דוגמה לעקרון אי-הוודאות של הייזנברג
עריכה
ניקח גל מישורי סטנדרטי:
y
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{\displaystyle \ y\left(x,t\right)=A\cdot \cos \left(\omega t-kx\right)=A\cdot Re\left(e^{-i\left(kx-\omega t\right)}\right)}
נגדיר :
y
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{\displaystyle \ {\tilde {\bar {y}}}(x,t)=A\cdot e^{-i\left(kx-\omega t\right)}}
k
{\displaystyle \ k}
k
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2
π
λ
{\displaystyle \ k={\frac {2\pi }{\lambda }}}
k
∈
[
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{\displaystyle \ k\in [k_{1},k_{2}]}
y
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{\displaystyle \ {\bar {y}}(x,t)={\frac {A}{k_{2}-k_{1}}}\underbrace {\int \limits _{k_{2}}^{k_{1}}dk} _{*}\cdot \underbrace {e^{i\left(kx-\omega t\right)}} _{**}}
∗
{\displaystyle \ *}
∗
∗
{\displaystyle \ **}
⇒
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{\displaystyle \ \Rightarrow {\bar {y}}(x,t)={\frac {A}{k_{2}-k_{1}}}\cdot e^{-i\omega t}\int \limits _{k_{2}}^{k_{1}}dk\cdot e^{ikx}=}
=
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1
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x
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{\displaystyle \ ={\frac {A}{k_{2}-k_{1}}}\cdot e^{-i\omega t}\cdot {\frac {1}{ix}}\left(e^{ik_{2}x}-e^{ik_{1}x}\right)=}
=
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∗
{\displaystyle \ ={\frac {A\cdot e^{i\omega t}}{k_{2}-k_{1}}}{\frac {1}{ix}}e^{i\left({\frac {k_{1}+k_{2}}{2}}\right)x}\underbrace {\left(e^{i{\frac {k_{2}-k_{1}}{2}}x}-e^{-i{\frac {k_{2}-k_{1}}{2}}x}\right)} _{*}}
∗
=
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x
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e
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k
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1
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=
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i
sin
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{\displaystyle \ *=\left(e^{i{\frac {k_{2}-k_{1}}{2}}x}-e^{-i{\frac {k_{2}-k_{1}}{2}}x}\right)=2i\sin \left({\frac {k_{2}-k_{1}}{2}}\right)}
Δ
K
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k
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{\displaystyle \ \Delta K=k_{2}-k_{1}}
y
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sin
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1
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{\displaystyle \ {\bar {y}}(x,t)=A\cdot e^{-i\omega t}e^{{\frac {i\left(k_{1}+k_{2}\right)}{2}}x}{\frac {\sin \left({\frac {1}{2}}\Delta kx\right)}{{\frac {1}{2}}\Delta kx}}}
⇒
y
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R
e
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y
^
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x
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sin
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k
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1
2
Δ
k
x
⋅
cos
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k
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k
2
2
x
−
ω
t
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{\displaystyle \ \Rightarrow y(x,t)=Re\left({\hat {y}}(x,t)\right)={\frac {\sin \left({\frac {1}{2}}\Delta kx\right)}{{\frac {1}{2}}\Delta kx}}\cdot \cos \left({\frac {k_{1}+k_{2}}{2}}x-\omega t\right)}
x
2
−
x
1
=
Δ
x
{\displaystyle \ x_{2}-x_{1}=\Delta x}
Δ
k
{\displaystyle \ \Delta k}
sin
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1
2
Δ
k
x
)
=
0
{\displaystyle \ \sin \left({\frac {1}{2}}\Delta kx\right)=0}
1
2
Δ
k
x
=
n
π
⇐
{\displaystyle \ {\frac {1}{2}}\Delta kx=n\pi \Leftarrow }
n
∈
Z
{\displaystyle \ n\in \mathbb {Z} }
x
1
{\displaystyle \ x_{1}}
1
2
Δ
k
x
1
=
−
π
{\displaystyle \ {\frac {1}{2}}\Delta kx_{1}=-\pi }
x
2
{\displaystyle \ x_{2}}
1
2
Δ
k
x
2
=
+
π
{\displaystyle \ {\frac {1}{2}}\Delta kx_{2}=+\pi }
1
2
Δ
k
Δ
x
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2
π
⇐
1
2
Δ
k
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2
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{\displaystyle \ {\frac {1}{2}}\Delta k\Delta x=2\pi \Leftarrow {\frac {1}{2}}\Delta k\left(x_{2}-x_{1}\right)=2\pi }
וקיבלנו:
Δ
x
Δ
k
=
4
π
{\displaystyle \ \Delta x\Delta k=4\pi }
כלומר: אם נרצה לקבל
Δ
x
{\displaystyle \ \Delta x}
Δ
k
{\displaystyle \ \Delta k}
הרכבה של גלים מישוריים (באופן כללי)
עריכה
כאמור, אסכמת (אינטגרל) של גלים מישוריים בתחום התדרים
[
k
1
,
k
2
]
{\displaystyle \ \left[k_{1},k_{2}\right]}
ψ
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x
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t
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=
A
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k
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∫
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d
k
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{\displaystyle \ \psi (x,t)={\frac {A}{k_{1}-k_{2}}}\int \limits _{k_{1}}^{k_{2}}dk\cdot e^{i\left(kx-\omega t\right)}}
k
∈
[
k
1
,
k
2
]
{\displaystyle \ k\in \left[k_{1},k_{2}\right]}
A
k
2
−
k
1
{\displaystyle \ {\frac {A}{k_{2}-k_{1}}}}
k
∈
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−
∞
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∞
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{\displaystyle \ k\in \left(-\infty ,\infty \right)}
g
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k
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{\displaystyle \ g\left(k\right)}
g
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k
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{\displaystyle \ g\left(k\right)}
k
{\displaystyle \ k}
k
{\displaystyle \ k}
k
{\displaystyle \ k}
g
(
k
)
{\displaystyle \ g\left(k\right)}
k
{\displaystyle \ k}
בתחילת ההרצאה הקודמת .
1
2
π
{\displaystyle \ {\frac {1}{\sqrt {2\pi }}}}
ψ
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{\displaystyle \ \psi (x,t)={\frac {1}{\sqrt {2\pi }}}\int \limits _{-\infty }^{\infty }dk\cdot g(k)\cdot e^{i\left(kx-\omega t\right)}}
t
=
o
{\displaystyle \ t=o}
ψ
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{\displaystyle \ \psi (x,t=0)=\psi (x,0)-{\frac {1}{\sqrt {2\pi }}}\int \limits _{-\infty }^{\infty }dk\cdot g(k)\cdot e^{ikx}}
g
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{\displaystyle \ g\left(k\right)={\frac {1}{\sqrt {2\pi }}}\int \limits _{-\infty }^{\infty }dx\cdot \psi (x,0)e^{-ikx}}
ניתוח החבורה
עריכה
הפונקציה
ψ
(
x
,
0
)
{\displaystyle \ \psi (x,0)}
g
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k
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{\displaystyle \ g(k)}
g
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k
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=
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g
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⋅
e
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α
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k
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{\displaystyle \ g(k)=\left|g(k)\right|\cdot e^{i\alpha (k)}}
Δ
k
=
∞
{\displaystyle \ \Delta k=\infty }
g
(
k
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{\displaystyle \ g(k)}
α
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k
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≃
α
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{\displaystyle \ \alpha (k)\simeq \alpha \left(k_{0}\right)+\underbrace {\left(k-k_{0}\right)} _{=\Delta k}\left.\left({\frac {\partial \alpha }{\partial k}}\right)\right|_{k=k_{0}}+...(*)}
⇒
ψ
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k
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{\displaystyle \ \Rightarrow \psi \left(x,0\right)\simeq {\frac {1}{\sqrt {2\pi }}}\int \limits _{-\infty }^{\infty }dk\left|g\left(k\right)\right|e^{i\left(\alpha \left(k_{0}\right)+\left(k-k_{o}\right)\left.\left({\frac {\partial \alpha }{\partial k}}\right)\right|_{k=k_{0}}\right)}e^{ikx}}
נסמן:
x
0
≡
−
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∂
α
∂
k
)
|
k
=
k
0
{\displaystyle \ x_{0}\equiv -\left.\left({\frac {\partial \alpha }{\partial k}}\right)\right|_{k=k_{0}}}
ψ
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x
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0
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≃
1
2
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∫
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∞
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k
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g
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α
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x
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=
{\displaystyle \ \psi \left(x,0\right)\simeq {\frac {1}{\sqrt {2\pi }}}\int \limits _{-\infty }^{\infty }dk\left|g\left(k\right)\right|e^{i\left(\alpha \left(k_{0}\right)+\left(k-k_{0}\right)x_{0}+kx\right)}=}
=
1
2
π
∫
−
∞
∞
d
k
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g
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k
)
|
e
i
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α
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k
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+
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x
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−
k
0
x
0
+
k
x
)
=
{\displaystyle \ ={\frac {1}{\sqrt {2\pi }}}\int \limits _{-\infty }^{\infty }dk\left|g\left(k\right)\right|e^{i\left(\alpha \left(k_{0}\right)+kx_{0}-k_{0}x_{0}+kx\right)}=}
1
2
π
∫
−
∞
∞
d
k
|
g
(
k
)
|
e
i
(
α
(
k
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)
+
k
x
0
−
k
0
x
0
+
k
x
+
k
0
x
−
k
0
x
)
=
{\displaystyle \ {\frac {1}{\sqrt {2\pi }}}\int \limits _{-\infty }^{\infty }dk\left|g\left(k\right)\right|e^{i\left(\alpha \left(k_{0}\right)+kx_{0}-k_{0}x_{0}+kx+k_{0}x-k_{0}x\right)}=}
1
2
π
∫
−
∞
∞
d
k
|
g
(
k
)
|
e
i
(
α
(
k
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)
+
k
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x
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+
i
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k
x
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−
k
0
x
0
+
k
x
−
k
0
x
)
{\displaystyle \ {\frac {1}{\sqrt {2\pi }}}\int \limits _{-\infty }^{\infty }dk\left|g\left(k\right)\right|e^{i\left(\alpha \left(k_{0}\right)+k_{0}x\right)+i\left(kx_{0}-k_{0}x_{0}+kx-k_{0}x\right)}}
α
(
k
0
)
{\displaystyle \ \alpha \left(k_{0}\right)}
k
0
x
{\displaystyle \ k_{0}x}
ψ
(
x
,
0
)
=≃
1
2
π
e
i
α
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k
0
)
+
k
0
x
∫
−
∞
∞
d
k
|
g
(
k
)
|
e
i
(
k
−
k
0
)
(
x
−
x
0
)
{\displaystyle \ \psi \left(x,0\right)=\simeq {\frac {1}{\sqrt {2\pi }}}e^{i\alpha \left(k_{0}\right)+k_{0}x}\int \limits _{-\infty }^{\infty }dk\left|g\left(k\right)\right|e^{i\left(k-k_{0}\right)\left(x-x_{0}\right)}}
A
=
∫
−
∞
∞
d
k
|
g
(
k
)
|
e
i
(
k
−
k
0
)
(
x
−
x
0
)
{\displaystyle \ A=\int \limits _{-\infty }^{\infty }dk\left|g\left(k\right)\right|e^{i\left(k-k_{0}\right)\left(x-x_{0}\right)}}
נסמן: ., ונרצה ללמוד משהו על הקשר בינו לבין (שהוא, כזכור, טווח בתדירויות שלנו). לשם כך, נבדוק מתי הסינוס מתאפס: 
![{\displaystyle \ k\in [k_{1},k_{2}]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/0575acb821bbf4cf6d1a484113352c46e3e090ff)
![{\displaystyle \ \left[k_{1},k_{2}\right]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/c594d1cc1c823d4d2ab131aaa400b280304c3a5a)
![{\displaystyle \ k\in \left[k_{1},k_{2}\right]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/0213db827413a1027cfa403cb9c0b50bfa06eb51)
