Second quantization notation - Hamiltonian on triplet state

Question

So I'm struggling quite a bit with dirac notation and second quantization and it seems like no one wants to really do calculations step-by-step to at least get the notation right. We were given the following Hamiltonian: $$ H=t\sum_\sigma c_{1,\sigma}^\dagger c_{2,\sigma} + c_{2,\sigma}^\dagger c_{1,\sigma} \tag{1}$$ with $c$ and $c^\dagger$ annihilation and creation operator with position and spin $\sigma=\uparrow,\downarrow$ as indices. Now I should apply this on the two-fermion triplet state $ \lvert \uparrow \uparrow \rangle $. To make this a little more straightforward I thought, I'd better write the states in terms of occupation number: $ \lvert n_{1\uparrow}, n_{2\uparrow}, n_{1\downarrow}, n_{2\downarrow} \rangle $

for my triplet state I now have: $ \lvert \uparrow \uparrow \rangle = \lvert 1,1,0,0\rangle $

Applying the Hamiltonian on this yields: $$ H \lvert 1,1,0,0\rangle = t(c_{1,\uparrow}^\dagger c_{2,\uparrow} + c_{2,\uparrow}^\dagger c_{1,\uparrow}+c_{1,\downarrow}^\dagger c_{2,\downarrow} + c_{2,\downarrow}^\dagger c_{1,\downarrow})\lvert 1,1,0,0\rangle=0+0+0+0 \tag{2}$$

By only using $a^\dagger \lvert 1 \rangle=0$ and $a \lvert 0 \rangle=0$. What's irritating: we should go on calculate more stuff with this "state". But there is no state anymore. [1.] Did I do something completly wrong?

Furthermore we should apply this Hamiltonian on other states. Like the singlet state: $ \frac{1}{\sqrt{2}}(\lvert \uparrow \downarrow\rangle - \lvert \downarrow \uparrow \rangle) = \frac{1}{\sqrt{2}} (\lvert 1,0,0,1\rangle-\lvert 0,1,1,0\rangle) $

Applying it the same way as above: $$ H \frac{1}{\sqrt{2}} \lvert 1,0,0,1\rangle = t (0+\lvert 0,1,0,1\rangle+\lvert 1,0,1,0\rangle+0) $$

$$H\frac{1}{\sqrt{2}}\lvert 0,1,1,0\rangle= t(\lvert 1,0,1,0\rangle+0+0+\lvert 0,1,0,1\rangle) $$

$$ H \frac{1}{\sqrt{2}}(\lvert \uparrow \downarrow\rangle - \lvert \downarrow \uparrow \rangle) =0 \tag{3}$$

Even if $0$ is also correct here, it is a very interchangeable result. Going on with this, there is only one triplet state which is non-zero.

[2.] Does that make sense/ is correct?

[3.] How would one write $\lvert 1,0,1,0\rangle$ in the "arrow notation" $\lvert 1,0,1,0\rangle=\lvert \uparrow \downarrow,0 \rangle$?

[4.] And are those notations: $\lvert \uparrow \downarrow,0 \rangle =\lvert \uparrow \downarrow\rangle \lvert0 \rangle$ or $\lvert \uparrow \uparrow \rangle=\lvert \uparrow \rangle\lvert \uparrow\rangle$ equivalent?

[5.] Is the complex conjugate of $\lvert \uparrow \uparrow\rangle^\dagger= \langle \uparrow \uparrow \lvert$ ?

[6.] Also on some other question I have seen the following line: $(\lvert\downarrow \rangle \lvert\uparrow\rangle - \lvert\uparrow \rangle \lvert\downarrow\rangle)/\sqrt{2} = \lvert 1_\downarrow 1_\uparrow \rangle $. How is the right-hand-side notation defined? I don't see how this can be only one term.

I'm sorry that this is more than one question, but it sorta belongs together for me. I thank you in advance :)

Answer 1

Your main problem seems to bet that you are mixing up state and eigenvalue of the Hamiltonian. Your states are eigenvectors of the Hamiltonian operator, s.t. one gets an eigenvalue (the total energy of the state), when applying $H$. This means that even when the Hamiltonian acting on a state gives zero, it doesn't mean that the state is zero. You can still calculate things with that state. This should take care of [1.] & [2.].

Also your notation of the states is somewhat confusing, as it makes it look as if you have four degrees of freedom, when you actually have only two (left spin and right spin). You can still use it, just be cautious with that. Knowing this [3.] isn't a problem anymore. The thing is that you have a system with two fermions at two positions without any kinetics. This means you can't have a state with both fermions in one position with different spins in this small example, even though describing such systems may be the point of Second Quantization.

For [5.]: As states aren't numbers but vectors in Hilbert space, where $\langle \Psi \rvert$ is the dual vector of $\lvert \Psi \rangle$, such that $\langle \Psi \vert \Psi \rangle$ is a number (generalized scalar product), the probability(-density). To make a scalar product work with complex numbers, we not only have to take the transpose of a vector to get the dual vector but the adjoint (also called conjugate transpose): $\lvert \Psi \rangle^\dagger = (\lvert \Psi \rangle^T)^* = (\lvert \Psi \rangle^*)^T = \langle \Psi \rvert$, where $^T$ marks a transpose and $^*$ marks the complex conjugate. So the answer is yes, but $^\dagger$ is not a simple complex conjugate.

I can't help you with all those notations. They should be defined somewhere. For this exercise the notation on your sheet with $\lvert \uparrow \uparrow \rangle$ etc. is all you need, as long as you now that the left arrow describes position one and the right arrow describes position two. You could translate e.g. "up" to $1$ and "down" to $-1$, but that doesn't really help, does it.

You may want to refresh your basics in Quantum mechanics before diving into Second Quantization.