Mixing
Why is the position expectation value for this wave function independent of the parameter?
Clash Royale CLAN TAG#URR8PPP
up vote
2
down vote
favorite
For a homework problem we are given the wave function
$$ Psi(x) = fracNx^2 + a^2, a > 0 $$
and asked to normalize it. Then we are to find the expectation value of $x$. To do so, I normalized the wave function
$$ Psi(x) = frac2 sqrta^3sqrtpi(x^2 + a^2) $$
Then, following my notes computed
$$ int_-infty^infty Psi(x)^* langle x rangle Psi(x) $$
Given that the wave equation is currently in the x basis I then computed
$$ int_-infty^infty
left(frac2 sqrta^3sqrtpi(x^2 + a^2)right)^*
x
left(frac2 sqrta^3sqrtpi(x^2 + a^2)right) $$
by plugging it into Wolfram Alpha. Though this results in a solution of $0$.
This does not seem correct. I feel like the solution should depend on $a$ in the very least. Is there some reason this is actually independent of $a$?
quantum-mechanics wavefunction
edited 19 mins ago
David Z♦
62.3k23135251
asked 1 hour ago
KDecker
4182717
add a comment |Â
up vote
2
down vote
favorite
For a homework problem we are given the wave function
$$ Psi(x) = fracNx^2 + a^2, a > 0 $$
and asked to normalize it. Then we are to find the expectation value of $x$. To do so, I normalized the wave function
$$ Psi(x) = frac2 sqrta^3sqrtpi(x^2 + a^2) $$
Then, following my notes computed
$$ int_-infty^infty Psi(x)^* langle x rangle Psi(x) $$
Given that the wave equation is currently in the x basis I then computed
$$ int_-infty^infty
left(frac2 sqrta^3sqrtpi(x^2 + a^2)right)^*
x
left(frac2 sqrta^3sqrtpi(x^2 + a^2)right) $$
by plugging it into Wolfram Alpha. Though this results in a solution of $0$.
This does not seem correct. I feel like the solution should depend on $a$ in the very least. Is there some reason this is actually independent of $a$?
quantum-mechanics wavefunction
edited 19 mins ago
David Z♦
62.3k23135251
asked 1 hour ago
KDecker
4182717
Hi KDecker; normally we don't handle homework questions that just ask if your work is correct, but you've done something good here by giving a reason to suspect your solution might be wrong. (i.e. your last paragraph) I think your question might be fine but could benefit from a better title and perhaps more precise wording in the body. Would you mind editing it to focus on the issue of whether the solution really should contain $a$? (If you prefer, I can try to make an edit along those lines for you, without changing the core of the question too much.)
– David Z♦
1 hour ago
@DavidZ That is perfectly fine with me. Just wasn't sure what else I could do add to show effort. Thank you for the information though, I'll be sure to format my questions along those lines in the future.
– KDecker
31 mins ago
Sounds good. I made an edit; have a look and please feel free to fix it up if it doesn't reflect what you wanted to ask well enough.
– David Z♦
19 mins ago
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
For a homework problem we are given the wave function
$$ Psi(x) = fracNx^2 + a^2, a > 0 $$
and asked to normalize it. Then we are to find the expectation value of $x$. To do so, I normalized the wave function
$$ Psi(x) = frac2 sqrta^3sqrtpi(x^2 + a^2) $$
Then, following my notes computed
$$ int_-infty^infty Psi(x)^* langle x rangle Psi(x) $$
Given that the wave equation is currently in the x basis I then computed
$$ int_-infty^infty
left(frac2 sqrta^3sqrtpi(x^2 + a^2)right)^*
x
left(frac2 sqrta^3sqrtpi(x^2 + a^2)right) $$
by plugging it into Wolfram Alpha. Though this results in a solution of $0$.
This does not seem correct. I feel like the solution should depend on $a$ in the very least. Is there some reason this is actually independent of $a$?
quantum-mechanics wavefunction
edited 19 mins ago
David Z♦
62.3k23135251
asked 1 hour ago
KDecker
4182717
For a homework problem we are given the wave function
$$ Psi(x) = fracNx^2 + a^2, a > 0 $$
and asked to normalize it. Then we are to find the expectation value of $x$. To do so, I normalized the wave function
$$ Psi(x) = frac2 sqrta^3sqrtpi(x^2 + a^2) $$
Then, following my notes computed
$$ int_-infty^infty Psi(x)^* langle x rangle Psi(x) $$
Given that the wave equation is currently in the x basis I then computed
$$ int_-infty^infty
left(frac2 sqrta^3sqrtpi(x^2 + a^2)right)^*
x
left(frac2 sqrta^3sqrtpi(x^2 + a^2)right) $$
by plugging it into Wolfram Alpha. Though this results in a solution of $0$.
This does not seem correct. I feel like the solution should depend on $a$ in the very least. Is there some reason this is actually independent of $a$?
quantum-mechanics wavefunction
quantum-mechanics wavefunction
edited 19 mins ago
David Z♦
62.3k23135251
asked 1 hour ago
KDecker
4182717
edited 19 mins ago
David Z♦
62.3k23135251
asked 1 hour ago
KDecker
4182717
edited 19 mins ago
David Z♦
62.3k23135251
edited 19 mins ago
David Z♦
62.3k23135251
edited 19 mins ago
David Z♦
62.3k23135251
62.3k23135251
asked 1 hour ago
KDecker
4182717
asked 1 hour ago
KDecker
4182717
asked 1 hour ago
KDecker
4182717
4182717
Hi KDecker; normally we don't handle homework questions that just ask if your work is correct, but you've done something good here by giving a reason to suspect your solution might be wrong. (i.e. your last paragraph) I think your question might be fine but could benefit from a better title and perhaps more precise wording in the body. Would you mind editing it to focus on the issue of whether the solution really should contain $a$? (If you prefer, I can try to make an edit along those lines for you, without changing the core of the question too much.)
– David Z♦
1 hour ago
@DavidZ That is perfectly fine with me. Just wasn't sure what else I could do add to show effort. Thank you for the information though, I'll be sure to format my questions along those lines in the future.
– KDecker
31 mins ago
Sounds good. I made an edit; have a look and please feel free to fix it up if it doesn't reflect what you wanted to ask well enough.
– David Z♦
19 mins ago
add a comment |Â
Hi KDecker; normally we don't handle homework questions that just ask if your work is correct, but you've done something good here by giving a reason to suspect your solution might be wrong. (i.e. your last paragraph) I think your question might be fine but could benefit from a better title and perhaps more precise wording in the body. Would you mind editing it to focus on the issue of whether the solution really should contain $a$? (If you prefer, I can try to make an edit along those lines for you, without changing the core of the question too much.)
– David Z♦
1 hour ago
@DavidZ That is perfectly fine with me. Just wasn't sure what else I could do add to show effort. Thank you for the information though, I'll be sure to format my questions along those lines in the future.
– KDecker
31 mins ago
Sounds good. I made an edit; have a look and please feel free to fix it up if it doesn't reflect what you wanted to ask well enough.
– David Z♦
19 mins ago
Hi KDecker; normally we don't handle homework questions that just ask if your work is correct, but you've done something good here by giving a reason to suspect your solution might be wrong. (i.e. your last paragraph) I think your question might be fine but could benefit from a better title and perhaps more precise wording in the body. Would you mind editing it to focus on the issue of whether the solution really should contain $a$? (If you prefer, I can try to make an edit along those lines for you, without changing the core of the question too much.)
– David Z♦
1 hour ago
Hi KDecker; normally we don't handle homework questions that just ask if your work is correct, but you've done something good here by giving a reason to suspect your solution might be wrong. (i.e. your last paragraph) I think your question might be fine but could benefit from a better title and perhaps more precise wording in the body. Would you mind editing it to focus on the issue of whether the solution really should contain $a$? (If you prefer, I can try to make an edit along those lines for you, without changing the core of the question too much.)
– David Z♦
1 hour ago
@DavidZ That is perfectly fine with me. Just wasn't sure what else I could do add to show effort. Thank you for the information though, I'll be sure to format my questions along those lines in the future.
– KDecker
31 mins ago
@DavidZ That is perfectly fine with me. Just wasn't sure what else I could do add to show effort. Thank you for the information though, I'll be sure to format my questions along those lines in the future.
– KDecker
31 mins ago
Sounds good. I made an edit; have a look and please feel free to fix it up if it doesn't reflect what you wanted to ask well enough.
– David Z♦
19 mins ago
Sounds good. I made an edit; have a look and please feel free to fix it up if it doesn't reflect what you wanted to ask well enough.
– David Z♦
19 mins ago
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
3
down vote
You've got the right answer! The wave-function is symmetric about $0$, while the operator $langle x rangle = x$ is odd. So you have the product of two even functions and an odd function in your integrand, which results in a composite odd function. Integrating an odd function over a symmetric interval will always give you $0$!
Checking the symmetry of your functions is always a good way to figure out if your answer makes sense with these kinds of problems.
answered 1 hour ago
Josh Hoffmann
434
add a comment |Â
up vote
0
down vote
Josh's answer is sufficient. While he uses the final expectation value integral in his answer, you can also just look at the wavefunction. Since it is even, and since $|psi|^2$ (also even) tells us the probability of measuring the particle to be within position $x$ and $x+dx$, we know that the particle "spends an equal amount of time" at positive and negative $x$. So it must be that the mean position is $0$.
I also wanted to address some of your notation and understanding of it.
It seems like you are familiar with bases. The expectation value of position can be written without specifying any basis:
$$langle Xrangle=langlepsi|X|psirangle$$
Since we are given the wavefunction in the position basis $psi(x)=langle x|psirangle$, it makes sense to work in the position basis.
Therefore, we can use the identity $int|xranglelangle x|dx=1$ to write our expectation value as
$$langle Xrangle=int_-infty^inftyint_-infty^inftylanglepsi|xranglelangle x|X|x'ranglelangle x'|psirangle dxdx'$$
Then, knowing that $X$ in its own eigenbasis is $langle x|X|x'rangle=x'delta(x'-x)$, the integral becomes
$$langle Xrangle=int_-infty^inftypsi^*(x)xpsi(x)dx$$
This is the integral you want. You don't want your expectation value inside the integral. Then you have an integral equation you don't want.
answered 18 mins ago
Aaron Stevens
4,2721624
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
You've got the right answer! The wave-function is symmetric about $0$, while the operator $langle x rangle = x$ is odd. So you have the product of two even functions and an odd function in your integrand, which results in a composite odd function. Integrating an odd function over a symmetric interval will always give you $0$!
Checking the symmetry of your functions is always a good way to figure out if your answer makes sense with these kinds of problems.
answered 1 hour ago
Josh Hoffmann
434
add a comment |Â
up vote
3
down vote
You've got the right answer! The wave-function is symmetric about $0$, while the operator $langle x rangle = x$ is odd. So you have the product of two even functions and an odd function in your integrand, which results in a composite odd function. Integrating an odd function over a symmetric interval will always give you $0$!
Checking the symmetry of your functions is always a good way to figure out if your answer makes sense with these kinds of problems.
answered 1 hour ago
Josh Hoffmann
434
add a comment |Â
up vote
3
down vote
up vote
3
down vote
You've got the right answer! The wave-function is symmetric about $0$, while the operator $langle x rangle = x$ is odd. So you have the product of two even functions and an odd function in your integrand, which results in a composite odd function. Integrating an odd function over a symmetric interval will always give you $0$!
Checking the symmetry of your functions is always a good way to figure out if your answer makes sense with these kinds of problems.
answered 1 hour ago
Josh Hoffmann
434
You've got the right answer! The wave-function is symmetric about $0$, while the operator $langle x rangle = x$ is odd. So you have the product of two even functions and an odd function in your integrand, which results in a composite odd function. Integrating an odd function over a symmetric interval will always give you $0$!
Checking the symmetry of your functions is always a good way to figure out if your answer makes sense with these kinds of problems.
answered 1 hour ago
Josh Hoffmann
434
answered 1 hour ago
Josh Hoffmann
434
answered 1 hour ago
Josh Hoffmann
434
answered 1 hour ago
Josh Hoffmann
434
434
add a comment |Â
add a comment |Â
up vote
0
down vote
Josh's answer is sufficient. While he uses the final expectation value integral in his answer, you can also just look at the wavefunction. Since it is even, and since $|psi|^2$ (also even) tells us the probability of measuring the particle to be within position $x$ and $x+dx$, we know that the particle "spends an equal amount of time" at positive and negative $x$. So it must be that the mean position is $0$.
I also wanted to address some of your notation and understanding of it.
It seems like you are familiar with bases. The expectation value of position can be written without specifying any basis:
$$langle Xrangle=langlepsi|X|psirangle$$
Since we are given the wavefunction in the position basis $psi(x)=langle x|psirangle$, it makes sense to work in the position basis.
Therefore, we can use the identity $int|xranglelangle x|dx=1$ to write our expectation value as
$$langle Xrangle=int_-infty^inftyint_-infty^inftylanglepsi|xranglelangle x|X|x'ranglelangle x'|psirangle dxdx'$$
Then, knowing that $X$ in its own eigenbasis is $langle x|X|x'rangle=x'delta(x'-x)$, the integral becomes
$$langle Xrangle=int_-infty^inftypsi^*(x)xpsi(x)dx$$
This is the integral you want. You don't want your expectation value inside the integral. Then you have an integral equation you don't want.
answered 18 mins ago
Aaron Stevens
4,2721624
add a comment |Â
up vote
0
down vote
Josh's answer is sufficient. While he uses the final expectation value integral in his answer, you can also just look at the wavefunction. Since it is even, and since $|psi|^2$ (also even) tells us the probability of measuring the particle to be within position $x$ and $x+dx$, we know that the particle "spends an equal amount of time" at positive and negative $x$. So it must be that the mean position is $0$.
I also wanted to address some of your notation and understanding of it.
It seems like you are familiar with bases. The expectation value of position can be written without specifying any basis:
$$langle Xrangle=langlepsi|X|psirangle$$
Since we are given the wavefunction in the position basis $psi(x)=langle x|psirangle$, it makes sense to work in the position basis.
Therefore, we can use the identity $int|xranglelangle x|dx=1$ to write our expectation value as
$$langle Xrangle=int_-infty^inftyint_-infty^inftylanglepsi|xranglelangle x|X|x'ranglelangle x'|psirangle dxdx'$$
Then, knowing that $X$ in its own eigenbasis is $langle x|X|x'rangle=x'delta(x'-x)$, the integral becomes
$$langle Xrangle=int_-infty^inftypsi^*(x)xpsi(x)dx$$
This is the integral you want. You don't want your expectation value inside the integral. Then you have an integral equation you don't want.
answered 18 mins ago
Aaron Stevens
4,2721624
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Josh's answer is sufficient. While he uses the final expectation value integral in his answer, you can also just look at the wavefunction. Since it is even, and since $|psi|^2$ (also even) tells us the probability of measuring the particle to be within position $x$ and $x+dx$, we know that the particle "spends an equal amount of time" at positive and negative $x$. So it must be that the mean position is $0$.
I also wanted to address some of your notation and understanding of it.
It seems like you are familiar with bases. The expectation value of position can be written without specifying any basis:
$$langle Xrangle=langlepsi|X|psirangle$$
Since we are given the wavefunction in the position basis $psi(x)=langle x|psirangle$, it makes sense to work in the position basis.
Therefore, we can use the identity $int|xranglelangle x|dx=1$ to write our expectation value as
$$langle Xrangle=int_-infty^inftyint_-infty^inftylanglepsi|xranglelangle x|X|x'ranglelangle x'|psirangle dxdx'$$
Then, knowing that $X$ in its own eigenbasis is $langle x|X|x'rangle=x'delta(x'-x)$, the integral becomes
$$langle Xrangle=int_-infty^inftypsi^*(x)xpsi(x)dx$$
This is the integral you want. You don't want your expectation value inside the integral. Then you have an integral equation you don't want.
answered 18 mins ago
Aaron Stevens
4,2721624
Josh's answer is sufficient. While he uses the final expectation value integral in his answer, you can also just look at the wavefunction. Since it is even, and since $|psi|^2$ (also even) tells us the probability of measuring the particle to be within position $x$ and $x+dx$, we know that the particle "spends an equal amount of time" at positive and negative $x$. So it must be that the mean position is $0$.
I also wanted to address some of your notation and understanding of it.
It seems like you are familiar with bases. The expectation value of position can be written without specifying any basis:
$$langle Xrangle=langlepsi|X|psirangle$$
Since we are given the wavefunction in the position basis $psi(x)=langle x|psirangle$, it makes sense to work in the position basis.
Therefore, we can use the identity $int|xranglelangle x|dx=1$ to write our expectation value as
$$langle Xrangle=int_-infty^inftyint_-infty^inftylanglepsi|xranglelangle x|X|x'ranglelangle x'|psirangle dxdx'$$
Then, knowing that $X$ in its own eigenbasis is $langle x|X|x'rangle=x'delta(x'-x)$, the integral becomes
$$langle Xrangle=int_-infty^inftypsi^*(x)xpsi(x)dx$$
This is the integral you want. You don't want your expectation value inside the integral. Then you have an integral equation you don't want.
answered 18 mins ago
Aaron Stevens
4,2721624
edited 1 min ago
answered 18 mins ago
Aaron Stevens
4,2721624
answered 18 mins ago
Aaron Stevens
4,2721624
answered 18 mins ago
Aaron Stevens
4,2721624
4,2721624
add a comment |Â
add a comment |Â
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fphysics.stackexchange.com%2fquestions%2f434376%2fwhy-is-the-position-expectation-value-for-this-wave-function-independent-of-the%23new-answer', 'question_page');
);
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Hi KDecker; normally we don't handle homework questions that just ask if your work is correct, but you've done something good here by giving a reason to suspect your solution might be wrong. (i.e. your last paragraph) I think your question might be fine but could benefit from a better title and perhaps more precise wording in the body. Would you mind editing it to focus on the issue of whether the solution really should contain $a$? (If you prefer, I can try to make an edit along those lines for you, without changing the core of the question too much.)
– David Z♦
1 hour ago
@DavidZ That is perfectly fine with me. Just wasn't sure what else I could do add to show effort. Thank you for the information though, I'll be sure to format my questions along those lines in the future.
– KDecker
31 mins ago
Sounds good. I made an edit; have a look and please feel free to fix it up if it doesn't reflect what you wanted to ask well enough.
– David Z♦
19 mins ago