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Normalization constant of a planar wave
Normalizing a set of eigenfunctions with different domainsUsing the Normalization Condition with WavefunctionImaginary Eigenvalue Of A Hermitian OperatorBox normalizationNormalization of time-independent Schroedinger equation in Spherical CoordinatesCompute the Momentum of the Wave FunctionWhy is the integral of $(nablapsi)^2$ the same as the integral of $(nabla|psi|)^2$?How to find the actual state vector in Quantum Mechanics?Impossible to decompose EM plane wave in spherical waves? (Normalization mismatch)
.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty margin-bottom:0;
$begingroup$
As we know for the plane waves ( $ae^i k x+b e^-i k x$), the normalization constant can be easily obtained from the integral $int^x_2_x_1psi^*psi dx=1$ by the relation $|a|^2+|b|^2=1$. But what happens if the parameter $k$ is imaginary, i.e. $k=i kappa$ where $kappa$ is real. Do we have the same relation for the normalization?
quantum-mechanics homework-and-exercises hilbert-space wavefunction schroedinger-equation
edited 14 hours ago
Qmechanic♦
112k13 gold badges218 silver badges1327 bronze badges
asked 14 hours ago
BaranBaran
365 bronze badges
$endgroup$
add a comment |
$begingroup$
As we know for the plane waves ( $ae^i k x+b e^-i k x$), the normalization constant can be easily obtained from the integral $int^x_2_x_1psi^*psi dx=1$ by the relation $|a|^2+|b|^2=1$. But what happens if the parameter $k$ is imaginary, i.e. $k=i kappa$ where $kappa$ is real. Do we have the same relation for the normalization?
quantum-mechanics homework-and-exercises hilbert-space wavefunction schroedinger-equation
edited 14 hours ago
Qmechanic♦
112k13 gold badges218 silver badges1327 bronze badges
asked 14 hours ago
BaranBaran
365 bronze badges
$endgroup$
$begingroup$
The question as posted is incomplete. In the question body you talk about plane waves without qualifiers. Then you claim that they can be normalized and exhibit the normalization over a restricted (but apparently arbitrary) range, but don't tell us what the boundary conditions are to be. In a comment on one answer you specify a range, but still don't specify the boundary conditions.
$endgroup$
– dmckee♦
4 hours ago
$begingroup$
I know that sounds unreasonably picky, but learning to specify your question s will help you at least two ways. First by helping you to notice what features of the theory are mathematically important, and secondly by teaching you to notice important things specified in problem problems (both prompts provided by teachers and in problems that come up naturally).
$endgroup$
– dmckee♦
4 hours ago
$begingroup$
Thanks, dmckee. The complete question is here: physics.stackexchange.com/questions/496440/…
$endgroup$
– Baran
3 hours ago
$begingroup$
boundary conditions have been included in the functions f and g in the link above
$endgroup$
– Baran
3 hours ago
add a comment |
$begingroup$
As we know for the plane waves ( $ae^i k x+b e^-i k x$), the normalization constant can be easily obtained from the integral $int^x_2_x_1psi^*psi dx=1$ by the relation $|a|^2+|b|^2=1$. But what happens if the parameter $k$ is imaginary, i.e. $k=i kappa$ where $kappa$ is real. Do we have the same relation for the normalization?
quantum-mechanics homework-and-exercises hilbert-space wavefunction schroedinger-equation
edited 14 hours ago
Qmechanic♦
112k13 gold badges218 silver badges1327 bronze badges
asked 14 hours ago
BaranBaran
365 bronze badges
$endgroup$
As we know for the plane waves ( $ae^i k x+b e^-i k x$), the normalization constant can be easily obtained from the integral $int^x_2_x_1psi^*psi dx=1$ by the relation $|a|^2+|b|^2=1$. But what happens if the parameter $k$ is imaginary, i.e. $k=i kappa$ where $kappa$ is real. Do we have the same relation for the normalization?
quantum-mechanics homework-and-exercises hilbert-space wavefunction schroedinger-equation
quantum-mechanics homework-and-exercises hilbert-space wavefunction schroedinger-equation
edited 14 hours ago
Qmechanic♦
112k13 gold badges218 silver badges1327 bronze badges
asked 14 hours ago
BaranBaran
365 bronze badges
edited 14 hours ago
Qmechanic♦
112k13 gold badges218 silver badges1327 bronze badges
asked 14 hours ago
BaranBaran
365 bronze badges
edited 14 hours ago
Qmechanic♦
112k13 gold badges218 silver badges1327 bronze badges
edited 14 hours ago
Qmechanic♦
112k13 gold badges218 silver badges1327 bronze badges
edited 14 hours ago
Qmechanic♦
112k13 gold badges218 silver badges1327 bronze badges
112k13 gold badges218 silver badges1327 bronze badges
asked 14 hours ago
BaranBaran
365 bronze badges
asked 14 hours ago
BaranBaran
365 bronze badges
asked 14 hours ago
BaranBaran
365 bronze badges
365 bronze badges
$begingroup$
The question as posted is incomplete. In the question body you talk about plane waves without qualifiers. Then you claim that they can be normalized and exhibit the normalization over a restricted (but apparently arbitrary) range, but don't tell us what the boundary conditions are to be. In a comment on one answer you specify a range, but still don't specify the boundary conditions.
$endgroup$
– dmckee♦
4 hours ago
$begingroup$
I know that sounds unreasonably picky, but learning to specify your question s will help you at least two ways. First by helping you to notice what features of the theory are mathematically important, and secondly by teaching you to notice important things specified in problem problems (both prompts provided by teachers and in problems that come up naturally).
$endgroup$
– dmckee♦
4 hours ago
$begingroup$
Thanks, dmckee. The complete question is here: physics.stackexchange.com/questions/496440/…
$endgroup$
– Baran
3 hours ago
$begingroup$
boundary conditions have been included in the functions f and g in the link above
$endgroup$
– Baran
3 hours ago
add a comment |
$begingroup$
The question as posted is incomplete. In the question body you talk about plane waves without qualifiers. Then you claim that they can be normalized and exhibit the normalization over a restricted (but apparently arbitrary) range, but don't tell us what the boundary conditions are to be. In a comment on one answer you specify a range, but still don't specify the boundary conditions.
$endgroup$
– dmckee♦
4 hours ago
$begingroup$
I know that sounds unreasonably picky, but learning to specify your question s will help you at least two ways. First by helping you to notice what features of the theory are mathematically important, and secondly by teaching you to notice important things specified in problem problems (both prompts provided by teachers and in problems that come up naturally).
$endgroup$
– dmckee♦
4 hours ago
$begingroup$
Thanks, dmckee. The complete question is here: physics.stackexchange.com/questions/496440/…
$endgroup$
– Baran
3 hours ago
$begingroup$
boundary conditions have been included in the functions f and g in the link above
$endgroup$
– Baran
3 hours ago
$begingroup$
The question as posted is incomplete. In the question body you talk about plane waves without qualifiers. Then you claim that they can be normalized and exhibit the normalization over a restricted (but apparently arbitrary) range, but don't tell us what the boundary conditions are to be. In a comment on one answer you specify a range, but still don't specify the boundary conditions.
$endgroup$
– dmckee♦
4 hours ago
$begingroup$
The question as posted is incomplete. In the question body you talk about plane waves without qualifiers. Then you claim that they can be normalized and exhibit the normalization over a restricted (but apparently arbitrary) range, but don't tell us what the boundary conditions are to be. In a comment on one answer you specify a range, but still don't specify the boundary conditions.
$endgroup$
– dmckee♦
4 hours ago
$begingroup$
I know that sounds unreasonably picky, but learning to specify your question s will help you at least two ways. First by helping you to notice what features of the theory are mathematically important, and secondly by teaching you to notice important things specified in problem problems (both prompts provided by teachers and in problems that come up naturally).
$endgroup$
– dmckee♦
4 hours ago
$begingroup$
I know that sounds unreasonably picky, but learning to specify your question s will help you at least two ways. First by helping you to notice what features of the theory are mathematically important, and secondly by teaching you to notice important things specified in problem problems (both prompts provided by teachers and in problems that come up naturally).
$endgroup$
– dmckee♦
4 hours ago
$begingroup$
Thanks, dmckee. The complete question is here: physics.stackexchange.com/questions/496440/…
$endgroup$
– Baran
3 hours ago
$begingroup$
Thanks, dmckee. The complete question is here: physics.stackexchange.com/questions/496440/…
$endgroup$
– Baran
3 hours ago
$begingroup$
boundary conditions have been included in the functions f and g in the link above
$endgroup$
– Baran
3 hours ago
$begingroup$
boundary conditions have been included in the functions f and g in the link above
$endgroup$
– Baran
3 hours ago
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Plane waves can't be normalised, because they don't represent physically realisable states. It doesn't make sense to normalise a function like $ psi = ae^ikx + be^-ikx $ over the boundary $(x_1, x_2)$ unless the particle is bounded, in which case the wavefunction will have a different solution. Another way to think about this: "There's no such thing as a free particle with a definite energy." See Griffiths intro to QM section 2.4
answered 14 hours ago
VisipiVisipi
513 bronze badges
New contributor
Visipi is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
1
$begingroup$
Thanks, yes, you are right. Plane wave can be normalized through Dirac delta function. But my question is how can I find the normalization constant $a$ for the wave $a(e^-kappa x+f(alpha) e^kappa x)$ where $kappa$ is real and $xin[0,fracpi2]$. This is a special problem related to negative energies. So, can I use the integral $int^pi/2_0psi^*psi dx=1$ to calculate $a$?
$endgroup$
– Baran
13 hours ago
add a comment |
$begingroup$
Using your parameterization, the wave is $ae^-kappa x+be^kappa x$. Note that this particular wavefunction blows up at $x=+infty,-infty$; so that it cannot be normalized unless we impose $a=0$ for $x<0$ and $b=0$ for $x>0$. If you do this, you can simply carry out an integration to find out the relation between $a$ and $b$ that will normalize the wave.
Remember that $k=sqrt2m(E-V)/hbar$, so that it will be imaginary in regions where $E<V$. In particular consider a wave incident in $x<0$ on a step potential barrier of height $V_0$ for all $x>0$. If $E<V_0$, it will have the form $ae^-kappa x$ at $x>0$, so that the wave actually exists inside the barrier even though the incident energy was less than the barrier height. This is how tunneling happens.
answered 14 hours ago
Mani JhaMani Jha
644 bronze badges
New contributor
Mani Jha is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
1
$begingroup$
Thanks. But here the domain of $x$ is limited, $xin[0,pi/2]$. So, divergence is not a problem here. Also, there is a relation between the two parameters $a$ and $b$. So, we can normalize the wave by the integral $int^pi/2_0psi^*psi dx=1$. Therefore, the relation $ |a|^2+|b|^2=1$ is not the case here, right?
$endgroup$
– Baran
14 hours ago
2
$begingroup$
Yes, because the exponential factors no longer cancel in $psi*psi$. You should also get conditions from the continuity and differentiability of the wave at the boundaries.
$endgroup$
– Mani Jha
13 hours ago
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Plane waves can't be normalised, because they don't represent physically realisable states. It doesn't make sense to normalise a function like $ psi = ae^ikx + be^-ikx $ over the boundary $(x_1, x_2)$ unless the particle is bounded, in which case the wavefunction will have a different solution. Another way to think about this: "There's no such thing as a free particle with a definite energy." See Griffiths intro to QM section 2.4
answered 14 hours ago
VisipiVisipi
513 bronze badges
New contributor
Visipi is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
1
$begingroup$
Thanks, yes, you are right. Plane wave can be normalized through Dirac delta function. But my question is how can I find the normalization constant $a$ for the wave $a(e^-kappa x+f(alpha) e^kappa x)$ where $kappa$ is real and $xin[0,fracpi2]$. This is a special problem related to negative energies. So, can I use the integral $int^pi/2_0psi^*psi dx=1$ to calculate $a$?
$endgroup$
– Baran
13 hours ago
add a comment |
$begingroup$
Plane waves can't be normalised, because they don't represent physically realisable states. It doesn't make sense to normalise a function like $ psi = ae^ikx + be^-ikx $ over the boundary $(x_1, x_2)$ unless the particle is bounded, in which case the wavefunction will have a different solution. Another way to think about this: "There's no such thing as a free particle with a definite energy." See Griffiths intro to QM section 2.4
answered 14 hours ago
VisipiVisipi
513 bronze badges
New contributor
Visipi is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
1
$begingroup$
Thanks, yes, you are right. Plane wave can be normalized through Dirac delta function. But my question is how can I find the normalization constant $a$ for the wave $a(e^-kappa x+f(alpha) e^kappa x)$ where $kappa$ is real and $xin[0,fracpi2]$. This is a special problem related to negative energies. So, can I use the integral $int^pi/2_0psi^*psi dx=1$ to calculate $a$?
$endgroup$
– Baran
13 hours ago
add a comment |
$begingroup$
Plane waves can't be normalised, because they don't represent physically realisable states. It doesn't make sense to normalise a function like $ psi = ae^ikx + be^-ikx $ over the boundary $(x_1, x_2)$ unless the particle is bounded, in which case the wavefunction will have a different solution. Another way to think about this: "There's no such thing as a free particle with a definite energy." See Griffiths intro to QM section 2.4
answered 14 hours ago
VisipiVisipi
513 bronze badges
New contributor
Visipi is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
Plane waves can't be normalised, because they don't represent physically realisable states. It doesn't make sense to normalise a function like $ psi = ae^ikx + be^-ikx $ over the boundary $(x_1, x_2)$ unless the particle is bounded, in which case the wavefunction will have a different solution. Another way to think about this: "There's no such thing as a free particle with a definite energy." See Griffiths intro to QM section 2.4
answered 14 hours ago
VisipiVisipi
513 bronze badges
New contributor
Visipi is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered 14 hours ago
VisipiVisipi
513 bronze badges
New contributor
Visipi is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered 14 hours ago
VisipiVisipi
513 bronze badges
answered 14 hours ago
VisipiVisipi
513 bronze badges
513 bronze badges
New contributor
Visipi is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Visipi is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
1
$begingroup$
Thanks, yes, you are right. Plane wave can be normalized through Dirac delta function. But my question is how can I find the normalization constant $a$ for the wave $a(e^-kappa x+f(alpha) e^kappa x)$ where $kappa$ is real and $xin[0,fracpi2]$. This is a special problem related to negative energies. So, can I use the integral $int^pi/2_0psi^*psi dx=1$ to calculate $a$?
$endgroup$
– Baran
13 hours ago
add a comment |
1
$begingroup$
Thanks, yes, you are right. Plane wave can be normalized through Dirac delta function. But my question is how can I find the normalization constant $a$ for the wave $a(e^-kappa x+f(alpha) e^kappa x)$ where $kappa$ is real and $xin[0,fracpi2]$. This is a special problem related to negative energies. So, can I use the integral $int^pi/2_0psi^*psi dx=1$ to calculate $a$?
$endgroup$
– Baran
13 hours ago
1
1
$begingroup$
Thanks, yes, you are right. Plane wave can be normalized through Dirac delta function. But my question is how can I find the normalization constant $a$ for the wave $a(e^-kappa x+f(alpha) e^kappa x)$ where $kappa$ is real and $xin[0,fracpi2]$. This is a special problem related to negative energies. So, can I use the integral $int^pi/2_0psi^*psi dx=1$ to calculate $a$?
$endgroup$
– Baran
13 hours ago
$begingroup$
Thanks, yes, you are right. Plane wave can be normalized through Dirac delta function. But my question is how can I find the normalization constant $a$ for the wave $a(e^-kappa x+f(alpha) e^kappa x)$ where $kappa$ is real and $xin[0,fracpi2]$. This is a special problem related to negative energies. So, can I use the integral $int^pi/2_0psi^*psi dx=1$ to calculate $a$?
$endgroup$
– Baran
13 hours ago
add a comment |
$begingroup$
Using your parameterization, the wave is $ae^-kappa x+be^kappa x$. Note that this particular wavefunction blows up at $x=+infty,-infty$; so that it cannot be normalized unless we impose $a=0$ for $x<0$ and $b=0$ for $x>0$. If you do this, you can simply carry out an integration to find out the relation between $a$ and $b$ that will normalize the wave.
Remember that $k=sqrt2m(E-V)/hbar$, so that it will be imaginary in regions where $E<V$. In particular consider a wave incident in $x<0$ on a step potential barrier of height $V_0$ for all $x>0$. If $E<V_0$, it will have the form $ae^-kappa x$ at $x>0$, so that the wave actually exists inside the barrier even though the incident energy was less than the barrier height. This is how tunneling happens.
answered 14 hours ago
Mani JhaMani Jha
644 bronze badges
New contributor
Mani Jha is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
1
$begingroup$
Thanks. But here the domain of $x$ is limited, $xin[0,pi/2]$. So, divergence is not a problem here. Also, there is a relation between the two parameters $a$ and $b$. So, we can normalize the wave by the integral $int^pi/2_0psi^*psi dx=1$. Therefore, the relation $ |a|^2+|b|^2=1$ is not the case here, right?
$endgroup$
– Baran
14 hours ago
2
$begingroup$
Yes, because the exponential factors no longer cancel in $psi*psi$. You should also get conditions from the continuity and differentiability of the wave at the boundaries.
$endgroup$
– Mani Jha
13 hours ago
add a comment |
$begingroup$
Using your parameterization, the wave is $ae^-kappa x+be^kappa x$. Note that this particular wavefunction blows up at $x=+infty,-infty$; so that it cannot be normalized unless we impose $a=0$ for $x<0$ and $b=0$ for $x>0$. If you do this, you can simply carry out an integration to find out the relation between $a$ and $b$ that will normalize the wave.
Remember that $k=sqrt2m(E-V)/hbar$, so that it will be imaginary in regions where $E<V$. In particular consider a wave incident in $x<0$ on a step potential barrier of height $V_0$ for all $x>0$. If $E<V_0$, it will have the form $ae^-kappa x$ at $x>0$, so that the wave actually exists inside the barrier even though the incident energy was less than the barrier height. This is how tunneling happens.
answered 14 hours ago
Mani JhaMani Jha
644 bronze badges
New contributor
Mani Jha is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
1
$begingroup$
Thanks. But here the domain of $x$ is limited, $xin[0,pi/2]$. So, divergence is not a problem here. Also, there is a relation between the two parameters $a$ and $b$. So, we can normalize the wave by the integral $int^pi/2_0psi^*psi dx=1$. Therefore, the relation $ |a|^2+|b|^2=1$ is not the case here, right?
$endgroup$
– Baran
14 hours ago
2
$begingroup$
Yes, because the exponential factors no longer cancel in $psi*psi$. You should also get conditions from the continuity and differentiability of the wave at the boundaries.
$endgroup$
– Mani Jha
13 hours ago
add a comment |
$begingroup$
Using your parameterization, the wave is $ae^-kappa x+be^kappa x$. Note that this particular wavefunction blows up at $x=+infty,-infty$; so that it cannot be normalized unless we impose $a=0$ for $x<0$ and $b=0$ for $x>0$. If you do this, you can simply carry out an integration to find out the relation between $a$ and $b$ that will normalize the wave.
Remember that $k=sqrt2m(E-V)/hbar$, so that it will be imaginary in regions where $E<V$. In particular consider a wave incident in $x<0$ on a step potential barrier of height $V_0$ for all $x>0$. If $E<V_0$, it will have the form $ae^-kappa x$ at $x>0$, so that the wave actually exists inside the barrier even though the incident energy was less than the barrier height. This is how tunneling happens.
answered 14 hours ago
Mani JhaMani Jha
644 bronze badges
New contributor
Mani Jha is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
Using your parameterization, the wave is $ae^-kappa x+be^kappa x$. Note that this particular wavefunction blows up at $x=+infty,-infty$; so that it cannot be normalized unless we impose $a=0$ for $x<0$ and $b=0$ for $x>0$. If you do this, you can simply carry out an integration to find out the relation between $a$ and $b$ that will normalize the wave.
Remember that $k=sqrt2m(E-V)/hbar$, so that it will be imaginary in regions where $E<V$. In particular consider a wave incident in $x<0$ on a step potential barrier of height $V_0$ for all $x>0$. If $E<V_0$, it will have the form $ae^-kappa x$ at $x>0$, so that the wave actually exists inside the barrier even though the incident energy was less than the barrier height. This is how tunneling happens.
answered 14 hours ago
Mani JhaMani Jha
644 bronze badges
New contributor
Mani Jha is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 13 hours ago
answered 14 hours ago
Mani JhaMani Jha
644 bronze badges
New contributor
Mani Jha is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered 14 hours ago
Mani JhaMani Jha
644 bronze badges
answered 14 hours ago
Mani JhaMani Jha
644 bronze badges
644 bronze badges
New contributor
Mani Jha is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Mani Jha is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
1
$begingroup$
Thanks. But here the domain of $x$ is limited, $xin[0,pi/2]$. So, divergence is not a problem here. Also, there is a relation between the two parameters $a$ and $b$. So, we can normalize the wave by the integral $int^pi/2_0psi^*psi dx=1$. Therefore, the relation $ |a|^2+|b|^2=1$ is not the case here, right?
$endgroup$
– Baran
14 hours ago
2
$begingroup$
Yes, because the exponential factors no longer cancel in $psi*psi$. You should also get conditions from the continuity and differentiability of the wave at the boundaries.
$endgroup$
– Mani Jha
13 hours ago
add a comment |
1
$begingroup$
Thanks. But here the domain of $x$ is limited, $xin[0,pi/2]$. So, divergence is not a problem here. Also, there is a relation between the two parameters $a$ and $b$. So, we can normalize the wave by the integral $int^pi/2_0psi^*psi dx=1$. Therefore, the relation $ |a|^2+|b|^2=1$ is not the case here, right?
$endgroup$
– Baran
14 hours ago
2
$begingroup$
Yes, because the exponential factors no longer cancel in $psi*psi$. You should also get conditions from the continuity and differentiability of the wave at the boundaries.
$endgroup$
– Mani Jha
13 hours ago
1
1
$begingroup$
Thanks. But here the domain of $x$ is limited, $xin[0,pi/2]$. So, divergence is not a problem here. Also, there is a relation between the two parameters $a$ and $b$. So, we can normalize the wave by the integral $int^pi/2_0psi^*psi dx=1$. Therefore, the relation $ |a|^2+|b|^2=1$ is not the case here, right?
$endgroup$
– Baran
14 hours ago
$begingroup$
Thanks. But here the domain of $x$ is limited, $xin[0,pi/2]$. So, divergence is not a problem here. Also, there is a relation between the two parameters $a$ and $b$. So, we can normalize the wave by the integral $int^pi/2_0psi^*psi dx=1$. Therefore, the relation $ |a|^2+|b|^2=1$ is not the case here, right?
$endgroup$
– Baran
14 hours ago
2
2
$begingroup$
Yes, because the exponential factors no longer cancel in $psi*psi$. You should also get conditions from the continuity and differentiability of the wave at the boundaries.
$endgroup$
– Mani Jha
13 hours ago
$begingroup$
Yes, because the exponential factors no longer cancel in $psi*psi$. You should also get conditions from the continuity and differentiability of the wave at the boundaries.
$endgroup$
– Mani Jha
13 hours ago
add a comment |
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$begingroup$
The question as posted is incomplete. In the question body you talk about plane waves without qualifiers. Then you claim that they can be normalized and exhibit the normalization over a restricted (but apparently arbitrary) range, but don't tell us what the boundary conditions are to be. In a comment on one answer you specify a range, but still don't specify the boundary conditions.
$endgroup$
– dmckee♦
4 hours ago
$begingroup$
I know that sounds unreasonably picky, but learning to specify your question s will help you at least two ways. First by helping you to notice what features of the theory are mathematically important, and secondly by teaching you to notice important things specified in problem problems (both prompts provided by teachers and in problems that come up naturally).
$endgroup$
– dmckee♦
4 hours ago
$begingroup$
Thanks, dmckee. The complete question is here: physics.stackexchange.com/questions/496440/…
$endgroup$
– Baran
3 hours ago
$begingroup$
boundary conditions have been included in the functions f and g in the link above
$endgroup$
– Baran
3 hours ago