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A formula for delta function in quantum mechanics

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2

$begingroup$

I met a formula for delta function in a QM book ( not in English). The formula is used in the scattering theory. Its form is
$$lim_alpharightarrow inftyexp[ialpha x]=2idelta(x), ~~(xgeq 0).$$ Before, I never saw this form. Is this formula correct? Where can I find relevant references?

quantum-mechanics dirac-delta-distributions

share|cite|improve this question

asked 12 hours ago

user10709user10709

562

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Which non-English QM book?
  $endgroup$
  – Qmechanic♦
  12 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Qmechanic The QM book is not written in English.
  $endgroup$
  – user10709
  12 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Even as a teacher of QM for 25 years I never saw this formula. I must check if it does make sense on terms of distributions...
  $endgroup$
  – Jhor
  11 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I don't think it can be an alternative definition of the Delta function. The Delta function needs to at least span in a neighborhood of $x=0$ for the integration properties of the Delta function to have a meaning. Note that I am not implying that the above relationship is wrong; it could be satisfied for $x>0$, but it won't serve as a definition of $delta (X)$.
  $endgroup$
  – Daniel Duque
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @user10709 Which QM book? (That's what Qmechanic asked).
  $endgroup$
  – Alex Strasser
  7 hours ago
  

  
  
  
  

 | 
show 1 more comment

2

$begingroup$

I met a formula for delta function in a QM book ( not in English). The formula is used in the scattering theory. Its form is
$$lim_alpharightarrow inftyexp[ialpha x]=2idelta(x), ~~(xgeq 0).$$ Before, I never saw this form. Is this formula correct? Where can I find relevant references?

quantum-mechanics dirac-delta-distributions

share|cite|improve this question

asked 12 hours ago

user10709user10709

562

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Which non-English QM book?
  $endgroup$
  – Qmechanic♦
  12 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Qmechanic The QM book is not written in English.
  $endgroup$
  – user10709
  12 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Even as a teacher of QM for 25 years I never saw this formula. I must check if it does make sense on terms of distributions...
  $endgroup$
  – Jhor
  11 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I don't think it can be an alternative definition of the Delta function. The Delta function needs to at least span in a neighborhood of $x=0$ for the integration properties of the Delta function to have a meaning. Note that I am not implying that the above relationship is wrong; it could be satisfied for $x>0$, but it won't serve as a definition of $delta (X)$.
  $endgroup$
  – Daniel Duque
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @user10709 Which QM book? (That's what Qmechanic asked).
  $endgroup$
  – Alex Strasser
  7 hours ago
  

  
  
  
  

 | 
show 1 more comment

$begingroup$

I met a formula for delta function in a QM book ( not in English). The formula is used in the scattering theory. Its form is
$$lim_alpharightarrow inftyexp[ialpha x]=2idelta(x), ~~(xgeq 0).$$ Before, I never saw this form. Is this formula correct? Where can I find relevant references?

quantum-mechanics dirac-delta-distributions

share|cite|improve this question

asked 12 hours ago

user10709user10709

562

$endgroup$

share|cite|improve this question

asked 12 hours ago

user10709user10709

562

share|cite|improve this question

asked 12 hours ago

user10709user10709

562

asked 12 hours ago

user10709user10709

562

asked 12 hours ago

user10709user10709

562

2 Answers
2

active

oldest

votes

2

$begingroup$

The Fourier transform of a smooth test function $f(x)$ defined on the whole real line will be tend to zero as $alphato infty$. If, however, it is non-zero only for $xge0$ and has a jump at the origin from $0$ on the negative axis to $f(0)$, and is thereafter smooth ,then it will go to zero as $-f(0)/ialpha$. So have you missed out a factor of $alpha$ in your formula?

Something like
$$
lim_alphato infty alpha e^ialpha x= i delta(x)?
$$
I don't know about the "$2$" though....

share|cite|improve this answer

answered 9 hours ago

mike stonemike stone

7,2251225

$endgroup$

add a comment | 

2

$begingroup$

First of all, it is the book's author's responsibility to prove the formula or give a reference. Second, I suspect the formula is wrong. My reasoning (which is not conclusive) is as follows.

As you know, delta-function is not a function, but a distribution, and for a test function $f(x)$ we have $int_-infty^infty f(x)delta(x)dx=f(0)$. Let us assume (and this may be controversial) that $int_0^infty f(x)delta(x)dx=frac12f(0)$. Then, assuming the formula is correct, we obtain $$int_0^infty f(x)frac12ilim_alphatoinftyexp(ialpha x)dx=frac12f(0).$$Let us try $f(x)=exp(-s x)$, where $s>0$. We obtain:$$int_0^infty exp(-s x)frac12ilim_alphatoinftyexp(ialpha x)dx=frac12ilim_alphatoinftyfrac-1ialpha-s=frac12.$$This does not look correct.

EDIT (3/16/2019): @mike stone's suggestion may make the formula correct: if we consider the following initial formula: $$lim_alphatoinftyalphaexp(ialpha x)=2idelta(x)$$ for $xgeq 0$, my calculation seems to give a correct result.

share|cite|improve this answer

edited 8 hours ago

answered 10 hours ago

akhmeteliakhmeteli

18.4k21843

$endgroup$

add a comment | 

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2

$begingroup$

The Fourier transform of a smooth test function $f(x)$ defined on the whole real line will be tend to zero as $alphato infty$. If, however, it is non-zero only for $xge0$ and has a jump at the origin from $0$ on the negative axis to $f(0)$, and is thereafter smooth ,then it will go to zero as $-f(0)/ialpha$. So have you missed out a factor of $alpha$ in your formula?

Something like
$$
lim_alphato infty alpha e^ialpha x= i delta(x)?
$$
I don't know about the "$2$" though....

share|cite|improve this answer

answered 9 hours ago

mike stonemike stone

7,2251225

$endgroup$

add a comment | 

2

$begingroup$

The Fourier transform of a smooth test function $f(x)$ defined on the whole real line will be tend to zero as $alphato infty$. If, however, it is non-zero only for $xge0$ and has a jump at the origin from $0$ on the negative axis to $f(0)$, and is thereafter smooth ,then it will go to zero as $-f(0)/ialpha$. So have you missed out a factor of $alpha$ in your formula?

Something like
$$
lim_alphato infty alpha e^ialpha x= i delta(x)?
$$
I don't know about the "$2$" though....

share|cite|improve this answer

answered 9 hours ago

mike stonemike stone

7,2251225

$endgroup$

add a comment | 

$begingroup$

The Fourier transform of a smooth test function $f(x)$ defined on the whole real line will be tend to zero as $alphato infty$. If, however, it is non-zero only for $xge0$ and has a jump at the origin from $0$ on the negative axis to $f(0)$, and is thereafter smooth ,then it will go to zero as $-f(0)/ialpha$. So have you missed out a factor of $alpha$ in your formula?

Something like
$$
lim_alphato infty alpha e^ialpha x= i delta(x)?
$$
I don't know about the "$2$" though....

share|cite|improve this answer

answered 9 hours ago

mike stonemike stone

7,2251225

$endgroup$

share|cite|improve this answer

answered 9 hours ago

mike stonemike stone

7,2251225

answered 9 hours ago

mike stonemike stone

7,2251225

answered 9 hours ago

mike stonemike stone

7,2251225

2

$begingroup$

First of all, it is the book's author's responsibility to prove the formula or give a reference. Second, I suspect the formula is wrong. My reasoning (which is not conclusive) is as follows.

As you know, delta-function is not a function, but a distribution, and for a test function $f(x)$ we have $int_-infty^infty f(x)delta(x)dx=f(0)$. Let us assume (and this may be controversial) that $int_0^infty f(x)delta(x)dx=frac12f(0)$. Then, assuming the formula is correct, we obtain $$int_0^infty f(x)frac12ilim_alphatoinftyexp(ialpha x)dx=frac12f(0).$$Let us try $f(x)=exp(-s x)$, where $s>0$. We obtain:$$int_0^infty exp(-s x)frac12ilim_alphatoinftyexp(ialpha x)dx=frac12ilim_alphatoinftyfrac-1ialpha-s=frac12.$$This does not look correct.

EDIT (3/16/2019): @mike stone's suggestion may make the formula correct: if we consider the following initial formula: $$lim_alphatoinftyalphaexp(ialpha x)=2idelta(x)$$ for $xgeq 0$, my calculation seems to give a correct result.

share|cite|improve this answer

edited 8 hours ago

answered 10 hours ago

akhmeteliakhmeteli

18.4k21843

$endgroup$

add a comment | 

2

$begingroup$

First of all, it is the book's author's responsibility to prove the formula or give a reference. Second, I suspect the formula is wrong. My reasoning (which is not conclusive) is as follows.

As you know, delta-function is not a function, but a distribution, and for a test function $f(x)$ we have $int_-infty^infty f(x)delta(x)dx=f(0)$. Let us assume (and this may be controversial) that $int_0^infty f(x)delta(x)dx=frac12f(0)$. Then, assuming the formula is correct, we obtain $$int_0^infty f(x)frac12ilim_alphatoinftyexp(ialpha x)dx=frac12f(0).$$Let us try $f(x)=exp(-s x)$, where $s>0$. We obtain:$$int_0^infty exp(-s x)frac12ilim_alphatoinftyexp(ialpha x)dx=frac12ilim_alphatoinftyfrac-1ialpha-s=frac12.$$This does not look correct.

EDIT (3/16/2019): @mike stone's suggestion may make the formula correct: if we consider the following initial formula: $$lim_alphatoinftyalphaexp(ialpha x)=2idelta(x)$$ for $xgeq 0$, my calculation seems to give a correct result.

share|cite|improve this answer

edited 8 hours ago

answered 10 hours ago

akhmeteliakhmeteli

18.4k21843

$endgroup$

add a comment | 

$begingroup$

First of all, it is the book's author's responsibility to prove the formula or give a reference. Second, I suspect the formula is wrong. My reasoning (which is not conclusive) is as follows.

As you know, delta-function is not a function, but a distribution, and for a test function $f(x)$ we have $int_-infty^infty f(x)delta(x)dx=f(0)$. Let us assume (and this may be controversial) that $int_0^infty f(x)delta(x)dx=frac12f(0)$. Then, assuming the formula is correct, we obtain $$int_0^infty f(x)frac12ilim_alphatoinftyexp(ialpha x)dx=frac12f(0).$$Let us try $f(x)=exp(-s x)$, where $s>0$. We obtain:$$int_0^infty exp(-s x)frac12ilim_alphatoinftyexp(ialpha x)dx=frac12ilim_alphatoinftyfrac-1ialpha-s=frac12.$$This does not look correct.

EDIT (3/16/2019): @mike stone's suggestion may make the formula correct: if we consider the following initial formula: $$lim_alphatoinftyalphaexp(ialpha x)=2idelta(x)$$ for $xgeq 0$, my calculation seems to give a correct result.

share|cite|improve this answer

edited 8 hours ago

answered 10 hours ago

akhmeteliakhmeteli

18.4k21843

$endgroup$

share|cite|improve this answer

edited 8 hours ago

answered 10 hours ago

akhmeteliakhmeteli

18.4k21843

answered 10 hours ago

akhmeteliakhmeteli

18.4k21843

answered 10 hours ago

akhmeteliakhmeteli

18.4k21843