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For a spin 1 system l = 1 and m = -1 , 0 , 1. For the operator <math>S_{z}</math> we have
'''Sakurai Excercise 4.11'''
 
The Hamiltonian for a spin 1 system is given by
 
<math>H = A  S_{z}^{2} + B ( S_{x}^{2} -  S_{y}^{2} )</math>
 
Solve this problem exactly to find the normalized energy eigenstates and eigenvalues. Is this Hamiltonian of this actually invariant under time reversal? How do the normalized eigenstates obtained transform under time reversal?
 
'''Solution:'''
 
 
For a spin 1 system l = 1 and m = -1 , 0 , 1. For the operator <math> S_{z} </math> we have


<math>S_z |l,m \rangle</math> <math>\Rightarrow</math>  <math> \langle l,n| S_z|l,m \rangle = m\hbar \langle l,n|l,m \rangle</math> <math>\Rightarrow ( S_z)_{nm} = m\hbar \delta _{nm}</math>
<math>S_z |l,m \rangle</math> <math>\Rightarrow</math>  <math> \langle l,n| S_z|l,m \rangle = m\hbar \langle l,n|l,m \rangle</math> <math>\Rightarrow ( S_z)_{nm} = m\hbar \delta _{nm}</math>
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\end{pmatrix} = 0</math>
\end{pmatrix} = 0</math>


<math>\Rightarrow</math> <math>\lambda_1</math> = 0, <math>\lambda_2 = {\hbar}^2(A + B)</math>, <math>\lambda_3 = {\hbar}^2(A - B)</math>
<math>\Rightarrow</math> <math> \lambda_{1} </math> = 0, <math>\lambda_2 = {\hbar}^2(A + B)</math>, <math>\lambda_3 = {\hbar}^2(A - B)</math>
 
To find the eigenstate <math>|n_(\lambda)</math> that corresponds to the eigenvalue <math>\lambda</math> we have to solve the following equation:
 
 
<math> {\hbar}^2\begin{pmatrix}
  A & 0 & B \\
  0 & 0 & 0 \\
  B & 0 & A\\
 
\end{pmatrix} \begin{pmatrix}
a\\
b\\
c
\end{pmatrix} \Rightarrow  \lambda _{c} \begin{pmatrix}
a\\
b\\
c
\end{pmatrix}</math>
 
For <math>\lambda_1 = 0</math>
 
<math> {\hbar}^2\begin{pmatrix}
  A & 0 & B \\
  0 & 0 & 0 \\
  B & 0 & A\\
 
\end{pmatrix} \begin{pmatrix}
a\\
b\\
c
\end{pmatrix}= 0 \Rightarrow a= 0 , c= 0</math> 
 
<math>|n_0 \rangle = \begin{pmatrix}
0\\
b \\
0
\end{pmatrix} (normalizing ) \Rightarrow |n_0 \rangle = \begin{pmatrix}
0\\
1 \\
0
\end{pmatrix}\Rightarrow  |n_0 \rangle = |1 , 0 \rangle</math>
 
In the same way for <math>\lambda_2 = {\hbar}^2(A + B)</math>
 
<math> \begin{pmatrix}
  A & 0 & B \\
  0 & 0 & 0 \\
  B & 0 & A\\
 
\end{pmatrix} \begin{pmatrix}
a\\
b\\
c
\end{pmatrix}= (A+B)\begin{pmatrix}
a\\
b\\
c
\end{pmatrix} \Rightarrow a= c , b= 0</math>
 
 
 
<math>|n_{A+B} \rangle = \begin{pmatrix}
c\\
0 \\
c
\end{pmatrix} (normalizing ) \Rightarrow |n_0 \rangle = \frac{1}{\sqrt{2}}\begin{pmatrix}
1\\
0 \\
1
\end{pmatrix}\Rightarrow  |n_{A+B} \rangle = \frac{1}{\sqrt{2}}|1 , +1 \rangle + \frac{1}{\sqrt{2}}|1 , -1 \rangle</math>
 
In the same way for <math>\lambda_2 = {\hbar}^2(A + B)</math>
 
<math> \begin{pmatrix}
  A & 0 & B \\
  0 & 0 & 0 \\
  B & 0 & A\\
 
\end{pmatrix} \begin{pmatrix}
a\\
b\\
c
\end{pmatrix}= (A+B)\begin{pmatrix}
a\\
b\\
c
\end{pmatrix} \Rightarrow a= c , b= 0</math> 
 
 
 
<math>|n_{A+B} \rangle = \begin{pmatrix}
c\\
0 \\
c
\end{pmatrix} (normalizing ) \Rightarrow |n_{A+B} \rangle = \frac{1}{\sqrt{2}}\begin{pmatrix}
1\\
0 \\
1
\end{pmatrix}\Rightarrow  |n_{A+B} \rangle = \frac{1}{\sqrt{2}}|1 , +1 \rangle + \frac{1}{\sqrt{2}}|1 , -1 \rangle</math>
 
For <math>\lambda_3 = {\hbar}^2(A - B)</math>
 
<math> \begin{pmatrix}
  A & 0 & B \\
  0 & 0 & 0 \\
  B & 0 & A\\
 
\end{pmatrix} \begin{pmatrix}
a\\
b\\
c
\end{pmatrix}= (A - B)\begin{pmatrix}
a\\
b\\
c
\end{pmatrix} \Rightarrow a= -c , b= 0</math> 
 
 
 
<math>|n_{A-B} \rangle = \begin{pmatrix}
c\\
0 \\
-c
\end{pmatrix} (normalizing ) \Rightarrow |n_{A-B}\rangle = \frac{1}{\sqrt{2}}\begin{pmatrix}
1\\
0 \\
-1
\end{pmatrix}\Rightarrow  |n_{A-B} \rangle = \frac{1}{\sqrt{2}}|1 , +1 \rangle - \frac{1}{\sqrt{2}}|1 , -1 \rangle</math>
 
Now we are going to check if the Hamiltonian is invariant under time reversal
 
<math>\Theta H\Theta ^{-1} = A\Theta  S_{z}^{2} \Theta^{-1} + B (\Theta  S_{x}^{2} \Theta^{-1}  - \Theta  S_{y}^{2} \Theta^{-1})</math>
 
 
 
 
<math> = A\Theta  S_{z} \Theta^{-1} \Theta S_{z} \Theta^{-1}  + B (\Theta  S_{x} \Theta^{-1} \Theta S_{x} \Theta^{-1}  - \Theta  S_{y} \Theta^{-1} \Theta S_{y} \Theta^{-1})</math>
 
 
<math> = A  S_{z}^{2} + B (  S_{x}^{2} -  S_{y}^{2}) = H</math>
 
To find the transformation of the eigenstates under time reversal we use the relation
 
<math> \Theta|l,m \rangle</math> = <math>(-1)^{m}\Theta |l, -m \rangle</math>
 
So <math> \Theta |n_0\rangle =  \Theta |1, 0\rangle =  \Theta |1,0\rangle = |n_0\rangle</math>
 
 
<math>\Theta |n_{A+B} \rangle = \frac{1}{\sqrt{2}}\Theta |1 , +1 \rangle + \frac{1}{\sqrt{2}}\Theta |1 , -1 \rangle</math>
 
<math>\Theta |n_{A+B} \rangle = -\frac{1}{\sqrt{2}}\Theta |1 , -1 \rangle - \frac{1}{\sqrt{2}} |1 , +1 \rangle</math> = <math> - |n_{A+B} \rangle</math>
 
 
<math>\Theta |n_{A-B} \rangle = \frac{1}{\sqrt{2}}\Theta |1 , +1 \rangle - \frac{1}{\sqrt{2}}\Theta |1 , -1 \rangle</math>
 
 
<math>\Theta |n_{A-B} \rangle = -\frac{1}{\sqrt{2}}\Theta |1 , -1 \rangle + \frac{1}{\sqrt{2}} |1 , +1 \rangle</math> = <math> |n_{A-B} \rangle</math>

Latest revision as of 00:22, 1 May 2010

Sakurai Excercise 4.11

The Hamiltonian for a spin 1 system is given by

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle H = A S_{z}^{2} + B ( S_{x}^{2} - S_{y}^{2} )}

Solve this problem exactly to find the normalized energy eigenstates and eigenvalues. Is this Hamiltonian of this actually invariant under time reversal? How do the normalized eigenstates obtained transform under time reversal?

Solution:


For a spin 1 system l = 1 and m = -1 , 0 , 1. For the operator Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle S_{z} } we have

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle S_z |l,m \rangle} Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \Rightarrow} Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \langle l,n| S_z|l,m \rangle = m\hbar \langle l,n|l,m \rangle} Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \Rightarrow ( S_z)_{nm} = m\hbar \delta _{nm}}

So


Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle S_{z} = \hbar \begin{pmatrix} 1 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & -1 \\ \end{pmatrix}} Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \Rightarrow} Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle S_{z}^{2} = {\hbar}^2\begin{pmatrix} 1 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 1 \\ \end{pmatrix}}

For the Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle S_x} operator we have



Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle S_{x} = \frac{\hbar}{2}\begin{pmatrix} 0 & \sqrt{2} & 0 \\ \sqrt{2} & 0 & \sqrt{2} \\ 0 & \sqrt{2} & 0\\ \end{pmatrix}} Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \Rightarrow} Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle S_{x}^{2} = {\hbar}^2\begin{pmatrix} \frac{1}{2} & 0 & \frac{1}{2} \\ 0 & 1 & 0 \\ \frac{1}{2}& 0 & \frac{1}{2}\\ \end{pmatrix}} Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \Rightarrow} Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle S_{y} = \frac{\hbar}{2i}\begin{pmatrix} 0 & \sqrt{2} & 0 \\ -\sqrt{2} & 0 & \sqrt{2} \\ 0 & -\sqrt{2} & 0\\ \end{pmatrix}} Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle S_{y}^{2} = {\hbar}^2\begin{pmatrix} \frac{1}{2} & 0 & -\frac{1}{2} \\ 0 & 1 & 0 \\ -\frac{1}{2}& 0 & \frac{1}{2}\\ \end{pmatrix}}

Thus the Hamiltonian can be represented by the matrix

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle H = {\hbar}^2\begin{pmatrix} A & 0 & B \\ 0 & 0 & 0 \\ B & 0 & A\\ \end{pmatrix}}

To find the energy eigenvalues we have to solve the secular equation

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle det(H - \lambda I) = 0 } Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \Rightarrow} Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle det \begin{pmatrix} A{\hbar}^2 - \lambda & 0 & B{\hbar}^2 \\ 0 & - \lambda & 0 \\ B{\hbar}^2 & 0 & A{\hbar}^2 - \lambda\\ \end{pmatrix} = 0}

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \Rightarrow} Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \lambda_{1} } = 0, Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \lambda_2 = {\hbar}^2(A + B)} , Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \lambda_3 = {\hbar}^2(A - B)}

To find the eigenstate Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle |n_(\lambda)} that corresponds to the eigenvalue Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \lambda} we have to solve the following equation:


Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle {\hbar}^2\begin{pmatrix} A & 0 & B \\ 0 & 0 & 0 \\ B & 0 & A\\ \end{pmatrix} \begin{pmatrix} a\\ b\\ c \end{pmatrix} \Rightarrow \lambda _{c} \begin{pmatrix} a\\ b\\ c \end{pmatrix}}

For Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \lambda_1 = 0}

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle {\hbar}^2\begin{pmatrix} A & 0 & B \\ 0 & 0 & 0 \\ B & 0 & A\\ \end{pmatrix} \begin{pmatrix} a\\ b\\ c \end{pmatrix}= 0 \Rightarrow a= 0 , c= 0}

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle |n_0 \rangle = \begin{pmatrix} 0\\ b \\ 0 \end{pmatrix} (normalizing ) \Rightarrow |n_0 \rangle = \begin{pmatrix} 0\\ 1 \\ 0 \end{pmatrix}\Rightarrow |n_0 \rangle = |1 , 0 \rangle}

In the same way for Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \lambda_2 = {\hbar}^2(A + B)}

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{pmatrix} A & 0 & B \\ 0 & 0 & 0 \\ B & 0 & A\\ \end{pmatrix} \begin{pmatrix} a\\ b\\ c \end{pmatrix}= (A+B)\begin{pmatrix} a\\ b\\ c \end{pmatrix} \Rightarrow a= c , b= 0}


Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle |n_{A+B} \rangle = \begin{pmatrix} c\\ 0 \\ c \end{pmatrix} (normalizing ) \Rightarrow |n_0 \rangle = \frac{1}{\sqrt{2}}\begin{pmatrix} 1\\ 0 \\ 1 \end{pmatrix}\Rightarrow |n_{A+B} \rangle = \frac{1}{\sqrt{2}}|1 , +1 \rangle + \frac{1}{\sqrt{2}}|1 , -1 \rangle}

In the same way for Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \lambda_2 = {\hbar}^2(A + B)}

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{pmatrix} A & 0 & B \\ 0 & 0 & 0 \\ B & 0 & A\\ \end{pmatrix} \begin{pmatrix} a\\ b\\ c \end{pmatrix}= (A+B)\begin{pmatrix} a\\ b\\ c \end{pmatrix} \Rightarrow a= c , b= 0}


Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle |n_{A+B} \rangle = \begin{pmatrix} c\\ 0 \\ c \end{pmatrix} (normalizing ) \Rightarrow |n_{A+B} \rangle = \frac{1}{\sqrt{2}}\begin{pmatrix} 1\\ 0 \\ 1 \end{pmatrix}\Rightarrow |n_{A+B} \rangle = \frac{1}{\sqrt{2}}|1 , +1 \rangle + \frac{1}{\sqrt{2}}|1 , -1 \rangle}

For Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \lambda_3 = {\hbar}^2(A - B)}

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{pmatrix} A & 0 & B \\ 0 & 0 & 0 \\ B & 0 & A\\ \end{pmatrix} \begin{pmatrix} a\\ b\\ c \end{pmatrix}= (A - B)\begin{pmatrix} a\\ b\\ c \end{pmatrix} \Rightarrow a= -c , b= 0}


Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle |n_{A-B} \rangle = \begin{pmatrix} c\\ 0 \\ -c \end{pmatrix} (normalizing ) \Rightarrow |n_{A-B}\rangle = \frac{1}{\sqrt{2}}\begin{pmatrix} 1\\ 0 \\ -1 \end{pmatrix}\Rightarrow |n_{A-B} \rangle = \frac{1}{\sqrt{2}}|1 , +1 \rangle - \frac{1}{\sqrt{2}}|1 , -1 \rangle}

Now we are going to check if the Hamiltonian is invariant under time reversal

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \Theta H\Theta ^{-1} = A\Theta S_{z}^{2} \Theta^{-1} + B (\Theta S_{x}^{2} \Theta^{-1} - \Theta S_{y}^{2} \Theta^{-1})}



Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle = A\Theta S_{z} \Theta^{-1} \Theta S_{z} \Theta^{-1} + B (\Theta S_{x} \Theta^{-1} \Theta S_{x} \Theta^{-1} - \Theta S_{y} \Theta^{-1} \Theta S_{y} \Theta^{-1})}


Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle = A S_{z}^{2} + B ( S_{x}^{2} - S_{y}^{2}) = H}

To find the transformation of the eigenstates under time reversal we use the relation

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \Theta|l,m \rangle} = Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle (-1)^{m}\Theta |l, -m \rangle}

So Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \Theta |n_0\rangle = \Theta |1, 0\rangle = \Theta |1,0\rangle = |n_0\rangle}


Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \Theta |n_{A+B} \rangle = \frac{1}{\sqrt{2}}\Theta |1 , +1 \rangle + \frac{1}{\sqrt{2}}\Theta |1 , -1 \rangle}

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \Theta |n_{A+B} \rangle = -\frac{1}{\sqrt{2}}\Theta |1 , -1 \rangle - \frac{1}{\sqrt{2}} |1 , +1 \rangle} = Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle - |n_{A+B} \rangle}


Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \Theta |n_{A-B} \rangle = \frac{1}{\sqrt{2}}\Theta |1 , +1 \rangle - \frac{1}{\sqrt{2}}\Theta |1 , -1 \rangle}


Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \Theta |n_{A-B} \rangle = -\frac{1}{\sqrt{2}}\Theta |1 , -1 \rangle + \frac{1}{\sqrt{2}} |1 , +1 \rangle} =

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