Consideremos 2 partículas idénticas (n y p) identificadas por sus valores propios respecto a un mismo CCOC.
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≡
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{\displaystyle |a'b'c'\ldots \rangle \equiv |a'\rangle }
A
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=
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{\displaystyle A|a'b'c'\ldots \rangle =a'|a'b'c'\ldots \rangle }
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=
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{\displaystyle |a''b''\ldots \rangle =|a''\rangle }
A
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=
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{\displaystyle A|a''b''\ldots \rangle =a''|a''\rangle }
El sistema de ambas partículas viene descrito por su producto directo
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=
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{\displaystyle |a'\rangle \otimes |a''\rangle =|a'\rangle |a''\rangle =|a';a''\rangle =|a'a''\rangle }
donde la posición que ocupan escritas en el ket (en el sentido de orden) indica si se trata de la primera o de la segunda partícula.
¿Cómo actúa el observable
A
=
A
1
+
A
2
=
A
1
⊗
I
2
+
I
1
⊗
A
2
{\displaystyle A=A_{1}+A_{2}=A_{1}\otimes {\mathcal {I}}_{2}+{\mathcal {I}}_{1}\otimes A_{2}}
A
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=
A
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⊗
I
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+
I
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⊗
A
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=
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{\displaystyle A|a'a''\rangle =A|a'\rangle \otimes {\mathcal {I}}|a''\rangle +{\mathcal {I}}|a'\rangle \otimes A|a''\rangle =(a'+a'')|a'a''\rangle }
Entonces
A
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=
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{\displaystyle A|a'a''\rangle =(a'+a'')|a'a''\rangle }
A
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{\displaystyle A|a''a'\rangle =(a''+a')|a''a'\rangle }
tenemos que estos dos estados matemáticamente distinguibles (ortogonales) son físicamente indistinguibles: los observables del CCOC da para ambos el mismo valor propio. Haciendo medidas no se pueden distinguir.
"Degeneración de intercambio"
c
1
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a
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+
c
2
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a
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a
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{\displaystyle c_{1}|a'a''\rangle +c_{2}|a''a'\rangle }
Es conveniente introducir el grupo
S
n
{\displaystyle S_{n}}
S
n
∋
p
{\displaystyle S_{n}\ni p}
S
n
∋
p
=
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1
2
⋯
n
p
1
p
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⋯
p
n
)
{\displaystyle S_{n}\ni p={\begin{pmatrix}1&2&\cdots &n\\p_{1}&p_{2}&\cdots &p_{n}\end{pmatrix}}}
Ejemplo:
S
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{
e
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,
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}
{\displaystyle S_{2}=\left\{e={\begin{pmatrix}1&2\\1&2\end{pmatrix}},{\begin{pmatrix}1&2\\2&1\end{pmatrix}}\right\}}
Ejemplo:
S
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{
e
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3
1
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)
,
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,
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}
{\displaystyle S_{3}=\left\{e={\begin{pmatrix}1&2&3\\1&2&3\end{pmatrix}},{\begin{pmatrix}1&2&3\\2&1&3\end{pmatrix}}{\begin{pmatrix}1&2&3\\3&2&1\end{pmatrix}},{\begin{pmatrix}1&2&3\\1&3&2\end{pmatrix}}{\begin{pmatrix}1&2&3\\2&3&1\end{pmatrix}},{\begin{pmatrix}1&2&3\\3&1&2\end{pmatrix}}\right\}}
Las tres primeras de este ejemplo son trasposiciones y las demás se componen con dos transpociones.
El orden de
S
n
{\displaystyle S_{n}}
n
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n
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⋯
=
n
!
{\displaystyle n(n-1)\cdots =n!}
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)
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=
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{\displaystyle {\begin{pmatrix}1&2&3\\2&1&3\end{pmatrix}}{\begin{pmatrix}1&2&3\\1&3&2\end{pmatrix}}={\begin{pmatrix}1&2&3\\2&3&1\end{pmatrix}}}
Notación "2-ado"
(
1
2
)
(
3
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{\displaystyle {\begin{pmatrix}1&2\end{pmatrix}}{\cancel {\begin{pmatrix}3\end{pmatrix}}}}
"1 va a 2, 2 va a 1"
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{\displaystyle {\begin{pmatrix}1&2&3\end{pmatrix}}}
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{\displaystyle {\begin{pmatrix}\ldots \end{pmatrix}}}
Y se multiplican leyendo...
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⋅
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=
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{\displaystyle {\begin{pmatrix}1&2\end{pmatrix}}\cdot {\begin{pmatrix}2&3\end{pmatrix}}={\begin{pmatrix}1&2&3\end{pmatrix}}}
Otro ejemplo..
(
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=
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2
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{\displaystyle {\begin{pmatrix}2&3\end{pmatrix}}{\begin{pmatrix}1&2\end{pmatrix}}={\begin{pmatrix}1&3\end{pmatrix}}{\cancel {\begin{pmatrix}2\end{pmatrix}}}}
Podemos definir la acción de una permutación de
S
n
{\displaystyle S_{n}}
Ejemplo:
n
=
2
{\displaystyle n=2}
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a
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{\displaystyle |a'a''\rangle }
S
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=
{
e
1
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}
{\displaystyle S_{2}=\{e_{1}{\begin{pmatrix}1&2\end{pmatrix}}\}}
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1
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)
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=
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{\displaystyle {\begin{pmatrix}1&2\end{pmatrix}}|a'a''\rangle =|a''a'\rangle }
Ejemplo
n
=
3
{\displaystyle n=3}
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{\displaystyle |a'a''a'''\rangle }
(
1
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)
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=
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{\displaystyle {\begin{pmatrix}1&2&3\end{pmatrix}}|a'a''a'''\rangle =|a'a''a'''\rangle }
Para el grupo
S
2
?
{
e
,
(
1
2
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}
{\displaystyle S_{2}?\{e,{\begin{pmatrix}1&2\end{pmatrix}}\}}
simetrizador
S
{\displaystyle S}
antisimetrizador
a
{\displaystyle a}
S
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{\displaystyle S={\frac {1}{2}}\left[e+{\begin{pmatrix}1&2\end{pmatrix}}\right]}
a
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{\displaystyle a={\frac {1}{2}}\left[e-{\begin{pmatrix}1&2\end{pmatrix}}\right]}
¿Cómo actúan
S
{\displaystyle S}
a
{\displaystyle a}
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{\displaystyle |a'a''\rangle }
S
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{\displaystyle S|a'a''\rangle ={\frac {1}{2}}\left(e+{\begin{pmatrix}1&2\end{pmatrix}}\right)|a'a''\rangle ={\frac {1}{2}}\left(|a'a''\rangle +|a''a'\rangle \right)}
Y normalizando
1
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+
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≡
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S
{\displaystyle {\frac {1}{\sqrt {2}}}{\frac {1}{2}}\left(|a'a''\rangle +|a''a'\rangle \right)\equiv |a'a''\rangle _{S}}
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S
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S
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{\displaystyle |a'a''\rangle _{S}={\begin{pmatrix}1&2\end{pmatrix}}|a'a''\rangle _{S}={\begin{pmatrix}1&2\end{pmatrix}}{\frac {1}{\sqrt {2}}}\left(|a'a''\rangle +|a''a'\rangle \right)={\frac {1}{\sqrt {2}}}\left(|a''a'\rangle +|a''a'\rangle \right)}
a
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{\displaystyle a|a'a''\rangle ={\frac {1}{2}}\left[e-{\begin{pmatrix}1&2\end{pmatrix}}|a'a''\rangle \right]={\frac {1}{2}}\left(|a'a''\rangle -|a''a'\rangle \right)}
Y normalizando
1
2
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−
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≡
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a
{\displaystyle {\frac {1}{\sqrt {2}}}\left(|a'a''\rangle -|a''a'\rangle \right)\equiv |a'a''\rangle _{a}}
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=
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{\displaystyle |a'a''\rangle _{a}={\begin{pmatrix}1&2\end{pmatrix}}|a'a''\rangle _{a}={\begin{pmatrix}1&2\end{pmatrix}}{\frac {1}{\sqrt {2}}}\left(|a'a''\rangle -|a''a'\rangle \right)={\frac {1}{\sqrt {2}}}\left(|a''a'\rangle -|a''a'\rangle \right)=\ldots }
Para n partículas
S
=
1
n
!
∑
p
p
{\displaystyle S={\frac {1}{n!}}\sum _{p}p}
S
3
→
s
=
1
6
[
e
+
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+
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+
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+
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+
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]
{\displaystyle S_{3}\to s={\frac {1}{6}}\left[e+{\begin{pmatrix}1&2\end{pmatrix}}+{\begin{pmatrix}1&3\end{pmatrix}}+{\begin{pmatrix}2&3\end{pmatrix}}+{\begin{pmatrix}1&2&3\end{pmatrix}}+{\begin{pmatrix}1&3&2\end{pmatrix}}\right]}
a
=
1
n
!
∑
p
(
−
1
)
p
p
{\displaystyle a={\frac {1}{n!}}\sum _{p}(-1)^{p}p}
S
3
→
a
=
1
6
[
e
−
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)
−
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3
)
−
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)
+
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+
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]
{\displaystyle S_{3}\to a={\frac {1}{6}}\left[e-{\begin{pmatrix}1&2\end{pmatrix}}-{\begin{pmatrix}1&3\end{pmatrix}}-{\begin{pmatrix}2&3\end{pmatrix}}+{\begin{pmatrix}1&2&3\end{pmatrix}}+{\begin{pmatrix}1&3&2\end{pmatrix}}\right]}
|
i
1
…
i
n
⟩
S
=
S
|
i
1
…
i
n
⟩
normalizado
{\displaystyle |i_{1}\ldots i_{n}\rangle _{S}=S|i_{1}\ldots i_{n}\rangle \quad {\text{normalizado}}}
es simétrico (par) bajo transposiciones.
|
i
1
…
i
n
⟩
a
=
a
|
i
1
…
i
n
⟩
normalizado
{\displaystyle |i_{1}\ldots i_{n}\rangle _{a}=a|i_{1}\ldots i_{n}\rangle \quad {\text{normalizado}}}
es antisimétrico (impar) bajo transposiciones.
Dadas 3 partículas idéticas en el estado
|
a
′
⟩
{\displaystyle |a'\rangle }
|
a
″
⟩
{\displaystyle |a''\rangle }
|
a
‴
⟩
{\displaystyle |a'''\rangle }
|
a
′
a
″
a
‴
⟩
S
{\displaystyle |a'a''a'''\rangle _{S}}
|
a
′
a
″
a
‴
⟩
a
{\displaystyle |a'a''a'''\rangle _{a}}
(Creo que esto va aquí)
Ejercicio para casa: Simetrizar y antisimetrizar este estado
|
a
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a
′
a
″
⟩
{\displaystyle |a'a'a''\rangle }
(
1
3
)
|
a
′
a
″
a
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⟩
s
=
+
|
a
′
a
″
a
‴
⟩
s
{\displaystyle {\begin{pmatrix}1&3\end{pmatrix}}|a'a''a'''\rangle _{s}=+|a'a''a'''\rangle _{s}}
(
1
3
)
|
a
′
a
″
a
‴
⟩
a
=
−
|
a
′
a
″
a
‴
⟩
a
{\displaystyle {\begin{pmatrix}1&3\end{pmatrix}}|a'a''a'''\rangle _{a}=-|a'a''a'''\rangle _{a}}
![{\displaystyle S={\frac {1}{2}}\left[e+{\begin{pmatrix}1&2\end{pmatrix}}\right]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/3a6db7232d0699f1160b69de7d961bf18ba00175)
![{\displaystyle a={\frac {1}{2}}\left[e-{\begin{pmatrix}1&2\end{pmatrix}}\right]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/dafdc4c01e448c7922cd04c10f573e1626f0c296)
![{\displaystyle a|a'a''\rangle ={\frac {1}{2}}\left[e-{\begin{pmatrix}1&2\end{pmatrix}}|a'a''\rangle \right]={\frac {1}{2}}\left(|a'a''\rangle -|a''a'\rangle \right)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/8f340329a76f6235f0408bdb22f6a69b9185e9f2)
![{\displaystyle S_{3}\to s={\frac {1}{6}}\left[e+{\begin{pmatrix}1&2\end{pmatrix}}+{\begin{pmatrix}1&3\end{pmatrix}}+{\begin{pmatrix}2&3\end{pmatrix}}+{\begin{pmatrix}1&2&3\end{pmatrix}}+{\begin{pmatrix}1&3&2\end{pmatrix}}\right]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/278fe359258a31b0507082e287c7f3e5f92372f5)
![{\displaystyle S_{3}\to a={\frac {1}{6}}\left[e-{\begin{pmatrix}1&2\end{pmatrix}}-{\begin{pmatrix}1&3\end{pmatrix}}-{\begin{pmatrix}2&3\end{pmatrix}}+{\begin{pmatrix}1&2&3\end{pmatrix}}+{\begin{pmatrix}1&3&2\end{pmatrix}}\right]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/51d28e84f071b560f9f7c59e4b1bbe7943967ed6)
