Write a question and its solution showing the probability of a particle being located in the bottom left quadrant of the ground state of the particle in a square.
edit
The particle in a square model is where a particle in confined to a restricted
region inside a 2D box. For this model we consider a particle which is free to move
anywhere inside the x,y dimensions but cannot exist outside of the 2D box where it would
have infinite potential energy.
The conditions for the potential energy are
ν
=
{
∞
,
x
<
0
,
y
<
0
0
,
0
≤
x
,
y
≤
L
∞
,
x
,
y
>
L
}
{\displaystyle \nu ={\begin{Bmatrix}\infty ,&x<0,y<0\\0,&0\leq x,y\leq L\\\infty ,&x,y>L\end{Bmatrix}}}
Since the particle in a square is a 2D model the wave function must take into account the x and y direction ;
(
Ψ
(
x
,
y
)
)
{\displaystyle (\Psi (x,y))}
The wave function for a particle in a square
Ψ
(
x
,
y
)
=
2
L
s
i
n
(
n
x
π
L
x
)
(
n
y
π
L
y
)
⋅
2
L
s
i
n
(
n
x
π
L
x
)
(
n
y
π
L
y
)
{\displaystyle \Psi (x,y)={\frac {2}{L}}sin\left({\frac {n_{x}\pi }{L}}x\right)\left({\frac {n_{y}\pi }{L}}y\right)\cdot {\frac {2}{L}}sin\left({\frac {n_{x}\pi }{L}}x\right)\left({\frac {n_{y}\pi }{L}}y\right)}
The states the quantum system is in are defined by the quantum numbers
n
x
,
n
y
{\displaystyle n_{x},n_{y}}
In quantum energy levels only discrete energy levels are possible. In this model the particle cannot be motionless meaning the lowest possible energy level must be non-zero (n=1). Since the particle is always moving even in the ground state the particle has a a non- zero energy level. Therefore the ground state of a particle in a box is
n
x
=
1
,
n
y
=
1
{\displaystyle n_{x}=1,n_{y}=1}
N
n
o
d
a
l
l
i
n
e
s
=
n
x
,
y
−
1
{\displaystyle Nnodallines=n_{x},_{y}-1}
Probability Distributions
edit
The probability (P) of a particle being found in an interval can be found by integrating the probability distribution (P(x))
The probability of finding a particle in the range
[
a
1
,
a
2
]
{\displaystyle \left[a_{1},a_{2}\right]}
[
b
1
,
b
2
]
{\displaystyle \left[b_{1},b_{2}\right]}
P
=
∫
a
1
a
2
∫
b
1
b
2
P
(
x
,
y
)
d
x
d
y
{\displaystyle P=\int \limits _{a_{1}}^{a_{2}}\int \limits _{b_{1}}^{b_{2}}P(x,y)dxdy}
The probability distribution of a a particle is the wave function squared.
For one particle in 2D (particle in a box):
P
(
x
,
y
)
=
|
Ψ
(
x
,
y
)
2
|
=
Ψ
(
x
,
y
)
∗
⋅
Ψ
(
x
,
y
)
{\displaystyle P(x,y)=\left\vert \Psi (x,y)^{2}\right\vert =\Psi (x,y)^{*}\cdot \Psi (x,y)}
Therefore the probability of finding a particle over a range can be re-written as:
P
=
∫
a
1
a
2
∫
b
1
b
2
Ψ
∗
(
x
,
y
)
Ψ
(
x
,
y
)
d
x
d
y
{\displaystyle P=\int \limits _{a_{1}}^{a_{2}}\int \limits _{b_{1}}^{b_{2}}\Psi ^{*}{\bigl (}x,y)\Psi {\bigl (}x,y)dxdy}
edit
If the particle is in the bottom left quadrant it is in the interval;
x
=
[
0
,
L
2
]
{\displaystyle x=\left[0,{\frac {L}{2}}\right]}
y
=
[
0
,
L
2
]
{\displaystyle y=\left[0,{\frac {L}{2}}\right]}
Therefore the probability density is:
P
=
∫
0
L
2
∫
0
L
2
Ψ
∗
(
x
,
y
)
Ψ
(
x
,
y
)
d
x
d
y
{\displaystyle P=\int \limits _{0}^{\tfrac {L}{2}}\int \limits _{0}^{\tfrac {L}{2}}\Psi ^{*}{\bigl (}x,y)\Psi {\bigl (}x,y)dxdy}
The wave-function for a particle in a square is:
Ψ
(
x
,
y
)
=
2
L
sin
(
n
x
π
L
x
)
(
n
y
π
L
y
)
⋅
2
L
sin
(
n
x
π
L
x
)
(
n
y
π
L
y
)
{\displaystyle \Psi (x,y)={\frac {2}{L}}\sin \left({\frac {n_{x}\pi }{L}}x\right)\left({\frac {n_{y}\pi }{L}}y\right)\cdot {\frac {2}{L}}\sin \left({\frac {n_{x}\pi }{L}}x\right)\left({\frac {n_{y}\pi }{L}}y\right)}
In the ground state
n
x
=
1
{\displaystyle n_{x}=1}
n
y
=
1
{\displaystyle n_{y}=1}
P
=
∫
0
L
2
∫
0
L
2
(
2
L
)
sin
(
π
L
x
)
sin
(
π
L
y
)
⋅
(
2
L
)
sin
(
π
L
x
)
sin
(
π
L
y
)
d
x
d
y
{\displaystyle P=\int \limits _{0}^{\frac {L}{2}}\int \limits _{0}^{\frac {L}{2}}{\biggl (}{2 \over L}{\biggr )}\sin \left({\frac {\pi }{L}}x\right)\sin \left({\frac {\pi }{L}}y\right)\centerdot {\biggl (}{2 \over L}{\biggr )}\sin \left({\frac {\pi }{L}}x\right)\sin \left({\frac {\pi }{L}}y\right)dxdy}
Then solving the integral,
P
=
(
4
L
2
)
∫
0
L
2
∫
0
L
2
sin
2
(
π
L
x
)
sin
2
(
π
L
y
)
d
x
d
y
{\displaystyle P=\left({\frac {4}{L^{2}}}\right)\int _{0}^{\frac {L}{2}}\int _{0}^{\frac {L}{2}}\sin ^{2}\left({\frac {\pi }{L}}x\right)\sin ^{2}\left({\frac {\pi }{L}}y\right)dxdy}
P
=
(
4
L
2
)
∫
0
L
2
sin
2
(
π
L
x
)
d
x
⋅
∫
0
L
2
sin
2
(
π
L
y
)
d
y
{\displaystyle P=\left({\frac {4}{L^{2}}}\right)\int _{0}^{\frac {L}{2}}\sin ^{2}\left({\frac {\pi }{L}}x\right)dx\cdot \int _{0}^{\frac {L}{2}}\sin ^{2}\left({\frac {\pi }{L}}y\right)dy}
From Table of Integrals
∫
sin
2
(
a
x
)
d
x
=
x
2
−
sin
(
2
a
x
)
4
a
{\displaystyle \int \sin ^{2}(ax)dx={\frac {x}{2}}-{\frac {\sin(2ax)}{4a}}}
P
=
(
4
L
2
)
[
y
2
−
sin
(
2
π
L
y
)
4
(
π
L
)
]
0
L
[
x
2
−
sin
(
2
π
L
x
)
4
(
π
L
)
]
0
L
{\displaystyle P=\left({\frac {4}{L^{2}}}\right)\left[{\frac {y}{2}}-{\frac {\sin {\bigl (}2{\frac {\pi }{L}}y{\bigr )}}{4\left({\frac {\pi }{L}}\right)}}\right]_{0}^{L}\left[{\frac {x}{2}}-{\frac {\sin {\bigl (}2{\frac {\pi }{L}}x{\bigr )}}{4\left({\frac {\pi }{L}}\right)}}\right]_{0}^{L}}
P
=
(
4
L
2
)
(
L
2
2
−
L
sin
(
2
π
L
L
2
)
4
(
π
)
)
(
L
2
2
−
L
sin
(
2
π
L
L
2
)
4
(
π
)
)
{\displaystyle P=\left({\frac {4}{L^{2}}}\right){\Biggl (}{\frac {\frac {L}{2}}{2}}-{\frac {L\sin {\bigl (}2{\frac {\pi }{L}}{\frac {L}{2}}{\bigr )}}{4\left(\pi \right)}}{\Biggr )}{\Biggl (}{\frac {\frac {L}{2}}{2}}-{\frac {L\sin {\bigl (}2{\frac {\pi }{L}}{\frac {L}{2}}{\bigr )}}{4\left(\pi \right)}}{\Biggr )}}
P
=
(
4
L
2
)
(
L
4
−
L
sin
(
π
)
4
(
π
)
)
(
L
4
−
L
sin
(
π
)
4
(
π
)
)
{\displaystyle P=\left({\frac {4}{L^{2}}}\right){\Biggl (}{\frac {L}{4}}-{\frac {L\sin {\bigl (}\pi {\bigr )}}{4\left(\pi \right)}}{\Biggr )}{\Biggl (}{\frac {L}{4}}-{\frac {L\sin {\bigl (}\pi {\bigr )}}{4\left(\pi \right)}}{\Biggr )}}
Since
sin
π
{\displaystyle \sin \pi }
P
=
(
4
L
2
)
(
L
4
−
0
)
(
L
4
−
0
)
{\displaystyle P=\left({\frac {4}{L^{2}}}\right){\Biggl (}{\frac {L}{4}}-0{\biggr )}{\Biggl (}{\frac {L}{4}}-0{\Biggr )}}
P
=
4
8
{\displaystyle P={\frac {4}{8}}}
P
=
1
4
{\displaystyle P={\frac {1}{4}}}
Therefore the the probability of finding a particle in the bottom left quadrant of a square in the ground state is 0.25 of 25%.
![{\displaystyle \left[a_{1},a_{2}\right]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/29e744bfc6fb9a2cc440c7fa30630afd29585603)
![{\displaystyle \left[b_{1},b_{2}\right]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/f36f1b59e1556b6112f26997eefa257fca5c69f0)
![{\displaystyle x=\left[0,{\frac {L}{2}}\right]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/626cf67fd71819374166958969dea01ba4fd8d16)
![{\displaystyle y=\left[0,{\frac {L}{2}}\right]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/e29af259b6c6c0a136a6670f0b5d3cf66ba3c178)
![{\displaystyle P=\left({\frac {4}{L^{2}}}\right)\left[{\frac {y}{2}}-{\frac {\sin {\bigl (}2{\frac {\pi }{L}}y{\bigr )}}{4\left({\frac {\pi }{L}}\right)}}\right]_{0}^{L}\left[{\frac {x}{2}}-{\frac {\sin {\bigl (}2{\frac {\pi }{L}}x{\bigr )}}{4\left({\frac {\pi }{L}}\right)}}\right]_{0}^{L}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/c62c66bf4fc87a9b70d0088e01b411fe1a32a29a)
