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Schrödinger's Equation with square potential
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$begingroup$
I have written some code to solve and plot the time independent Schrödinger's equation with potential x^2, which has a bound state with odd integral energy eigenvalues. My code plots the graphs up to some finite x value. When I input the energy eigenvalue to be 2.9995, the wavefunction seems to increase to infinity but when I input 3.0005, it decreases to negative infinity.
I have justified this as being expected by the 'tail wag' argument - i.e. in between we expect a normalisable bound state, i.e. when E=3.
But how can I be sure that the graph will continue to increase/decrease monotonically?
Any help would be much appreciated!
matlab quantum-mechanics
edited Mar 13 at 15:44
Bernard
123k741117
asked Mar 13 at 15:00
MathematicianPMathematicianP
3416
$endgroup$
add a comment |
$begingroup$
I have written some code to solve and plot the time independent Schrödinger's equation with potential x^2, which has a bound state with odd integral energy eigenvalues. My code plots the graphs up to some finite x value. When I input the energy eigenvalue to be 2.9995, the wavefunction seems to increase to infinity but when I input 3.0005, it decreases to negative infinity.
I have justified this as being expected by the 'tail wag' argument - i.e. in between we expect a normalisable bound state, i.e. when E=3.
But how can I be sure that the graph will continue to increase/decrease monotonically?
Any help would be much appreciated!
matlab quantum-mechanics
edited Mar 13 at 15:44
Bernard
123k741117
asked Mar 13 at 15:00
MathematicianPMathematicianP
3416
$endgroup$
$begingroup$
It probably depends at least a little on the algorithm you use to invent a fictional wavefunction from an incorrect eigenenergy. For starters, it sounds like you've ensured the wavefunction will be real (which isn't the case in general), but haven't forced it to be $ge 0$. Without knowing more about how you do that, it's unclear what if anything its sign means. Have you looked at what the square modulus does?
$endgroup$
– J.G.
Mar 13 at 15:04
$begingroup$
I guess my question is : why is the 'wag the tail' argument valid? How do we know there is a bound state solution between the two energy eigenvalues?
$endgroup$
– MathematicianP
Mar 13 at 15:45
add a comment |
$begingroup$
I have written some code to solve and plot the time independent Schrödinger's equation with potential x^2, which has a bound state with odd integral energy eigenvalues. My code plots the graphs up to some finite x value. When I input the energy eigenvalue to be 2.9995, the wavefunction seems to increase to infinity but when I input 3.0005, it decreases to negative infinity.
I have justified this as being expected by the 'tail wag' argument - i.e. in between we expect a normalisable bound state, i.e. when E=3.
But how can I be sure that the graph will continue to increase/decrease monotonically?
Any help would be much appreciated!
matlab quantum-mechanics
edited Mar 13 at 15:44
Bernard
123k741117
asked Mar 13 at 15:00
MathematicianPMathematicianP
3416
$endgroup$
I have written some code to solve and plot the time independent Schrödinger's equation with potential x^2, which has a bound state with odd integral energy eigenvalues. My code plots the graphs up to some finite x value. When I input the energy eigenvalue to be 2.9995, the wavefunction seems to increase to infinity but when I input 3.0005, it decreases to negative infinity.
I have justified this as being expected by the 'tail wag' argument - i.e. in between we expect a normalisable bound state, i.e. when E=3.
But how can I be sure that the graph will continue to increase/decrease monotonically?
Any help would be much appreciated!
matlab quantum-mechanics
matlab quantum-mechanics
edited Mar 13 at 15:44
Bernard
123k741117
asked Mar 13 at 15:00
MathematicianPMathematicianP
3416
edited Mar 13 at 15:44
Bernard
123k741117
asked Mar 13 at 15:00
MathematicianPMathematicianP
3416
edited Mar 13 at 15:44
Bernard
123k741117
edited Mar 13 at 15:44
Bernard
123k741117
edited Mar 13 at 15:44
Bernard
123k741117
123k741117
asked Mar 13 at 15:00
MathematicianPMathematicianP
3416
asked Mar 13 at 15:00
MathematicianPMathematicianP
3416
asked Mar 13 at 15:00
MathematicianPMathematicianP
3416
3416
$begingroup$
It probably depends at least a little on the algorithm you use to invent a fictional wavefunction from an incorrect eigenenergy. For starters, it sounds like you've ensured the wavefunction will be real (which isn't the case in general), but haven't forced it to be $ge 0$. Without knowing more about how you do that, it's unclear what if anything its sign means. Have you looked at what the square modulus does?
$endgroup$
– J.G.
Mar 13 at 15:04
$begingroup$
I guess my question is : why is the 'wag the tail' argument valid? How do we know there is a bound state solution between the two energy eigenvalues?
$endgroup$
– MathematicianP
Mar 13 at 15:45
add a comment |
$begingroup$
It probably depends at least a little on the algorithm you use to invent a fictional wavefunction from an incorrect eigenenergy. For starters, it sounds like you've ensured the wavefunction will be real (which isn't the case in general), but haven't forced it to be $ge 0$. Without knowing more about how you do that, it's unclear what if anything its sign means. Have you looked at what the square modulus does?
$endgroup$
– J.G.
Mar 13 at 15:04
$begingroup$
I guess my question is : why is the 'wag the tail' argument valid? How do we know there is a bound state solution between the two energy eigenvalues?
$endgroup$
– MathematicianP
Mar 13 at 15:45
$begingroup$
It probably depends at least a little on the algorithm you use to invent a fictional wavefunction from an incorrect eigenenergy. For starters, it sounds like you've ensured the wavefunction will be real (which isn't the case in general), but haven't forced it to be $ge 0$. Without knowing more about how you do that, it's unclear what if anything its sign means. Have you looked at what the square modulus does?
$endgroup$
– J.G.
Mar 13 at 15:04
$begingroup$
It probably depends at least a little on the algorithm you use to invent a fictional wavefunction from an incorrect eigenenergy. For starters, it sounds like you've ensured the wavefunction will be real (which isn't the case in general), but haven't forced it to be $ge 0$. Without knowing more about how you do that, it's unclear what if anything its sign means. Have you looked at what the square modulus does?
$endgroup$
– J.G.
Mar 13 at 15:04
$begingroup$
I guess my question is : why is the 'wag the tail' argument valid? How do we know there is a bound state solution between the two energy eigenvalues?
$endgroup$
– MathematicianP
Mar 13 at 15:45
$begingroup$
I guess my question is : why is the 'wag the tail' argument valid? How do we know there is a bound state solution between the two energy eigenvalues?
$endgroup$
– MathematicianP
Mar 13 at 15:45
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
If you have studied the finite square well, you know that only solutions with energy exactly equal to the eigenvalue go to $0$ at $infty$ . For other energies they blow up exponentially (in at least one direction). This is the same phenomenon.
Instead of trying to calculate the eigenfunction at $t = 0$ by solving the equation, you could use the analytic form of the function (available in most QM texts).
answered Mar 13 at 18:58
Keith McClaryKeith McClary
8381412
$endgroup$
$begingroup$
I'm not sure a finite square well was intended here; I think the SHO is considered, in units of $hbaromega/2$. However, the outcome is, in the details you mentioned, basically the same.
$endgroup$
– J.G.
Mar 14 at 11:58
add a comment |
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1 Answer
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1 Answer
1
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oldest
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$begingroup$
If you have studied the finite square well, you know that only solutions with energy exactly equal to the eigenvalue go to $0$ at $infty$ . For other energies they blow up exponentially (in at least one direction). This is the same phenomenon.
Instead of trying to calculate the eigenfunction at $t = 0$ by solving the equation, you could use the analytic form of the function (available in most QM texts).
answered Mar 13 at 18:58
Keith McClaryKeith McClary
8381412
$endgroup$
$begingroup$
I'm not sure a finite square well was intended here; I think the SHO is considered, in units of $hbaromega/2$. However, the outcome is, in the details you mentioned, basically the same.
$endgroup$
– J.G.
Mar 14 at 11:58
add a comment |
$begingroup$
If you have studied the finite square well, you know that only solutions with energy exactly equal to the eigenvalue go to $0$ at $infty$ . For other energies they blow up exponentially (in at least one direction). This is the same phenomenon.
Instead of trying to calculate the eigenfunction at $t = 0$ by solving the equation, you could use the analytic form of the function (available in most QM texts).
answered Mar 13 at 18:58
Keith McClaryKeith McClary
8381412
$endgroup$
$begingroup$
I'm not sure a finite square well was intended here; I think the SHO is considered, in units of $hbaromega/2$. However, the outcome is, in the details you mentioned, basically the same.
$endgroup$
– J.G.
Mar 14 at 11:58
add a comment |
$begingroup$
If you have studied the finite square well, you know that only solutions with energy exactly equal to the eigenvalue go to $0$ at $infty$ . For other energies they blow up exponentially (in at least one direction). This is the same phenomenon.
Instead of trying to calculate the eigenfunction at $t = 0$ by solving the equation, you could use the analytic form of the function (available in most QM texts).
answered Mar 13 at 18:58
Keith McClaryKeith McClary
8381412
$endgroup$
If you have studied the finite square well, you know that only solutions with energy exactly equal to the eigenvalue go to $0$ at $infty$ . For other energies they blow up exponentially (in at least one direction). This is the same phenomenon.
Instead of trying to calculate the eigenfunction at $t = 0$ by solving the equation, you could use the analytic form of the function (available in most QM texts).
answered Mar 13 at 18:58
Keith McClaryKeith McClary
8381412
answered Mar 13 at 18:58
Keith McClaryKeith McClary
8381412
answered Mar 13 at 18:58
Keith McClaryKeith McClary
8381412
answered Mar 13 at 18:58
Keith McClaryKeith McClary
8381412
8381412
$begingroup$
I'm not sure a finite square well was intended here; I think the SHO is considered, in units of $hbaromega/2$. However, the outcome is, in the details you mentioned, basically the same.
$endgroup$
– J.G.
Mar 14 at 11:58
add a comment |
$begingroup$
I'm not sure a finite square well was intended here; I think the SHO is considered, in units of $hbaromega/2$. However, the outcome is, in the details you mentioned, basically the same.
$endgroup$
– J.G.
Mar 14 at 11:58
$begingroup$
I'm not sure a finite square well was intended here; I think the SHO is considered, in units of $hbaromega/2$. However, the outcome is, in the details you mentioned, basically the same.
$endgroup$
– J.G.
Mar 14 at 11:58
$begingroup$
I'm not sure a finite square well was intended here; I think the SHO is considered, in units of $hbaromega/2$. However, the outcome is, in the details you mentioned, basically the same.
$endgroup$
– J.G.
Mar 14 at 11:58
add a comment |
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$begingroup$
It probably depends at least a little on the algorithm you use to invent a fictional wavefunction from an incorrect eigenenergy. For starters, it sounds like you've ensured the wavefunction will be real (which isn't the case in general), but haven't forced it to be $ge 0$. Without knowing more about how you do that, it's unclear what if anything its sign means. Have you looked at what the square modulus does?
$endgroup$
– J.G.
Mar 13 at 15:04
$begingroup$
I guess my question is : why is the 'wag the tail' argument valid? How do we know there is a bound state solution between the two energy eigenvalues?
$endgroup$
– MathematicianP
Mar 13 at 15:45