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3

$begingroup$

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

share|cite|improve this question

edited Aug 12 at 11:26

Qmechanic♦

113k1313 gold badges220220 silver badges13371337 bronze badges

asked Aug 12 at 10:59

BaranBaran

3655 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♦
  Aug 12 at 21:09
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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♦
  Aug 12 at 21:15
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thanks, dmckee. The complete question is here: physics.stackexchange.com/questions/496440/…
  $endgroup$
  – Baran
  Aug 12 at 22:52
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  boundary conditions have been included in the functions f and g in the link above
  $endgroup$
  – Baran
  Aug 12 at 22:53
  

  
  
  
  

add a comment | 

3

$begingroup$

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

share|cite|improve this question

edited Aug 12 at 11:26

Qmechanic♦

113k1313 gold badges220220 silver badges13371337 bronze badges

asked Aug 12 at 10:59

BaranBaran

3655 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♦
  Aug 12 at 21:09
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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♦
  Aug 12 at 21:15
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thanks, dmckee. The complete question is here: physics.stackexchange.com/questions/496440/…
  $endgroup$
  – Baran
  Aug 12 at 22:52
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  boundary conditions have been included in the functions f and g in the link above
  $endgroup$
  – Baran
  Aug 12 at 22:53
  

  
  
  
  

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

share|cite|improve this question

edited Aug 12 at 11:26

Qmechanic♦

113k1313 gold badges220220 silver badges13371337 bronze badges

asked Aug 12 at 10:59

BaranBaran

3655 bronze badges

$endgroup$

share|cite|improve this question

edited Aug 12 at 11:26

Qmechanic♦

113k1313 gold badges220220 silver badges13371337 bronze badges

asked Aug 12 at 10:59

BaranBaran

3655 bronze badges

share|cite|improve this question

edited Aug 12 at 11:26

Qmechanic♦

113k1313 gold badges220220 silver badges13371337 bronze badges

asked Aug 12 at 10:59

BaranBaran

3655 bronze badges

edited Aug 12 at 11:26

Qmechanic♦

113k1313 gold badges220220 silver badges13371337 bronze badges

edited Aug 12 at 11:26

Qmechanic♦

113k1313 gold badges220220 silver badges13371337 bronze badges

asked Aug 12 at 10:59

BaranBaran

3655 bronze badges

asked Aug 12 at 10:59

BaranBaran

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2 Answers
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5

$begingroup$

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

share|cite|improve this answer

answered Aug 12 at 11:49

VisipiVisipi

7966 bronze badges

$endgroup$

• 
  
  
  

  
  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
  Aug 12 at 12:00
  
  

  
  
  
  

add a comment | 

5

$begingroup$

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.

share|cite|improve this answer

edited Aug 12 at 12:04

answered Aug 12 at 11:29

Mani JhaMani Jha

8455 bronze badges

$endgroup$

• 
  
  
  

  
  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
  Aug 12 at 11:53
  
  

  
  
  
  


• 
  
  
  

  
  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
  Aug 12 at 11:59
  
  

  
  
  
  

add a comment | 

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5

$begingroup$

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

share|cite|improve this answer

answered Aug 12 at 11:49

VisipiVisipi

7966 bronze badges

$endgroup$

• 
  
  
  

  
  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
  Aug 12 at 12:00
  
  

  
  
  
  

add a comment | 

5

$begingroup$

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

share|cite|improve this answer

answered Aug 12 at 11:49

VisipiVisipi

7966 bronze badges

$endgroup$

• 
  
  
  

  
  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
  Aug 12 at 12:00
  
  

  
  
  
  

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

share|cite|improve this answer

answered Aug 12 at 11:49

VisipiVisipi

7966 bronze badges

$endgroup$

share|cite|improve this answer

answered Aug 12 at 11:49

VisipiVisipi

7966 bronze badges

answered Aug 12 at 11:49

VisipiVisipi

7966 bronze badges

answered Aug 12 at 11:49

VisipiVisipi

7966 bronze badges

5

$begingroup$

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.

share|cite|improve this answer

edited Aug 12 at 12:04

answered Aug 12 at 11:29

Mani JhaMani Jha

8455 bronze badges

$endgroup$

• 
  
  
  

  
  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
  Aug 12 at 11:53
  
  

  
  
  
  


• 
  
  
  

  
  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
  Aug 12 at 11:59
  
  

  
  
  
  

add a comment | 

5

$begingroup$

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.

share|cite|improve this answer

edited Aug 12 at 12:04

answered Aug 12 at 11:29

Mani JhaMani Jha

8455 bronze badges

$endgroup$

• 
  
  
  

  
  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
  Aug 12 at 11:53
  
  

  
  
  
  


• 
  
  
  

  
  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
  Aug 12 at 11:59
  
  

  
  
  
  

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.

share|cite|improve this answer

edited Aug 12 at 12:04

answered Aug 12 at 11:29

Mani JhaMani Jha

8455 bronze badges

$endgroup$

share|cite|improve this answer

edited Aug 12 at 12:04

answered Aug 12 at 11:29

Mani JhaMani Jha

8455 bronze badges

answered Aug 12 at 11:29

Mani JhaMani Jha

8455 bronze badges

answered Aug 12 at 11:29

Mani JhaMani Jha

8455 bronze badges