A Representation in second quantization

[hokhani]

TL;DR Summary

to represent the states without ambiguity in second quantized form

For setting the fermionic state $|n_1 n_2\rangle$ where $n_i$ is the number of particles in the orbital $i$ we can use the following both representations for the state $|1 1\rangle$: $$ \hat {\mathbf c_1} ^{\dagger} \hat {\mathbf c_2} ^{\dagger} |vac\rangle $$ or $$ \hat {\mathbf c_2} ^{\dagger} \hat {\mathbf c_1} ^{\dagger} |vac\rangle $$ where $|vac\rangle$ is the vaccum state. But these two representations are different in a minus sign, because $\hat {\mathbf c_1} ^{\dagger} \hat {\mathbf c_2} ^{\dagger} |vac\rangle=-\hat {\mathbf c_2} ^{\dagger} \hat {\mathbf c_1} ^{\dagger} |vac\rangle$, and this issue cause problem in calculations. Which representation is correct?

 

[pines-demon]

TL;DR Summary: to represent the states without ambiguity in second quantized form

For setting the fermionic state $|n_1 n_2\rangle$ where $n_i$ is the number of particles in the orbital $i$ we can use the following both representations for the state $|1 1\rangle$: $$ \hat {\mathbf c_1} ^{\dagger} \hat {\mathbf c_2} ^{\dagger} |vac\rangle $$ or $$ \hat {\mathbf c_2} ^{\dagger} \hat {\mathbf c_1} ^{\dagger} |vac\rangle $$ where $|vac\rangle$ is the vaccum state. But these two representations are different in a minus sign, because $\hat {\mathbf c_1} ^{\dagger} \hat {\mathbf c_2} ^{\dagger} |vac\rangle=-\hat {\mathbf c_2} ^{\dagger} \hat {\mathbf c_1} ^{\dagger} |vac\rangle$, and this issue cause problem in calculations. Which representation is correct?

It's just a global phase, one gives you $|1_11_2\rangle$ and the other is $-|1_11_2\rangle$. Just stick to one ordering (I prefer ascending) and you'll be fine.

 

[PeterDonis]

this issue cause problem in calculations

It shouldn't; it should lead to correct calculations, since the sign change when you flip the ordering is just one way that the Pauli exclusion principle shows up in the math. And if your math doesn't correctly reflect the Pauli exclusion principle for fermions, it will give you wrong answers.

 

[hokhani]

It's just a global phase, one gives you $|1_11_2\rangle$ and the other is $-|1_11_2\rangle$. Just stick to one ordering (I prefer ascending) and you'll be fine.

How about calculation of states with more than two orbitals like $|n_1 n_2 n_3 n_4 \rangle$? To be specific consider for example $\mathbf c_1^\dagger \mathbf c_2 |0110\rangle$. Is the following way, correct?
$$\mathbf c_1^\dagger \mathbf c_2 |0110\rangle \\
=\mathbf c_1^\dagger \mathbf c_2 \mathbf c_2^\dagger \mathbf c_3^\dagger |vac\rangle

=\mathbf c_1^\dagger(1-\mathbf c_2^\dagger \mathbf c_2) \mathbf c_3^\dagger |vac\rangle
=\mathbf c_1^\dagger \mathbf c_3^\dagger |vac\rangle
=|1010\rangle
$$
I am stuck solving this formula and appreciate any help.

 

[pines-demon]

How about calculation of states with more than two orbitals like $|n_1 n_2 n_3 n_4 \rangle$? To be specific consider for example $\mathbf c_1^\dagger \mathbf c_2 |0110\rangle$. Is the following way, correct?
$$\mathbf c_1^\dagger \mathbf c_2 |0110\rangle \\
=\mathbf c_1^\dagger \mathbf c_2 \mathbf c_2^\dagger \mathbf c_3^\dagger |vac\rangle

=\mathbf c_1^\dagger(1-\mathbf c_2^\dagger \mathbf c_2) \mathbf c_3^\dagger |vac\rangle
=\mathbf c_1^\dagger \mathbf c_3^\dagger |vac\rangle
=|1010\rangle
$$
I am stuck solving this formula and appreciate any help.

The calculation is right, what step do you not understand?

 

[hokhani]

The calculation is right, what step do you not understand?

Thanks, I have to explain the problem exactly. In a Hilbert space including two orbitals $f$ and $c$ we have the states as $|n_f^{\uparrow} n_f^{\downarrow} n_c^{\uparrow} n_c^{\downarrow} \rangle$ where $n$ indicates the number of electron in each orbital. The total-spin operator for this system is defined as: $$\vec S =\frac {1} {2} \sum_{\mu \nu} (f_{\mu} ^{\dagger} \vec {\sigma}_{_{\mu \nu} } f_\nu+ c_{\mu} ^{\dagger} \vec {\sigma}_{_{\mu \nu} } c_\nu),$$ where $\vec {\sigma}$ is the vector of Pauli matrices and $\mu, \nu$ the spin indices.

With expanding this formula, we have:

\begin{align}

s_x=\frac{1}{2} ({f^{\dagger}_{\uparrow} f_{\downarrow} + c^{\dagger}_{\uparrow} c_{\downarrow} + f^{\dagger}_{\downarrow} f_{\uparrow} + c^{\dagger}_{\downarrow} c_{\uparrow} }) \\

s_y=\frac{1}{2} ({(-i)(f^{\dagger}_{\uparrow} f_{\downarrow} + c^{\dagger}_{\uparrow} c_{\downarrow}) +i(f^{\dagger}_{\downarrow} f_{\uparrow} + c^{\dagger}_{\downarrow} c_{\uparrow}) }) \\

s_z=\frac{1}{2} ({f^{\dagger}_{\uparrow} f_{\uparrow} + c^{\dagger}_{\uparrow} c_{\uparrow} - f^{\dagger}_{\downarrow} f_{\downarrow} - c^{\dagger}_{\downarrow} c_{\downarrow} }).

\end{align}

Now, I want to calculate $S^2 |0 1 1 0\rangle = (S_x^2+ S_y^2 + S_z^2 )|0 1 1 0\rangle$ which is expected to be zero because we have two opposite spins, one in $f$ and the other in $c$ orbitals. However, my direct calculation doesn’t result in this expectation. My calculations are briefly as: $$s_z |0110\rangle =0$$ and

\begin{align}



s_x |0110\rangle=\frac{1}{2} (|1010\rangle +|0101\rangle) \\

s_y |0110\rangle=\frac{1}{2} (-i|1010\rangle +i|0101\rangle) \\



s_x |1010\rangle=\frac{1}{2} (|0110\rangle +|1001\rangle) \\

s_y |1010\rangle=\frac{1}{2} i (|0110\rangle +|1001\rangle)\\





s_x |0101\rangle=\frac{1}{2} (|1001\rangle +|0110\rangle) \\

s_y |0101\rangle=\frac{1}{2} (-i)(|1001\rangle +|0110\rangle)\\

\end{align}



then we have:

\begin{align}



s^2_x|0110\rangle=\frac{1}{2} s_x(|1010\rangle +|0101\rangle) =\frac{1}{2} (|1001\rangle +|0110\rangle) \\
\end{align}
and similarly:
\begin{align}

s^2_y|0110\rangle=\frac{1}{2} (|0110\rangle +|1001\rangle) \\
\end{align}
and
\begin{align}

s^2_z|0110\rangle=0

\end{align}



Finally, we have:

$$s^2 |0110\rangle = (s^2_x+ s^2_y+ s^2_z)|0110\rangle =|0110\rangle +|0110\rangle$$ which obviously is not zero.
I would be grateful if you could please provide any help on this.

 

[pines-demon]

Thanks, I have to explain the problem exactly. In a Hilbert space including two orbitals $f$ and $c$ we have the states as $|n_f^{\uparrow} n_f^{\downarrow} n_c^{\uparrow} n_c^{\downarrow} \rangle$ where $n$ indicates the number of electron in each orbital. The total-spin operator for this system is defined as: $$\vec S =\frac {1} {2} \sum_{\mu \nu} (f_{\mu} ^{\dagger} \vec {\sigma}_{_{\mu \nu} } f_\nu+ c_{\mu} ^{\dagger} \vec {\sigma}_{_{\mu \nu} } c_\nu),$$ where $\vec {\sigma}$ is the vector of Pauli matrices and $\mu, \nu$ the spin indices.

With expanding this formula, we have:

\begin{align}

s_x=\frac{1}{2} ({f^{\dagger}_{\uparrow} f_{\downarrow} + c^{\dagger}_{\uparrow} c_{\downarrow} + f^{\dagger}_{\downarrow} f_{\uparrow} + c^{\dagger}_{\downarrow} c_{\uparrow} }) \\

s_y=\frac{1}{2} ({(-i)(f^{\dagger}_{\uparrow} f_{\downarrow} + c^{\dagger}_{\uparrow} c_{\downarrow}) +i(f^{\dagger}_{\downarrow} f_{\uparrow} + c^{\dagger}_{\downarrow} c_{\uparrow}) }) \\

s_z=\frac{1}{2} ({f^{\dagger}_{\uparrow} f_{\uparrow} + c^{\dagger}_{\uparrow} c_{\uparrow} - f^{\dagger}_{\downarrow} f_{\downarrow} - c^{\dagger}_{\downarrow} c_{\downarrow} }).

\end{align}

Now, I want to calculate $S^2 |0 1 1 0\rangle = (S_x^2+ S_y^2 + S_z^2 )|0 1 1 0\rangle$ which is expected to be zero because we have two opposite spins, one in $f$ and the other in $c$ orbitals. However, my direct calculation doesn’t result in this expectation. My calculations are briefly as: $$s_z |0110\rangle =0$$ and

\begin{align}



s_x |0110\rangle=\frac{1}{2} (|1010\rangle +|0101\rangle) \\

s_y |0110\rangle=\frac{1}{2} (-i|1010\rangle +i|0101\rangle) \\



s_x |1010\rangle=\frac{1}{2} (|0110\rangle +|1001\rangle) \\

s_y |1010\rangle=\frac{1}{2} i (|0110\rangle +|1001\rangle)\\





s_x |0101\rangle=\frac{1}{2} (|1001\rangle +|0110\rangle) \\

s_y |0101\rangle=\frac{1}{2} (-i)(|1001\rangle +|0110\rangle)\\

\end{align}



then we have:

\begin{align}



s^2_x|0110\rangle=\frac{1}{2} s_x(|1010\rangle +|0101\rangle) =\frac{1}{2} (|1001\rangle +|0110\rangle) \\
\end{align}
and similarly:
\begin{align}

s^2_y|0110\rangle=\frac{1}{2} (|0110\rangle +|1001\rangle) \\
\end{align}
and
\begin{align}

s^2_z|0110\rangle=0

\end{align}



Finally, we have:

$$s^2 |0110\rangle = (s^2_x+ s^2_y+ s^2_z)|0110\rangle =|0110\rangle +|0110\rangle$$ which obviously is not zero.
I would be grateful if you could please provide any help on this.

You changed the question what was the problem with the previous question? Also you posted a calculation once again what step do you not understand?

 

[hokhani]

You changed the question what was the problem with the previous question? Also you posted a calculation once again what step do you not understand?

My main problem was this last post (#6), but I thought that the problem might be with my way of calculations. So, first I expressed the problem as in the previous question (post #4) which you approved my calculations (post #5). In the above calculations (#6) I followed the same way, but the final result is not convincing.