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Find the value of the constant $A$ given the particle has the wavefunction $psi(x,t=0)=Acos^3left(fracpi x2aright)$
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Consider a particle in an infinite square well with $$V(x)
begincases
= 0 & −a lt x lt a\
to infty & textotherwise
endcases$$
At $t = 0$, the particle has the wavefunction (defined over $−a lt x lt a$):
$$psi(x,t=0)=Acos^3left(fracpi x2aright)$$
Find the value of the constant $A$.
Using De Moivre's theorem I was able to show that $$psi(x,t=0)=Acos^3left(fracpi x2aright)=frac3A4cosleft(fracpi x2aright)+fracA4cosleft(frac3pi x2aright)tag1$$
& using the fact that $$psi=sum_na_nphi_ntag2$$
Using $mathrm(1)$ & $mathrm(2)$ I find that
$$psi(x, t=0)=sum_na_nphi_n=a_1phi_1+a_3phi_3$$
so $a_1=dfrac3A4$, $,,phi_1=cosleft(dfracpi x2aright)$, $,,a_3=dfracA4$, $,,phi_3=cosleft(dfrac3pi x2aright)$
Now using
$$sum_n=-infty^inftylvert a_n rvert^2=1$$
So $$left(frac3A4right)^2+left(fracA4right)^2=1$$
Which, on rearranging, gives $$A=sqrtfrac85$$
The problem is that the correct answer is $$A=sqrtfrac85a$$
I am very curious about why that factor of $a$ is part of the normalization constant $mathrmA$ since it is also half the width of the well and appears in the argument of the cosine eigenstate. How can this normalization constant have a property of the system?
I have applied what I thought was the correct logic. But it seems I am missing something. Does anyone have any idea how the author reached that answer?
EDIT:
I have been given two answers to this question, thank you to those that took the time to answer.
There is still one thing I can't understand, and unfortuanately I will have to upload the full question and solution in order to get the point across. Below is a 2nd year undergraduate physics assessed problem:
Here is the professors solution to the first part of the question:
and here is the solution to the second part of the question:
Many thanks.
proof-verification mathematical-physics quantum-mechanics
asked Jul 20 at 23:40
BLAZE
5,88992653
add a comment |Â
up vote
2
down vote
favorite
Consider a particle in an infinite square well with $$V(x)
begincases
= 0 & −a lt x lt a\
to infty & textotherwise
endcases$$
At $t = 0$, the particle has the wavefunction (defined over $−a lt x lt a$):
$$psi(x,t=0)=Acos^3left(fracpi x2aright)$$
Find the value of the constant $A$.
Using De Moivre's theorem I was able to show that $$psi(x,t=0)=Acos^3left(fracpi x2aright)=frac3A4cosleft(fracpi x2aright)+fracA4cosleft(frac3pi x2aright)tag1$$
& using the fact that $$psi=sum_na_nphi_ntag2$$
Using $mathrm(1)$ & $mathrm(2)$ I find that
$$psi(x, t=0)=sum_na_nphi_n=a_1phi_1+a_3phi_3$$
so $a_1=dfrac3A4$, $,,phi_1=cosleft(dfracpi x2aright)$, $,,a_3=dfracA4$, $,,phi_3=cosleft(dfrac3pi x2aright)$
Now using
$$sum_n=-infty^inftylvert a_n rvert^2=1$$
So $$left(frac3A4right)^2+left(fracA4right)^2=1$$
Which, on rearranging, gives $$A=sqrtfrac85$$
The problem is that the correct answer is $$A=sqrtfrac85a$$
I am very curious about why that factor of $a$ is part of the normalization constant $mathrmA$ since it is also half the width of the well and appears in the argument of the cosine eigenstate. How can this normalization constant have a property of the system?
I have applied what I thought was the correct logic. But it seems I am missing something. Does anyone have any idea how the author reached that answer?
EDIT:
I have been given two answers to this question, thank you to those that took the time to answer.
There is still one thing I can't understand, and unfortuanately I will have to upload the full question and solution in order to get the point across. Below is a 2nd year undergraduate physics assessed problem:
Here is the professors solution to the first part of the question:
and here is the solution to the second part of the question:
Many thanks.
proof-verification mathematical-physics quantum-mechanics
asked Jul 20 at 23:40
BLAZE
5,88992653
1
@ArnaudMortier The voltage function $V$ is a part of the differential (Schrodinger's) equation to which $psi$ is a solution. For more, see the particle in a box wiki page.
– Omnomnomnom
Jul 20 at 23:58
@BLAZE "Irrelavant to the problem" isn't quite the right way to put it, but hopefully you see what I mean. Finally, the error in your approach is that your $phi_1$ and $phi_3$ are not normalized eigenstates; they need to be scaled so that $int phi_j^2 dx= 1$
– Omnomnomnom
Jul 21 at 0:25
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
Consider a particle in an infinite square well with $$V(x)
begincases
= 0 & −a lt x lt a\
to infty & textotherwise
endcases$$
At $t = 0$, the particle has the wavefunction (defined over $−a lt x lt a$):
$$psi(x,t=0)=Acos^3left(fracpi x2aright)$$
Find the value of the constant $A$.
Using De Moivre's theorem I was able to show that $$psi(x,t=0)=Acos^3left(fracpi x2aright)=frac3A4cosleft(fracpi x2aright)+fracA4cosleft(frac3pi x2aright)tag1$$
& using the fact that $$psi=sum_na_nphi_ntag2$$
Using $mathrm(1)$ & $mathrm(2)$ I find that
$$psi(x, t=0)=sum_na_nphi_n=a_1phi_1+a_3phi_3$$
so $a_1=dfrac3A4$, $,,phi_1=cosleft(dfracpi x2aright)$, $,,a_3=dfracA4$, $,,phi_3=cosleft(dfrac3pi x2aright)$
Now using
$$sum_n=-infty^inftylvert a_n rvert^2=1$$
So $$left(frac3A4right)^2+left(fracA4right)^2=1$$
Which, on rearranging, gives $$A=sqrtfrac85$$
The problem is that the correct answer is $$A=sqrtfrac85a$$
I am very curious about why that factor of $a$ is part of the normalization constant $mathrmA$ since it is also half the width of the well and appears in the argument of the cosine eigenstate. How can this normalization constant have a property of the system?
I have applied what I thought was the correct logic. But it seems I am missing something. Does anyone have any idea how the author reached that answer?
EDIT:
I have been given two answers to this question, thank you to those that took the time to answer.
There is still one thing I can't understand, and unfortuanately I will have to upload the full question and solution in order to get the point across. Below is a 2nd year undergraduate physics assessed problem:
Here is the professors solution to the first part of the question:
and here is the solution to the second part of the question:
Many thanks.
proof-verification mathematical-physics quantum-mechanics
asked Jul 20 at 23:40
BLAZE
5,88992653
Consider a particle in an infinite square well with $$V(x)
begincases
= 0 & −a lt x lt a\
to infty & textotherwise
endcases$$
At $t = 0$, the particle has the wavefunction (defined over $−a lt x lt a$):
$$psi(x,t=0)=Acos^3left(fracpi x2aright)$$
Find the value of the constant $A$.
Using De Moivre's theorem I was able to show that $$psi(x,t=0)=Acos^3left(fracpi x2aright)=frac3A4cosleft(fracpi x2aright)+fracA4cosleft(frac3pi x2aright)tag1$$
& using the fact that $$psi=sum_na_nphi_ntag2$$
Using $mathrm(1)$ & $mathrm(2)$ I find that
$$psi(x, t=0)=sum_na_nphi_n=a_1phi_1+a_3phi_3$$
so $a_1=dfrac3A4$, $,,phi_1=cosleft(dfracpi x2aright)$, $,,a_3=dfracA4$, $,,phi_3=cosleft(dfrac3pi x2aright)$
Now using
$$sum_n=-infty^inftylvert a_n rvert^2=1$$
So $$left(frac3A4right)^2+left(fracA4right)^2=1$$
Which, on rearranging, gives $$A=sqrtfrac85$$
The problem is that the correct answer is $$A=sqrtfrac85a$$
I am very curious about why that factor of $a$ is part of the normalization constant $mathrmA$ since it is also half the width of the well and appears in the argument of the cosine eigenstate. How can this normalization constant have a property of the system?
I have applied what I thought was the correct logic. But it seems I am missing something. Does anyone have any idea how the author reached that answer?
EDIT:
I have been given two answers to this question, thank you to those that took the time to answer.
There is still one thing I can't understand, and unfortuanately I will have to upload the full question and solution in order to get the point across. Below is a 2nd year undergraduate physics assessed problem:
Here is the professors solution to the first part of the question:
and here is the solution to the second part of the question:
Many thanks.
proof-verification mathematical-physics quantum-mechanics
asked Jul 20 at 23:40
BLAZE
5,88992653
edited Aug 2 at 6:16
asked Jul 20 at 23:40
BLAZE
5,88992653
asked Jul 20 at 23:40
BLAZE
5,88992653
asked Jul 20 at 23:40
BLAZE
5,88992653
5,88992653
1
@ArnaudMortier The voltage function $V$ is a part of the differential (Schrodinger's) equation to which $psi$ is a solution. For more, see the particle in a box wiki page.
– Omnomnomnom
Jul 20 at 23:58
@BLAZE "Irrelavant to the problem" isn't quite the right way to put it, but hopefully you see what I mean. Finally, the error in your approach is that your $phi_1$ and $phi_3$ are not normalized eigenstates; they need to be scaled so that $int phi_j^2 dx= 1$
– Omnomnomnom
Jul 21 at 0:25
add a comment |Â
1
@ArnaudMortier The voltage function $V$ is a part of the differential (Schrodinger's) equation to which $psi$ is a solution. For more, see the particle in a box wiki page.
– Omnomnomnom
Jul 20 at 23:58
@BLAZE "Irrelavant to the problem" isn't quite the right way to put it, but hopefully you see what I mean. Finally, the error in your approach is that your $phi_1$ and $phi_3$ are not normalized eigenstates; they need to be scaled so that $int phi_j^2 dx= 1$
– Omnomnomnom
Jul 21 at 0:25
1
1
@ArnaudMortier The voltage function $V$ is a part of the differential (Schrodinger's) equation to which $psi$ is a solution. For more, see the particle in a box wiki page.
– Omnomnomnom
Jul 20 at 23:58
@ArnaudMortier The voltage function $V$ is a part of the differential (Schrodinger's) equation to which $psi$ is a solution. For more, see the particle in a box wiki page.
– Omnomnomnom
Jul 20 at 23:58
@BLAZE "Irrelavant to the problem" isn't quite the right way to put it, but hopefully you see what I mean. Finally, the error in your approach is that your $phi_1$ and $phi_3$ are not normalized eigenstates; they need to be scaled so that $int phi_j^2 dx= 1$
– Omnomnomnom
Jul 21 at 0:25
@BLAZE "Irrelavant to the problem" isn't quite the right way to put it, but hopefully you see what I mean. Finally, the error in your approach is that your $phi_1$ and $phi_3$ are not normalized eigenstates; they need to be scaled so that $int phi_j^2 dx= 1$
– Omnomnomnom
Jul 21 at 0:25
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
1
down vote
accepted
I would like to state that you have a mistake. The (normalized) eigenstate function $phi_n$ should look like
$$phi_n(x)=frac1sqrta,cosleft(fracnpi x2aright)text for xin[-a,+a],.$$
This can be easily checked since
$$int_-a^+a,cos^2left(fracnpi x2aright),textdx=a,.$$
Therefore, from (1) and (2) in your question, you should get $$a_1=frac3Asqrta4text and a_3=fracAsqrta4,.$$
answered Jul 21 at 16:30
Batominovski
23.2k22777
That's excellent! (+1) This is exactly what I was hoping to see
– BLAZE
Jul 21 at 16:35
Could you please explain why $1/sqrta$ normalizes the eigenstates
– BLAZE
Jul 21 at 16:38
1
From your study (or wherever you got this problem from), you should get that the $n$-th eigenstate $phi_n$ satisfies $phi_n(x)=alpha_n cosleft(fracnpi x2aright)$ for $xin[-a,+a]$, right? You want to normalize it, using $int_-a^+a,big|phi_n(x)big|^2,textdx=1$. This is where the second equation of my answer comes into play, and you then get $alpha_n^2=frac1a$.
– Batominovski
Jul 21 at 16:41
add a comment |Â
up vote
2
down vote
You're working far too hard. Just normalize your wave function:
$$intlimits_-a^a left| A cos^3 left( pi x over 2 a right) right|^2 dx = A^2 intlimits_-a^a cos^6 left( pi x over 2 aright) dx = 1$$
where the integral is computed symbolically instantly in Mathematica or can be found in a table or can be computed through the trigonometric substitution given by Omnomnomnom, below.
Regardless, this method gives $A = sqrt8 over 5 a$ very quickly indeed.
Incidentally, indeed your (incorrect) answer must depend upon $a$. Qualitatively, if $a$ is large (the normalized wave function is spread out), then its amplitude $A$ must be small. In fact, if you think about the physics, the the amplitude must go down as $1/sqrta$.
answered Jul 20 at 23:49
David G. Stork
7,6672929
1
I think you might be the one working too hard if you're going to try and do it like that.
– BLAZE
Jul 21 at 0:03
1
@BLAZE: Huh? The integral is simple, and the answer appears immediately, avoiding all the series expansions and deMoivre's Theorem you employ. It also gives the correct answer whereas yours does not. Why the down-vote?
– David G. Stork
Jul 21 at 0:06
1
@BLAZE the usual (without series or complex analysis) method of evaluating $int cos^6(u),du$ is to use trig identities: $$ cos^6(u) = frac 12^3[1 + cos(2u)]^3 = frac 18[1 + 3 cos(2u) + 3 cos^2(2u) + cos^3(2u)] $$ the only terms above which have a non-zero integral over a period are $1$ and $3 cos^2(2u)$. The latter can be integrated by applying the same identity again.
– Omnomnomnom
Jul 21 at 0:11
2
@BLAZE: I solved your entire problem, beginning to end, in less than five minutes--I'll bet less time than it took you to write out the MatJax. (Granted, I have a PhD in physics and was a physics professor for many years.) But the core idea of $int | psi(x)|^2 dx = 1$ is so basic I don't know why anyone would use a different method. Regardless, I'll look forward to your "faster" method in 9 or so days.
– David G. Stork
Jul 21 at 0:23
1
@DavidG.Stork I think that your answer would benefit from a quick explanation of how one should compute the integral in question, especially since this is the asker's apparent sticking point. +1 regardless.
– Omnomnomnom
Jul 21 at 0:31
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
I would like to state that you have a mistake. The (normalized) eigenstate function $phi_n$ should look like
$$phi_n(x)=frac1sqrta,cosleft(fracnpi x2aright)text for xin[-a,+a],.$$
This can be easily checked since
$$int_-a^+a,cos^2left(fracnpi x2aright),textdx=a,.$$
Therefore, from (1) and (2) in your question, you should get $$a_1=frac3Asqrta4text and a_3=fracAsqrta4,.$$
answered Jul 21 at 16:30
Batominovski
23.2k22777
That's excellent! (+1) This is exactly what I was hoping to see
– BLAZE
Jul 21 at 16:35
Could you please explain why $1/sqrta$ normalizes the eigenstates
– BLAZE
Jul 21 at 16:38
1
From your study (or wherever you got this problem from), you should get that the $n$-th eigenstate $phi_n$ satisfies $phi_n(x)=alpha_n cosleft(fracnpi x2aright)$ for $xin[-a,+a]$, right? You want to normalize it, using $int_-a^+a,big|phi_n(x)big|^2,textdx=1$. This is where the second equation of my answer comes into play, and you then get $alpha_n^2=frac1a$.
– Batominovski
Jul 21 at 16:41
add a comment |Â
up vote
1
down vote
accepted
I would like to state that you have a mistake. The (normalized) eigenstate function $phi_n$ should look like
$$phi_n(x)=frac1sqrta,cosleft(fracnpi x2aright)text for xin[-a,+a],.$$
This can be easily checked since
$$int_-a^+a,cos^2left(fracnpi x2aright),textdx=a,.$$
Therefore, from (1) and (2) in your question, you should get $$a_1=frac3Asqrta4text and a_3=fracAsqrta4,.$$
answered Jul 21 at 16:30
Batominovski
23.2k22777
That's excellent! (+1) This is exactly what I was hoping to see
– BLAZE
Jul 21 at 16:35
Could you please explain why $1/sqrta$ normalizes the eigenstates
– BLAZE
Jul 21 at 16:38
1
From your study (or wherever you got this problem from), you should get that the $n$-th eigenstate $phi_n$ satisfies $phi_n(x)=alpha_n cosleft(fracnpi x2aright)$ for $xin[-a,+a]$, right? You want to normalize it, using $int_-a^+a,big|phi_n(x)big|^2,textdx=1$. This is where the second equation of my answer comes into play, and you then get $alpha_n^2=frac1a$.
– Batominovski
Jul 21 at 16:41
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
I would like to state that you have a mistake. The (normalized) eigenstate function $phi_n$ should look like
$$phi_n(x)=frac1sqrta,cosleft(fracnpi x2aright)text for xin[-a,+a],.$$
This can be easily checked since
$$int_-a^+a,cos^2left(fracnpi x2aright),textdx=a,.$$
Therefore, from (1) and (2) in your question, you should get $$a_1=frac3Asqrta4text and a_3=fracAsqrta4,.$$
answered Jul 21 at 16:30
Batominovski
23.2k22777
I would like to state that you have a mistake. The (normalized) eigenstate function $phi_n$ should look like
$$phi_n(x)=frac1sqrta,cosleft(fracnpi x2aright)text for xin[-a,+a],.$$
This can be easily checked since
$$int_-a^+a,cos^2left(fracnpi x2aright),textdx=a,.$$
Therefore, from (1) and (2) in your question, you should get $$a_1=frac3Asqrta4text and a_3=fracAsqrta4,.$$
answered Jul 21 at 16:30
Batominovski
23.2k22777
answered Jul 21 at 16:30
Batominovski
23.2k22777
answered Jul 21 at 16:30
Batominovski
23.2k22777
answered Jul 21 at 16:30
Batominovski
23.2k22777
23.2k22777
That's excellent! (+1) This is exactly what I was hoping to see
– BLAZE
Jul 21 at 16:35
Could you please explain why $1/sqrta$ normalizes the eigenstates
– BLAZE
Jul 21 at 16:38
1
From your study (or wherever you got this problem from), you should get that the $n$-th eigenstate $phi_n$ satisfies $phi_n(x)=alpha_n cosleft(fracnpi x2aright)$ for $xin[-a,+a]$, right? You want to normalize it, using $int_-a^+a,big|phi_n(x)big|^2,textdx=1$. This is where the second equation of my answer comes into play, and you then get $alpha_n^2=frac1a$.
– Batominovski
Jul 21 at 16:41
add a comment |Â
That's excellent! (+1) This is exactly what I was hoping to see
– BLAZE
Jul 21 at 16:35
Could you please explain why $1/sqrta$ normalizes the eigenstates
– BLAZE
Jul 21 at 16:38
1
From your study (or wherever you got this problem from), you should get that the $n$-th eigenstate $phi_n$ satisfies $phi_n(x)=alpha_n cosleft(fracnpi x2aright)$ for $xin[-a,+a]$, right? You want to normalize it, using $int_-a^+a,big|phi_n(x)big|^2,textdx=1$. This is where the second equation of my answer comes into play, and you then get $alpha_n^2=frac1a$.
– Batominovski
Jul 21 at 16:41
That's excellent! (+1) This is exactly what I was hoping to see
– BLAZE
Jul 21 at 16:35
That's excellent! (+1) This is exactly what I was hoping to see
– BLAZE
Jul 21 at 16:35
Could you please explain why $1/sqrta$ normalizes the eigenstates
– BLAZE
Jul 21 at 16:38
Could you please explain why $1/sqrta$ normalizes the eigenstates
– BLAZE
Jul 21 at 16:38
1
1
From your study (or wherever you got this problem from), you should get that the $n$-th eigenstate $phi_n$ satisfies $phi_n(x)=alpha_n cosleft(fracnpi x2aright)$ for $xin[-a,+a]$, right? You want to normalize it, using $int_-a^+a,big|phi_n(x)big|^2,textdx=1$. This is where the second equation of my answer comes into play, and you then get $alpha_n^2=frac1a$.
– Batominovski
Jul 21 at 16:41
From your study (or wherever you got this problem from), you should get that the $n$-th eigenstate $phi_n$ satisfies $phi_n(x)=alpha_n cosleft(fracnpi x2aright)$ for $xin[-a,+a]$, right? You want to normalize it, using $int_-a^+a,big|phi_n(x)big|^2,textdx=1$. This is where the second equation of my answer comes into play, and you then get $alpha_n^2=frac1a$.
– Batominovski
Jul 21 at 16:41
add a comment |Â
up vote
2
down vote
You're working far too hard. Just normalize your wave function:
$$intlimits_-a^a left| A cos^3 left( pi x over 2 a right) right|^2 dx = A^2 intlimits_-a^a cos^6 left( pi x over 2 aright) dx = 1$$
where the integral is computed symbolically instantly in Mathematica or can be found in a table or can be computed through the trigonometric substitution given by Omnomnomnom, below.
Regardless, this method gives $A = sqrt8 over 5 a$ very quickly indeed.
Incidentally, indeed your (incorrect) answer must depend upon $a$. Qualitatively, if $a$ is large (the normalized wave function is spread out), then its amplitude $A$ must be small. In fact, if you think about the physics, the the amplitude must go down as $1/sqrta$.
answered Jul 20 at 23:49
David G. Stork
7,6672929
1
I think you might be the one working too hard if you're going to try and do it like that.
– BLAZE
Jul 21 at 0:03
1
@BLAZE: Huh? The integral is simple, and the answer appears immediately, avoiding all the series expansions and deMoivre's Theorem you employ. It also gives the correct answer whereas yours does not. Why the down-vote?
– David G. Stork
Jul 21 at 0:06
1
@BLAZE the usual (without series or complex analysis) method of evaluating $int cos^6(u),du$ is to use trig identities: $$ cos^6(u) = frac 12^3[1 + cos(2u)]^3 = frac 18[1 + 3 cos(2u) + 3 cos^2(2u) + cos^3(2u)] $$ the only terms above which have a non-zero integral over a period are $1$ and $3 cos^2(2u)$. The latter can be integrated by applying the same identity again.
– Omnomnomnom
Jul 21 at 0:11
2
@BLAZE: I solved your entire problem, beginning to end, in less than five minutes--I'll bet less time than it took you to write out the MatJax. (Granted, I have a PhD in physics and was a physics professor for many years.) But the core idea of $int | psi(x)|^2 dx = 1$ is so basic I don't know why anyone would use a different method. Regardless, I'll look forward to your "faster" method in 9 or so days.
– David G. Stork
Jul 21 at 0:23
1
@DavidG.Stork I think that your answer would benefit from a quick explanation of how one should compute the integral in question, especially since this is the asker's apparent sticking point. +1 regardless.
– Omnomnomnom
Jul 21 at 0:31
add a comment |Â
up vote
2
down vote
You're working far too hard. Just normalize your wave function:
$$intlimits_-a^a left| A cos^3 left( pi x over 2 a right) right|^2 dx = A^2 intlimits_-a^a cos^6 left( pi x over 2 aright) dx = 1$$
where the integral is computed symbolically instantly in Mathematica or can be found in a table or can be computed through the trigonometric substitution given by Omnomnomnom, below.
Regardless, this method gives $A = sqrt8 over 5 a$ very quickly indeed.
Incidentally, indeed your (incorrect) answer must depend upon $a$. Qualitatively, if $a$ is large (the normalized wave function is spread out), then its amplitude $A$ must be small. In fact, if you think about the physics, the the amplitude must go down as $1/sqrta$.
answered Jul 20 at 23:49
David G. Stork
7,6672929
1
I think you might be the one working too hard if you're going to try and do it like that.
– BLAZE
Jul 21 at 0:03
1
@BLAZE: Huh? The integral is simple, and the answer appears immediately, avoiding all the series expansions and deMoivre's Theorem you employ. It also gives the correct answer whereas yours does not. Why the down-vote?
– David G. Stork
Jul 21 at 0:06
1
@BLAZE the usual (without series or complex analysis) method of evaluating $int cos^6(u),du$ is to use trig identities: $$ cos^6(u) = frac 12^3[1 + cos(2u)]^3 = frac 18[1 + 3 cos(2u) + 3 cos^2(2u) + cos^3(2u)] $$ the only terms above which have a non-zero integral over a period are $1$ and $3 cos^2(2u)$. The latter can be integrated by applying the same identity again.
– Omnomnomnom
Jul 21 at 0:11
2
@BLAZE: I solved your entire problem, beginning to end, in less than five minutes--I'll bet less time than it took you to write out the MatJax. (Granted, I have a PhD in physics and was a physics professor for many years.) But the core idea of $int | psi(x)|^2 dx = 1$ is so basic I don't know why anyone would use a different method. Regardless, I'll look forward to your "faster" method in 9 or so days.
– David G. Stork
Jul 21 at 0:23
1
@DavidG.Stork I think that your answer would benefit from a quick explanation of how one should compute the integral in question, especially since this is the asker's apparent sticking point. +1 regardless.
– Omnomnomnom
Jul 21 at 0:31
add a comment |Â
up vote
2
down vote
up vote
2
down vote
You're working far too hard. Just normalize your wave function:
$$intlimits_-a^a left| A cos^3 left( pi x over 2 a right) right|^2 dx = A^2 intlimits_-a^a cos^6 left( pi x over 2 aright) dx = 1$$
where the integral is computed symbolically instantly in Mathematica or can be found in a table or can be computed through the trigonometric substitution given by Omnomnomnom, below.
Regardless, this method gives $A = sqrt8 over 5 a$ very quickly indeed.
Incidentally, indeed your (incorrect) answer must depend upon $a$. Qualitatively, if $a$ is large (the normalized wave function is spread out), then its amplitude $A$ must be small. In fact, if you think about the physics, the the amplitude must go down as $1/sqrta$.
answered Jul 20 at 23:49
David G. Stork
7,6672929
You're working far too hard. Just normalize your wave function:
$$intlimits_-a^a left| A cos^3 left( pi x over 2 a right) right|^2 dx = A^2 intlimits_-a^a cos^6 left( pi x over 2 aright) dx = 1$$
where the integral is computed symbolically instantly in Mathematica or can be found in a table or can be computed through the trigonometric substitution given by Omnomnomnom, below.
Regardless, this method gives $A = sqrt8 over 5 a$ very quickly indeed.
Incidentally, indeed your (incorrect) answer must depend upon $a$. Qualitatively, if $a$ is large (the normalized wave function is spread out), then its amplitude $A$ must be small. In fact, if you think about the physics, the the amplitude must go down as $1/sqrta$.
answered Jul 20 at 23:49
David G. Stork
7,6672929
edited Jul 21 at 16:27
answered Jul 20 at 23:49
David G. Stork
7,6672929
answered Jul 20 at 23:49
David G. Stork
7,6672929
answered Jul 20 at 23:49
David G. Stork
7,6672929
7,6672929
1
I think you might be the one working too hard if you're going to try and do it like that.
– BLAZE
Jul 21 at 0:03
1
@BLAZE: Huh? The integral is simple, and the answer appears immediately, avoiding all the series expansions and deMoivre's Theorem you employ. It also gives the correct answer whereas yours does not. Why the down-vote?
– David G. Stork
Jul 21 at 0:06
1
@BLAZE the usual (without series or complex analysis) method of evaluating $int cos^6(u),du$ is to use trig identities: $$ cos^6(u) = frac 12^3[1 + cos(2u)]^3 = frac 18[1 + 3 cos(2u) + 3 cos^2(2u) + cos^3(2u)] $$ the only terms above which have a non-zero integral over a period are $1$ and $3 cos^2(2u)$. The latter can be integrated by applying the same identity again.
– Omnomnomnom
Jul 21 at 0:11
2
@BLAZE: I solved your entire problem, beginning to end, in less than five minutes--I'll bet less time than it took you to write out the MatJax. (Granted, I have a PhD in physics and was a physics professor for many years.) But the core idea of $int | psi(x)|^2 dx = 1$ is so basic I don't know why anyone would use a different method. Regardless, I'll look forward to your "faster" method in 9 or so days.
– David G. Stork
Jul 21 at 0:23
1
@DavidG.Stork I think that your answer would benefit from a quick explanation of how one should compute the integral in question, especially since this is the asker's apparent sticking point. +1 regardless.
– Omnomnomnom
Jul 21 at 0:31
add a comment |Â
1
I think you might be the one working too hard if you're going to try and do it like that.
– BLAZE
Jul 21 at 0:03
1
@BLAZE: Huh? The integral is simple, and the answer appears immediately, avoiding all the series expansions and deMoivre's Theorem you employ. It also gives the correct answer whereas yours does not. Why the down-vote?
– David G. Stork
Jul 21 at 0:06
1
@BLAZE the usual (without series or complex analysis) method of evaluating $int cos^6(u),du$ is to use trig identities: $$ cos^6(u) = frac 12^3[1 + cos(2u)]^3 = frac 18[1 + 3 cos(2u) + 3 cos^2(2u) + cos^3(2u)] $$ the only terms above which have a non-zero integral over a period are $1$ and $3 cos^2(2u)$. The latter can be integrated by applying the same identity again.
– Omnomnomnom
Jul 21 at 0:11
2
@BLAZE: I solved your entire problem, beginning to end, in less than five minutes--I'll bet less time than it took you to write out the MatJax. (Granted, I have a PhD in physics and was a physics professor for many years.) But the core idea of $int | psi(x)|^2 dx = 1$ is so basic I don't know why anyone would use a different method. Regardless, I'll look forward to your "faster" method in 9 or so days.
– David G. Stork
Jul 21 at 0:23
1
@DavidG.Stork I think that your answer would benefit from a quick explanation of how one should compute the integral in question, especially since this is the asker's apparent sticking point. +1 regardless.
– Omnomnomnom
Jul 21 at 0:31
1
1
I think you might be the one working too hard if you're going to try and do it like that.
– BLAZE
Jul 21 at 0:03
I think you might be the one working too hard if you're going to try and do it like that.
– BLAZE
Jul 21 at 0:03
1
1
@BLAZE: Huh? The integral is simple, and the answer appears immediately, avoiding all the series expansions and deMoivre's Theorem you employ. It also gives the correct answer whereas yours does not. Why the down-vote?
– David G. Stork
Jul 21 at 0:06
@BLAZE: Huh? The integral is simple, and the answer appears immediately, avoiding all the series expansions and deMoivre's Theorem you employ. It also gives the correct answer whereas yours does not. Why the down-vote?
– David G. Stork
Jul 21 at 0:06
1
1
@BLAZE the usual (without series or complex analysis) method of evaluating $int cos^6(u),du$ is to use trig identities: $$ cos^6(u) = frac 12^3[1 + cos(2u)]^3 = frac 18[1 + 3 cos(2u) + 3 cos^2(2u) + cos^3(2u)] $$ the only terms above which have a non-zero integral over a period are $1$ and $3 cos^2(2u)$. The latter can be integrated by applying the same identity again.
– Omnomnomnom
Jul 21 at 0:11
@BLAZE the usual (without series or complex analysis) method of evaluating $int cos^6(u),du$ is to use trig identities: $$ cos^6(u) = frac 12^3[1 + cos(2u)]^3 = frac 18[1 + 3 cos(2u) + 3 cos^2(2u) + cos^3(2u)] $$ the only terms above which have a non-zero integral over a period are $1$ and $3 cos^2(2u)$. The latter can be integrated by applying the same identity again.
– Omnomnomnom
Jul 21 at 0:11
2
2
@BLAZE: I solved your entire problem, beginning to end, in less than five minutes--I'll bet less time than it took you to write out the MatJax. (Granted, I have a PhD in physics and was a physics professor for many years.) But the core idea of $int | psi(x)|^2 dx = 1$ is so basic I don't know why anyone would use a different method. Regardless, I'll look forward to your "faster" method in 9 or so days.
– David G. Stork
Jul 21 at 0:23
@BLAZE: I solved your entire problem, beginning to end, in less than five minutes--I'll bet less time than it took you to write out the MatJax. (Granted, I have a PhD in physics and was a physics professor for many years.) But the core idea of $int | psi(x)|^2 dx = 1$ is so basic I don't know why anyone would use a different method. Regardless, I'll look forward to your "faster" method in 9 or so days.
– David G. Stork
Jul 21 at 0:23
1
1
@DavidG.Stork I think that your answer would benefit from a quick explanation of how one should compute the integral in question, especially since this is the asker's apparent sticking point. +1 regardless.
– Omnomnomnom
Jul 21 at 0:31
@DavidG.Stork I think that your answer would benefit from a quick explanation of how one should compute the integral in question, especially since this is the asker's apparent sticking point. +1 regardless.
– Omnomnomnom
Jul 21 at 0:31
add a comment |Â
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1
@ArnaudMortier The voltage function $V$ is a part of the differential (Schrodinger's) equation to which $psi$ is a solution. For more, see the particle in a box wiki page.
– Omnomnomnom
Jul 20 at 23:58
@BLAZE "Irrelavant to the problem" isn't quite the right way to put it, but hopefully you see what I mean. Finally, the error in your approach is that your $phi_1$ and $phi_3$ are not normalized eigenstates; they need to be scaled so that $int phi_j^2 dx= 1$
– Omnomnomnom
Jul 21 at 0:25