Substitution - mistake
0
$begingroup$
Where do I mistake please? My computation differ from the result in the text about red terms. Thank you
substitution quantum-mechanics
edited Dec 31 '18 at 21:12
Andrei
13.9k21330
asked Dec 31 '18 at 20:35
ElisabethElisabeth
1066
$endgroup$
add a comment |
0
$begingroup$
Where do I mistake please? My computation differ from the result in the text about red terms. Thank you
substitution quantum-mechanics
edited Dec 31 '18 at 21:12
Andrei
13.9k21330
asked Dec 31 '18 at 20:35
ElisabethElisabeth
1066
$endgroup$
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
add a comment |
0
0
0
$begingroup$
Where do I mistake please? My computation differ from the result in the text about red terms. Thank you
substitution quantum-mechanics
edited Dec 31 '18 at 21:12
Andrei
13.9k21330
asked Dec 31 '18 at 20:35
ElisabethElisabeth
1066
$endgroup$
Where do I mistake please? My computation differ from the result in the text about red terms. Thank you
substitution quantum-mechanics
substitution quantum-mechanics
edited Dec 31 '18 at 21:12
Andrei
13.9k21330
asked Dec 31 '18 at 20:35
ElisabethElisabeth
1066
edited Dec 31 '18 at 21:12
Andrei
13.9k21330
asked Dec 31 '18 at 20:35
ElisabethElisabeth
1066
edited Dec 31 '18 at 21:12
Andrei
13.9k21330
edited Dec 31 '18 at 21:12
Andrei
13.9k21330
edited Dec 31 '18 at 21:12
Andrei
13.9k21330
13.9k21330
asked Dec 31 '18 at 20:35
ElisabethElisabeth
1066
asked Dec 31 '18 at 20:35
ElisabethElisabeth
1066
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
add a comment |
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
add a comment |
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.
answered Dec 31 '18 at 20:54
AndreiAndrei
13.9k21330
$endgroup$
$begingroup$
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
|
show 3 more comments
Your Answer
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
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active
oldest
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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.
answered Dec 31 '18 at 20:54
AndreiAndrei
13.9k21330
$endgroup$
$begingroup$
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
|
show 3 more comments
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.
answered Dec 31 '18 at 20:54
AndreiAndrei
13.9k21330
$endgroup$
$begingroup$
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
|
show 3 more comments
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.
answered Dec 31 '18 at 20:54
AndreiAndrei
13.9k21330
$endgroup$
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.
answered Dec 31 '18 at 20:54
AndreiAndrei
13.9k21330
answered Dec 31 '18 at 20:54
AndreiAndrei
13.9k21330
answered Dec 31 '18 at 20:54
AndreiAndrei
13.9k21330
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
|
show 3 more comments
$begingroup$
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
|
show 3 more comments
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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