# Dirac notation: what if we display tensor products vertically?

Ok hear me out.

One thing I have always liked about Dirac notation is that it visually displays where expressions expect inputs/outputs. For example $$\langle\psi|$$ expects an input to the right to form a complete expression like $$\langle\psi|\phi\rangle$$. The angle brackets indicate the expression is 'complete'. Things like $$|\psi\rangle\langle\phi|$$ require input both on the left and the right as indicated by the vertical bars. Now for tensor products this behaviour kind of breaks down. Expressions like $$|n\rangle|\phi\rangle\equiv|n\rangle\otimes|\phi\rangle$$ have two inputs but the input of the second ket is stuck behind the first ket. If you calculate something like $$\langle m|\langle\psi|\langle\alpha||n\rangle|\phi\rangle|\beta\rangle=\langle m| n\rangle \langle\psi |\phi\rangle \langle\alpha |\beta\rangle$$ the corresponding inputs and outputs are separated really far.

This might not sound like a big a deal (it isn't a big deal) but what if we stacked tensor products vertically? Like $$|n\rangle|\phi\rangle\equiv\begin{array}{c}|n\rangle\\|\phi\rangle\end{array}$$ Then the triple dot product from earlier becomes $$\begin{array}{c}\langle m|\\\langle\psi|\\\langle\alpha|\end{array}\cdot\begin{array}{c}|n\rangle\\|\phi\rangle\\|\beta\rangle\end{array}= \begin{array}{c}\langle m| n\rangle\\ \langle\psi |\phi\rangle \\\langle\alpha |\beta\rangle \end{array}= \langle m| n\rangle \langle\psi |\phi\rangle \langle\alpha |\beta\rangle$$ and products of general operators become $$(A\otimes B)\cdot(C\otimes D)=\begin{array}{c}A\\B\end{array}\cdot \begin{array}{c}C\\D\end{array}=\begin{array}{c}A\cdot C\\B\cdot D\end{array}=(A\cdot B)\otimes(C\cdot D)$$ So there are two benefits:

1. Inputs/outputs are visually closer. So this is subjectively easier to understand.
2. Expression with tensor products tend to work 'in parallel', that is each term in a tensor product often works separately from the others. The $$|n\rangle$$ and $$|\phi\rangle$$ in my example could be from entirely different Hilbert spaces. This vertical notation displays this visually because each tensor product term gets a separate row and the rows never talk to eachother.

Any thoughts? Improvements? I don't expect this notation will ever catch on because it is quite unwieldy but I hope this question gives you a fresh view on tensor products

• I'm not sure this if this is on topic by my two cents are that I like that this mimics the way you'd lay out multiple registers in a quantum circuit (corresponding to multiple qubits) but dislike that it seems like it will take up lots of space. – jacob1729 Jun 17 '20 at 22:24
• I really like this question but I fear it might be off topic, because it is just asking for opinions. – Javier Jun 17 '20 at 23:07
• Your notation can quickly become misleading: You might well visualize the one vector over the other as a concatenation of two vectors into one vector of dimensionality being the sum of those of the two. But that is the direct sum $\oplus$, and not the direct product $\otimes$. Multiplying the direct product by a number amounts to multiplying either factor by said number, but I defy you to intuit that in your vertical notation. Sooner or later, somebody clueless will slip and break his neck... – Cosmas Zachos Jun 17 '20 at 23:11
• @Javier Yeah I was afraid of that. I hope that having soft-question as a tag prevents this question from being closed. – AccidentalTaylorExpansion Jun 17 '20 at 23:31
• There is no reason to stop in 2D, some operators are easier to understand in 3D. But it is a hassle to write in 3D so it usually stops at proof of concept... – Emil Jun 18 '20 at 6:58

Suppose the dimension of $$|n\rangle$$ is a, take it to be 3, and that of $$|\phi\rangle$$ is b, take it to be 4.
Then the dimension of the vector $$|n\rangle|\phi\rangle$$ is ab, so, 12. However, the proposed notation, $$|n\rangle|\phi\rangle\equiv\begin{array}{c}|n\rangle\\|\phi\rangle\end{array}$$ all but invites the reader to think of the r.h.s. as the pileup of the two vectors, i.e. an a+b dimensional vector, here 7-d, (so $$|n\rangle \oplus|\phi\rangle$$), and not a 12-d vector.
You might beat your readers to not misunderstand the obvious, but somebody always will. (I am reminded of the trick-question calculus wag, who'd use the $$x(f)$$, $$~~dx/df$$, $$~~d^2x/df^2$$,... notation in trivial questions and stump his graders.) After virtually a century of Clebsching, it is a tall order to rescind collective public expectations.