.
answered 50 mins ago
E. Bellec
311
New contributor
E. Bellec is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered 50 mins ago
E. Bellec
311
New contributor
E. Bellec is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered 50 mins ago
E. Bellec
311
answered 50 mins ago
E. Bellec
311
311
New contributor
E. Bellec is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
E. Bellec is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
E. Bellec is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
First, thank you for taking your time. However, I am completely new to quantum mechanics and I fail to see why the two equation for psi(-x) and psi(x) are the same when the potential is the same. Could you clarify this, or give a hint?
– Jacob Kaldau
41 mins ago
@JacobKaldau This shows that $hat Hpsi(x)=Epsi(x)$ and $hat Hpsi(-x)=Epsi(-x)$, where $hat H$ and $E$ is the same in either case. So $psi(x)$ and $psi(-x)$ have the same eigenvalues $E$ of the same operator $hat H$
– Aaron Stevens
32 mins ago
Ahhh, yes of course. Now I understand it. Thanks a lot! :)
– Jacob Kaldau
24 mins ago
add a comment |Â
First, thank you for taking your time. However, I am completely new to quantum mechanics and I fail to see why the two equation for psi(-x) and psi(x) are the same when the potential is the same. Could you clarify this, or give a hint?
– Jacob Kaldau
41 mins ago
@JacobKaldau This shows that $hat Hpsi(x)=Epsi(x)$ and $hat Hpsi(-x)=Epsi(-x)$, where $hat H$ and $E$ is the same in either case. So $psi(x)$ and $psi(-x)$ have the same eigenvalues $E$ of the same operator $hat H$
– Aaron Stevens
32 mins ago
Ahhh, yes of course. Now I understand it. Thanks a lot! :)
– Jacob Kaldau
24 mins ago
First, thank you for taking your time. However, I am completely new to quantum mechanics and I fail to see why the two equation for psi(-x) and psi(x) are the same when the potential is the same. Could you clarify this, or give a hint?
– Jacob Kaldau
41 mins ago
First, thank you for taking your time. However, I am completely new to quantum mechanics and I fail to see why the two equation for psi(-x) and psi(x) are the same when the potential is the same. Could you clarify this, or give a hint?
– Jacob Kaldau
41 mins ago
@JacobKaldau This shows that $hat Hpsi(x)=Epsi(x)$ and $hat Hpsi(-x)=Epsi(-x)$, where $hat H$ and $E$ is the same in either case. So $psi(x)$ and $psi(-x)$ have the same eigenvalues $E$ of the same operator $hat H$
– Aaron Stevens
32 mins ago
@JacobKaldau This shows that $hat Hpsi(x)=Epsi(x)$ and $hat Hpsi(-x)=Epsi(-x)$, where $hat H$ and $E$ is the same in either case. So $psi(x)$ and $psi(-x)$ have the same eigenvalues $E$ of the same operator $hat H$
– Aaron Stevens
32 mins ago
Ahhh, yes of course. Now I understand it. Thanks a lot! :)
– Jacob Kaldau
24 mins ago
Ahhh, yes of course. Now I understand it. Thanks a lot! :)
– Jacob Kaldau
24 mins ago
add a comment |Â
up vote
2
down vote
The step you take between your two equations is an illegal operation. This is because dividing $d^2 psi/dx^2$ by $psi(x)$ does not yield $d^2/dx^2$; it yields $psi''(x)/psi(x)$.
An analogous statement would be if you had a scalar $lambda$, a vector $vecv$, and a matrix $A$ such that $A vecv = lambda vecv$. You can't "divide by $vecv$" to yield $A = lambda$.
answered 46 mins ago
Michael Seifert
13.7k12651
add a comment |Â
up vote
2
down vote
The step you take between your two equations is an illegal operation. This is because dividing $d^2 psi/dx^2$ by $psi(x)$ does not yield $d^2/dx^2$; it yields $psi''(x)/psi(x)$.
An analogous statement would be if you had a scalar $lambda$, a vector $vecv$, and a matrix $A$ such that $A vecv = lambda vecv$. You can't "divide by $vecv$" to yield $A = lambda$.
answered 46 mins ago
Michael Seifert
13.7k12651
add a comment |Â
up vote
2
down vote
up vote
2
down vote
The step you take between your two equations is an illegal operation. This is because dividing $d^2 psi/dx^2$ by $psi(x)$ does not yield $d^2/dx^2$; it yields $psi''(x)/psi(x)$.
An analogous statement would be if you had a scalar $lambda$, a vector $vecv$, and a matrix $A$ such that $A vecv = lambda vecv$. You can't "divide by $vecv$" to yield $A = lambda$.
answered 46 mins ago
Michael Seifert
13.7k12651
The step you take between your two equations is an illegal operation. This is because dividing $d^2 psi/dx^2$ by $psi(x)$ does not yield $d^2/dx^2$; it yields $psi''(x)/psi(x)$.
An analogous statement would be if you had a scalar $lambda$, a vector $vecv$, and a matrix $A$ such that $A vecv = lambda vecv$. You can't "divide by $vecv$" to yield $A = lambda$.
answered 46 mins ago
Michael Seifert
13.7k12651
answered 46 mins ago
Michael Seifert
13.7k12651
answered 46 mins ago
Michael Seifert
13.7k12651
answered 46 mins ago
Michael Seifert
13.7k12651
13.7k12651
add a comment |Â
add a comment |Â
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Jacob Kaldau is a new contributor. Be nice, and check out our Code of Conduct.
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