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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 :)

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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.

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  • $\begingroup$ In my understanding I can just successively apply the operators in my Hamiltonian on my state to get my result. If I do it on this first example the annihilation/creation operator will act on 0/1 states and give a bare 0.... no state (according to my notes). How would you do eq2 if you don't get 0 ? Why shouldn't it have 4 degrees of freedom? An electron can change places or switch spin if it's allowed to, can't it? [5] as most of the other questions is just a question on notation "how would you write the adjoint of the given state?" $\endgroup$ – Kanaa Apr 11 at 9:15
  • $\begingroup$ @Kanaa Zero is the right answer. But that means that the energy of the state is zero, not that the state itself is zero/trivial or whatever. $\endgroup$ – Paul Apr 11 at 9:44
  • $\begingroup$ So what is the state then after applying the H? If I read our script on commutation relations correctly then it should be 0 since the operations are "not allowed". $\endgroup$ – Kanaa Apr 11 at 11:41
  • $\begingroup$ I don't know what Skript you reading, but assuming your state is an eigenstate of the Hamiltonian operator, applying the operator gives you eigenvalue times state. So at zero energy you get zero times the state which is a zero vector. It's all linear algebra. $\endgroup$ – Paul Apr 11 at 13:51
  • $\begingroup$ Ok I found a script on multi-body problems where they are doing more or less the same thing with just n-particles. They move to k-space to find an eigenfunction. Those I have stated are not eigenvectors thus it is perfectly fine if they change under this Hamiltonian. $\endgroup$ – Kanaa Apr 12 at 12:10

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