Constructing a Hamiltonian for $N$-qubits

Let us assume we have a qubit with an internal Hamiltonian $$H_0 = \sum_i \varepsilon_i |i\rangle\langle i|$$.

Now let's assume we have 2 such qubits. How would their joint Hamiltonian look like?

I have here a formula for N-qubits which I don't fully understand.

$$\begin{equation} \begin{split} H_0^{N} = \sum_k |1\rangle_k\langle 1|_k \otimes_{j \neq k} \mathbb{I}^j \end{split} \end{equation}$$

So for N=2 I'd get

$$\begin{equation} \begin{split} H_0^{2} = |1\rangle_1\langle 1|_1 \otimes_{2} \mathbb{I}^2 + |1\rangle_2\langle 1|_2 \otimes_{1} \mathbb{I}^1 \end{split} \end{equation}$$

Which I don't understand.

• Do you understand tensor products of Hilbert spaces and operators? – Norbert Schuch Dec 28 '18 at 12:09
• About the notation you used for tensor product in your last equation. Whereas $\otimes_{j \neq k}$ is OK, writing $\otimes_1$ is meaningless. The first notation is like the one you would use for a sum: $\sum_{j\neq k}$. It means that the tensor product is extended to all spaces except $\#k$. But what could possibly $\otimes_1$ mean? You have just 2 component spaces. In the first you choose the operator $|1\rangle_1\langle1|_1$, in the second you choose $\Bbb I$. Then make the tensor product - only one is allowed, so the index is useless. – Elio Fabri Dec 31 '18 at 20:58

Any Hamiltonian can be written in the form you give $$H=\sum_i\varepsilon_i|i\rangle\langle i|$$ as long as the eigenstates form a basis. This is still true for a many-body system. So for a single qubit you'd have $$H_1=\varepsilon_0|0\rangle\langle 0|+\varepsilon_1|1\rangle\langle 1|$$ And for two it'd look like $$H_2=\varepsilon_{00}|00\rangle\langle 00|+\varepsilon_{01}|01\rangle\langle 01|+\varepsilon_{10}|10\rangle\langle 10|+\varepsilon_{11}|11\rangle\langle 11|$$ This representation is completely general and so offers no real insight into anything. If you assume all of your qubits are governed by the same Hamiltonian and independent of the others (noninteracting) then you can write the joint Hamiltonian as a sum over each individual one like you have above. This is generally the situation of interest. $$H=^{(*)}\sum_j H_j=\sum_j\sum_i\varepsilon_i|i\rangle_j\langle i|_j$$

*Here $$H_j$$ means $$H_j\otimes_{k\neq j} \mathbb{I}_k$$ and $$|i\rangle_j\langle i|_j=|i\rangle_j\langle i|_j\otimes_{k\neq j} \mathbb{I}_k$$ i.e. they just act on qubit $$j$$

So for a two qubit system you'd have, fully written out: $$H_2=\varepsilon_0|0\rangle\langle 0|\otimes\mathbb{I}+\varepsilon_1|1\rangle\langle 1|\otimes\mathbb{I}+\varepsilon_0\mathbb{I}\otimes|0\rangle\langle 0|+\varepsilon_1\mathbb{I}\otimes|1\rangle\langle 1|$$

Or to more closely match your notation (which is not useful when written out in full, I'd only use it to compactify expressions) $$H_2=\varepsilon_0|0\rangle_1\langle 0|_1\otimes_2\mathbb{I}_2+\varepsilon_1|1\rangle_1\langle 1|_1\otimes_2\mathbb{I}_2+\varepsilon_0|0\rangle_2\langle 0|_2\otimes_1\mathbb{I}_1+\varepsilon_1|1\rangle_2\langle 1|_2\otimes_1\mathbb{I}_1$$

• Why the downvote? – bRost03 Dec 28 '18 at 4:37
• I see thank you. So if my calculations are correct this would mean that in this example for a 2 qubit system we have following relations: $\varepsilon_{00} = \varepsilon_{0}+\varepsilon_{0}$, $\varepsilon_{01}=\varepsilon_{0}+\varepsilon_{1}$ etc. Is this correct? Edit: I have no idea why or who downvoted it. I found it very helpful! – Benjamin Jabl Dec 28 '18 at 13:28
• Yeah, with $\varepsilon_{01}=\varepsilon_{10}$ – bRost03 Dec 28 '18 at 13:35
• Oh sorry I did not know, I have to accept. Also I did upvote but cause my account is new it's not shown to the public – Benjamin Jabl Dec 28 '18 at 16:52
• Easy to understand if you think of them as matrices. $(A+B)|v\rangle$ means apply $A$ to $|v\rangle$ then $B$ to $|v\rangle$ then add the resulting states. $A\cdot B|v\rangle$ means apply $B$ to $|v\rangle$ then apply $A$ to $B|v\rangle$. You might want to work through some relevant chapters of a linear algebra or QM textbook before continuing, these ideas are an absolute basic necessity to be able to understand the QC literature. Cheers! – bRost03 Dec 28 '18 at 18:26