Is My Quantum Well Probability Calculation Correct?

[Theodore0101]

Homework Statement

An electron is in state n=2 in an infinite quantum well with width L. What is the probability to find the electron if you measure within the well's central third? Can someone cofirm whether or not my solution is correct

Relevant Equations

Y=Asin(n*pi*x/L)

Homework Statement:: An electron is in state n=2 in an infinite quantum well with width L. What is the probability to find the electron if you measure within the well's central third? Can someone cofirm whether or not my solution is correct
Homework Equations:: Y=Asin(n*pi*x/L)

I use the wave funtion Asin(n*pi*x/L) for a wave in a well of the interval 0 to L and square the absolute value of it, getting abs(Asin(n*pi*x/k))^2 = abs(A)^2 *sin^2(n*pi*x/L) (since sin^2(x) is greater or equal to 0) (I use n=2 of course)

I then take the abs(Y)^2 function and integrate it from 0 to L, (using that sin^2(x)`= 1/2(1-cos(2x)) and define the expression as equal to 1 in order to be able to normate it. I get that abs(A)^2*L/2=1 => abs(A)^2 = 2/L

I use 2/L in place of the abs(A)^2 in the integral expression and integrate from L/3 to 2L/3 so that it is the third in the middle. I get that it is equal to 0.229959166. Is this the correct way to solve it? Also, if anyone knows their significant figures, I'm also unsure about how many I am supposed to use in my final answer, since the problem only gives us n=2 which is discrete and that we are calculating for 1/3 of L, and doesn't give any regular values

Thanks

 

[PeroK]

Homework Statement:: An electron is in state n=2 in an infinite quantum well with width L. What is the probability to find the electron if you measure within the well's central third? Can someone cofirm whether or not my solution is correct
Homework Equations:: Y=Asin(n*pi*x/L)

I use the wave funtion Asin(n*pi*x/L) for a wave in a well of the interval 0 to L and square the absolute value of it, getting abs(Asin(n*pi*x/k))^2 = abs(A)^2 *sin^2(n*pi*x/L) (since sin^2(x) is greater or equal to 0) (I use n=2 of course)

I then take the abs(Y)^2 function and integrate it from 0 to L, (using that sin^2(x)`= 1/2(1-cos(2x)) and define the expression as equal to 1 in order to be able to normate it. I get that abs(A)^2*L/2=1 => abs(A)^2 = 2/L

I use 2/L in place of the abs(A)^2 in the integral expression and integrate from L/3 to 2L/3 so that it is the third in the middle. I get that it is equal to 0.229959166. Is this the correct way to solve it? Also, if anyone knows their significant figures, I'm also unsure about how many I am supposed to use in my final answer, since the problem only gives us n=2 which is discrete and that we are calculating for 1/3 of L, and doesn't give any regular values

Thanks

That answer seems not quite right. Can you check it? Personally, I would give a precise answer, in terms of $\pi$, and give an approximation to at most 3 decimal places.

 

[Theodore0101]

That answer seems not quite right. Can you check it? Personally, I would give a precise answer, in terms of $\pi$, and give an approximation to at most 3 decimal places.

Thanks for your reply. I think I might have done some calculating mistake when I was doing the final step, and that it should instead be 0.33333... or 1/3. Is this the correct answer, or were you referring to something else as being wrong?

 

[PeroK]

Thanks for your reply. I think I might have done some calculating mistake when I was doing the final step, and that it should instead be 0.33333... or 1/3. Is this the correct answer, or were you referring to something else as being wrong?

It's $\frac 1 3 - \Delta$, where $\Delta$ is some factor that depends on $n$.

What was the result of your integral?

 

[Theodore0101]

It's $\frac 1 3 - \Delta$, where $\Delta$ is some factor that depends on $n$.

What was the result of your integral?

Using 2/L*int(sin^2(2pi*x/L))dx (from L/3 to 2L/3) I got that it is 1/3 (with 2/L=abs(A)^2 and 2=n). Is it possible for Δ to be 0, or am I still doing something wrong somewhere?

 

[PeroK]

Using 2/L*int(sin^2(2pi*x/L))dx (from L/3 to 2L/3) I got that it is 1/3 (with 2/L=abs(A)^2 and 2=n). Is it possible for Δ to be 0, or am I still doing something wrong somewhere?

$\Delta = 0$ for $n = 3$, but not for $n =2$.

You can look up the integral of $\sin^2$ to check, but you should get:
$$\int_{L/3}^{2L/3} \frac 2 L \sin^2(\frac{2 \pi x}{L})dx = (\frac 2 L)(\frac{L}{8\pi})\bigg[ \frac{4\pi x}{L} - \sin(\frac{4\pi x}{L}) \bigg ] _{L/3}^{2L/3}$$

 

[Theodore0101]

$\Delta = 0$ for $n = 3$, but not for $n =2$.

You can look up the integral of $\sin^2$ to check, but you should get:
$$\int_{L/3}^{2L/3} \frac 2 L \sin^2(\frac{2 \pi x}{L})dx = (\frac 2 L)(\frac{L}{8\pi})\bigg[ \frac{4\pi x}{L} - \sin(\frac{4\pi x}{L}) \bigg ] _{L/3}^{2L/3}$$

Thank you! My mistake seems to have been when doing the integral, and I wrote cos instead of sin in the final expression. But I think I'm getting the correct answer now.

 

[PeroK]

Thank you! My mistake seems to have been when doing the integral, and I wrote cos instead of sin in the final expression. But I think I'm getting the correct answer now.

Which is?

 

[Theodore0101]

Which is?

About 19.6%

 

[PeroK]

About 19.6%

That would be $\frac 1 3 - \frac{\sqrt{3}}{4\pi}$