Substitution - mistake












0












$begingroup$


Where do I mistake please? My computation differ from the result in the text about red terms. Thank you
enter image description here






substitution quantum-mechanics







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$endgroup$








  • 1




    $begingroup$
    Doesn't look like you made a mistake
    $endgroup$
    – Shubham Johri
    Dec 31 '18 at 20:49










  • $begingroup$
    once again x in equation (85)?
    $endgroup$
    – Elisabeth
    Dec 31 '18 at 20:54






  • 1




    $begingroup$
    In equation (83) should be $y$.
    $endgroup$
    – Andrei
    Dec 31 '18 at 20:56










  • $begingroup$
    Do you appreciate the normalized wavefunctions squared have to integrate to 1 in both cases?
    $endgroup$
    – Cosmas Zachos
    Jan 4 at 17:10
















0












$begingroup$


Where do I mistake please? My computation differ from the result in the text about red terms. Thank you
enter image description here






substitution quantum-mechanics







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Doesn't look like you made a mistake
    $endgroup$
    – Shubham Johri
    Dec 31 '18 at 20:49










  • $begingroup$
    once again x in equation (85)?
    $endgroup$
    – Elisabeth
    Dec 31 '18 at 20:54






  • 1




    $begingroup$
    In equation (83) should be $y$.
    $endgroup$
    – Andrei
    Dec 31 '18 at 20:56










  • $begingroup$
    Do you appreciate the normalized wavefunctions squared have to integrate to 1 in both cases?
    $endgroup$
    – Cosmas Zachos
    Jan 4 at 17:10














0












0








0





$begingroup$


Where do I mistake please? My computation differ from the result in the text about red terms. Thank you
enter image description here






substitution quantum-mechanics







share|cite|improve this question











$endgroup$




Where do I mistake please? My computation differ from the result in the text about red terms. Thank you
enter image description here






substitution quantum-mechanics




substitution quantum-mechanics






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 31 '18 at 21:12









Andrei

13.9k21330




13.9k21330










asked Dec 31 '18 at 20:35









ElisabethElisabeth

1066




1066








  • 1




    $begingroup$
    Doesn't look like you made a mistake
    $endgroup$
    – Shubham Johri
    Dec 31 '18 at 20:49










  • $begingroup$
    once again x in equation (85)?
    $endgroup$
    – Elisabeth
    Dec 31 '18 at 20:54






  • 1




    $begingroup$
    In equation (83) should be $y$.
    $endgroup$
    – Andrei
    Dec 31 '18 at 20:56










  • $begingroup$
    Do you appreciate the normalized wavefunctions squared have to integrate to 1 in both cases?
    $endgroup$
    – Cosmas Zachos
    Jan 4 at 17:10














  • 1




    $begingroup$
    Doesn't look like you made a mistake
    $endgroup$
    – Shubham Johri
    Dec 31 '18 at 20:49










  • $begingroup$
    once again x in equation (85)?
    $endgroup$
    – Elisabeth
    Dec 31 '18 at 20:54






  • 1




    $begingroup$
    In equation (83) should be $y$.
    $endgroup$
    – Andrei
    Dec 31 '18 at 20:56










  • $begingroup$
    Do you appreciate the normalized wavefunctions squared have to integrate to 1 in both cases?
    $endgroup$
    – Cosmas Zachos
    Jan 4 at 17:10








1




1




$begingroup$
Doesn't look like you made a mistake
$endgroup$
– Shubham Johri
Dec 31 '18 at 20:49




$begingroup$
Doesn't look like you made a mistake
$endgroup$
– Shubham Johri
Dec 31 '18 at 20:49












$begingroup$
once again x in equation (85)?
$endgroup$
– Elisabeth
Dec 31 '18 at 20:54




$begingroup$
once again x in equation (85)?
$endgroup$
– Elisabeth
Dec 31 '18 at 20:54




1




1




$begingroup$
In equation (83) should be $y$.
$endgroup$
– Andrei
Dec 31 '18 at 20:56




$begingroup$
In equation (83) should be $y$.
$endgroup$
– Andrei
Dec 31 '18 at 20:56












$begingroup$
Do you appreciate the normalized wavefunctions squared have to integrate to 1 in both cases?
$endgroup$
– Cosmas Zachos
Jan 4 at 17:10




$begingroup$
Do you appreciate the normalized wavefunctions squared have to integrate to 1 in both cases?
$endgroup$
– Cosmas Zachos
Jan 4 at 17:10










1 Answer
1






active

oldest

votes


















2












$begingroup$

When you do the substitution, you do it so that the integral is easier to calculate. $$int_X|Psi_0(x)|^2dx=<0|0>=int_Y|Psi_0(y)|^2dy$$
That means that you need to account for $frac{dx}{dy}$ term, or in fact the square root of that.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    thank you and where is a diferential when I compute $psi$? I thought about dx = x dy but I don't know the reason for that
    $endgroup$
    – Elisabeth
    Dec 31 '18 at 20:58












  • $begingroup$
    When you change variables $int f(x) dx=int f(y(x))frac{dx}{dy(x)}dy$
    $endgroup$
    – Andrei
    Dec 31 '18 at 21:02












  • $begingroup$
    How to think of the integral please?
    $endgroup$
    – Elisabeth
    Dec 31 '18 at 21:05












  • $begingroup$
    Sorry, I don't understand the question. What exactly are you asking?
    $endgroup$
    – Andrei
    Dec 31 '18 at 21:06






  • 1




    $begingroup$
    Let us continue this discussion in chat.
    $endgroup$
    – Andrei
    Dec 31 '18 at 21:18












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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









2












$begingroup$

When you do the substitution, you do it so that the integral is easier to calculate. $$int_X|Psi_0(x)|^2dx=<0|0>=int_Y|Psi_0(y)|^2dy$$
That means that you need to account for $frac{dx}{dy}$ term, or in fact the square root of that.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    thank you and where is a diferential when I compute $psi$? I thought about dx = x dy but I don't know the reason for that
    $endgroup$
    – Elisabeth
    Dec 31 '18 at 20:58












  • $begingroup$
    When you change variables $int f(x) dx=int f(y(x))frac{dx}{dy(x)}dy$
    $endgroup$
    – Andrei
    Dec 31 '18 at 21:02












  • $begingroup$
    How to think of the integral please?
    $endgroup$
    – Elisabeth
    Dec 31 '18 at 21:05












  • $begingroup$
    Sorry, I don't understand the question. What exactly are you asking?
    $endgroup$
    – Andrei
    Dec 31 '18 at 21:06






  • 1




    $begingroup$
    Let us continue this discussion in chat.
    $endgroup$
    – Andrei
    Dec 31 '18 at 21:18
















2












$begingroup$

When you do the substitution, you do it so that the integral is easier to calculate. $$int_X|Psi_0(x)|^2dx=<0|0>=int_Y|Psi_0(y)|^2dy$$
That means that you need to account for $frac{dx}{dy}$ term, or in fact the square root of that.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    thank you and where is a diferential when I compute $psi$? I thought about dx = x dy but I don't know the reason for that
    $endgroup$
    – Elisabeth
    Dec 31 '18 at 20:58












  • $begingroup$
    When you change variables $int f(x) dx=int f(y(x))frac{dx}{dy(x)}dy$
    $endgroup$
    – Andrei
    Dec 31 '18 at 21:02












  • $begingroup$
    How to think of the integral please?
    $endgroup$
    – Elisabeth
    Dec 31 '18 at 21:05












  • $begingroup$
    Sorry, I don't understand the question. What exactly are you asking?
    $endgroup$
    – Andrei
    Dec 31 '18 at 21:06






  • 1




    $begingroup$
    Let us continue this discussion in chat.
    $endgroup$
    – Andrei
    Dec 31 '18 at 21:18














2












2








2





$begingroup$

When you do the substitution, you do it so that the integral is easier to calculate. $$int_X|Psi_0(x)|^2dx=<0|0>=int_Y|Psi_0(y)|^2dy$$
That means that you need to account for $frac{dx}{dy}$ term, or in fact the square root of that.






share|cite|improve this answer









$endgroup$



When you do the substitution, you do it so that the integral is easier to calculate. $$int_X|Psi_0(x)|^2dx=<0|0>=int_Y|Psi_0(y)|^2dy$$
That means that you need to account for $frac{dx}{dy}$ term, or in fact the square root of that.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 31 '18 at 20:54









AndreiAndrei

13.9k21330




13.9k21330












  • $begingroup$
    thank you and where is a diferential when I compute $psi$? I thought about dx = x dy but I don't know the reason for that
    $endgroup$
    – Elisabeth
    Dec 31 '18 at 20:58












  • $begingroup$
    When you change variables $int f(x) dx=int f(y(x))frac{dx}{dy(x)}dy$
    $endgroup$
    – Andrei
    Dec 31 '18 at 21:02












  • $begingroup$
    How to think of the integral please?
    $endgroup$
    – Elisabeth
    Dec 31 '18 at 21:05












  • $begingroup$
    Sorry, I don't understand the question. What exactly are you asking?
    $endgroup$
    – Andrei
    Dec 31 '18 at 21:06






  • 1




    $begingroup$
    Let us continue this discussion in chat.
    $endgroup$
    – Andrei
    Dec 31 '18 at 21:18


















  • $begingroup$
    thank you and where is a diferential when I compute $psi$? I thought about dx = x dy but I don't know the reason for that
    $endgroup$
    – Elisabeth
    Dec 31 '18 at 20:58












  • $begingroup$
    When you change variables $int f(x) dx=int f(y(x))frac{dx}{dy(x)}dy$
    $endgroup$
    – Andrei
    Dec 31 '18 at 21:02












  • $begingroup$
    How to think of the integral please?
    $endgroup$
    – Elisabeth
    Dec 31 '18 at 21:05












  • $begingroup$
    Sorry, I don't understand the question. What exactly are you asking?
    $endgroup$
    – Andrei
    Dec 31 '18 at 21:06






  • 1




    $begingroup$
    Let us continue this discussion in chat.
    $endgroup$
    – Andrei
    Dec 31 '18 at 21:18
















$begingroup$
thank you and where is a diferential when I compute $psi$? I thought about dx = x dy but I don't know the reason for that
$endgroup$
– Elisabeth
Dec 31 '18 at 20:58






$begingroup$
thank you and where is a diferential when I compute $psi$? I thought about dx = x dy but I don't know the reason for that
$endgroup$
– Elisabeth
Dec 31 '18 at 20:58














$begingroup$
When you change variables $int f(x) dx=int f(y(x))frac{dx}{dy(x)}dy$
$endgroup$
– Andrei
Dec 31 '18 at 21:02






$begingroup$
When you change variables $int f(x) dx=int f(y(x))frac{dx}{dy(x)}dy$
$endgroup$
– Andrei
Dec 31 '18 at 21:02














$begingroup$
How to think of the integral please?
$endgroup$
– Elisabeth
Dec 31 '18 at 21:05






$begingroup$
How to think of the integral please?
$endgroup$
– Elisabeth
Dec 31 '18 at 21:05














$begingroup$
Sorry, I don't understand the question. What exactly are you asking?
$endgroup$
– Andrei
Dec 31 '18 at 21:06




$begingroup$
Sorry, I don't understand the question. What exactly are you asking?
$endgroup$
– Andrei
Dec 31 '18 at 21:06




1




1




$begingroup$
Let us continue this discussion in chat.
$endgroup$
– Andrei
Dec 31 '18 at 21:18




$begingroup$
Let us continue this discussion in chat.
$endgroup$
– Andrei
Dec 31 '18 at 21:18


















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