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I was starting to learn Dirac notation with MIT's notes on QM. The introduction states that Dirac notation starts from turning inner products from: $$ \langle{u}, v\rangle $$ to, substituting the comma with a bar: $$ \langle{u}| v\rangle$$ Then it is said that we separate $\langle{u}|$ (bras) and $| v\rangle$ (kets) as objects by themselves, so that, if the vector space under consideration is denoted $V$, $v \in V$ (in this notation) doesn't make sense anymore, but instead, $| v\rangle \in V$ is true, as now the kets are the elements of the space now: the letter inside $| \space \space\rangle$ is just a label, so operations, etc. must be carried out with the kets as a whole. However, consider the linearity condition for the second slot in inner products: $$ \langle u, a_1 v_1 +a_2 v_2 \rangle = a_1 \langle u, v_1 \rangle + a_2 \langle u, v_2 \rangle$$ This makes sense, as the quantity $a_1 v_1 + a_2 v_2$ is a vector, but $v_1$ and $v_2$ are themselves vectors as well. Then it is stated that, in Dirac notation, this is translated to: $$ \langle u| a_1 v_1 +a_2 v_2 \rangle = a_1 \langle u| v_1 \rangle + a_2 \langle u| v_2 \rangle$$ This looks very wrong to me: now $a_1 v_1 +a_2 v_2$ is not a vector anymore, but $| a_1 v_1 +a_2 v_2 \rangle$ is. Therefore, it doesn't even make sense to say that $|v_1 \rangle$ is a vector, since the place where is is superposed is inside the label of the ket. So, is this condition actually stated correctly and there is something I'm not getting, or is there another way to state such axiom correctly?

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    $\begingroup$ What you've encountered is an example of this, but it's not a bad thing. $\endgroup$
    – J.G.
    Jul 8, 2021 at 6:03
  • $\begingroup$ At the end of the day, this is all just notation, so as long as you understand what that notation means, there is no problem. In your case, we have $| a_1 v_1 + a_2 v_2 \rangle = a_1 | v_1 \rangle + a_2 | v_2 \rangle$ and $\langle a_1 v_1 + a_2 v_2 | = a_1^* \langle v_1 | + a_2^* \langle v_2 |$. $\endgroup$
    – Prahar
    Jul 8, 2021 at 10:16

4 Answers 4

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The key with Dirac notation is, as you mention, that the thing within the bra or the ket is a label, not a vector in a vector space. I always say that writing $|\psi\rangle\in V$ as an element of the vector space is, in the notation more common in mathematics, $v_\psi\in V$. Note that $\psi$ is not an element of the vector space. It is simply a symbol we have decided to label the vector by the same way we might attach subscripts to vectors to keep track of them.

With this in mind, one never needs to employ any notion of linearity to the symbols "inside" the bra or ket, though you'll often see this sloppy notation used in introductory sources. In all such cases, the definition of linearity in the label is linearity in the vector. By that, I mean authors tend to write $|a\psi_1 + b\psi_2\rangle \equiv a|\psi_1\rangle + b|\psi_2\rangle$. Again, this is sloppy notation and I recommend avoiding it as much as possible: the only reason to write $|a\psi_1+b\psi_2\rangle$ rather than $a|\psi_1\rangle + b|\psi_2\rangle$ is laziness (as writing things correctly would often necessitate an additional set of parentheses be drawn).

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    $\begingroup$ What you call "laziness" others might call "clarity". Why make the reader parse an extra set of parentheses when those parentheses serve no purpose but pedantry? $\endgroup$
    – Nobody
    Jul 8, 2021 at 12:00
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    $\begingroup$ @Nobody if it were truly nothing more than pedantry, then questions like OP's surely would not exist. Symbols and operations should have consistent meaning. $\endgroup$ Jul 8, 2021 at 18:46
  • $\begingroup$ I don't think that's "surely" the case at all. There is no notation so clear that there won't be somebody somewhere that's confused by it, and for all we know the source that the OP was working from simply did a bad job of explaining this usage. Symbols and operations should be used in the way that communicates the underlying ideas most clearly, even if that means the occasional benign inconsistency. $\endgroup$
    – Nobody
    Jul 8, 2021 at 20:05
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It sounds like your question is essentially semantic, whether it's "proper" to understand

$$|a_1 v_1 + a_2 v_2 \rangle \equiv a_1 |v_1 \rangle + a_2 |v_2 \rangle$$.

To me, it's not, but this seems to be the level of notational rigor that physicists often use. For example, we might refer to the Fourier transform of $f(t)$ as $f(\omega)$ rather than $\tilde{f}(\omega)$ and rely on the argument to indicate which is which. It can cause some confusion, but it can also prove convenient.

(Incidentally, I've been trying to get my student to stop using this questionable notation. It struck him as intuitive, but I think he's willing to give it up.)

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It's a shame the notes began with $u,\,v$ instead of $\phi,\,\psi$. With the latter, they could then have emphasized that $\sum_ia_i\psi_i$, with scalars $a_i$ (note the Roman symbol) and vectors $\psi_i$ (note the Greek symbol) in the pre-(bra-ket) formalism, becomes $\sum_ia_i|\psi_i\rangle$ in the bra-ket formalism, wherein this vector can also be written as $\big|\sum_ia_i\psi_i\big\rangle$. (In theory, two colours could have achieved the same thing.)

This is a slightly more general vector-label relationship that acknowledges and harnesses the linearity of kets, not to mention the antilinearity of bras and sesquilinearity of inner products once we do the same thing with bras.

As one answer to a question on abuse of notation notes, human readers

are capable of using context, guessing, and all sorts of other information when decoding what we write/say. It is generally immensely more efficient to take advantage of this.

In this case it wasn't perfectly exploited, but hopefully it's clear now.

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The notation that is used by the author makes this unnecessarily confusing. The following statements are just relabellings and indicate the same thing. From left to right they are vector notation, vector component notation and Dirac notation. \begin{align} v&\leftrightarrow\begin{pmatrix}v_1\\v_2\end{pmatrix}\leftrightarrow|v\rangle\\ \\ v^\dagger&\leftrightarrow \begin{pmatrix}v_1^*&v_2^*\end{pmatrix}\leftrightarrow\langle v|\\ \\ a\, u+b\,v&\leftrightarrow \begin{pmatrix}au_1+bv_1\\au_2+bv_2\end{pmatrix}\leftrightarrow a|u\rangle+b|v\rangle \end{align} Here I used $u,v$ instead of $v_1,v_2$ to avoid confusion with vector components. So the more explicit way to write the statement would be $$\langle u|\bigg(a|v\rangle+b|w\rangle\bigg)=a\langle u|v\rangle+b\langle u|w\rangle$$ which is equivalent to the shorthand notation $$\langle u|av+bw\rangle=a\langle u|v\rangle+b\langle u|w\rangle$$

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    $\begingroup$ This made it much clearer, thanks! It made it even more confusing that the notes themselves mention afterwards that, for position states, you cannot say |ax>=a|x>, as the first one represents a position eigenstate at "ax" and the second is a position state at "x" with amplitude "a". $\endgroup$
    – Johnn.27
    Jul 8, 2021 at 22:08
  • $\begingroup$ @Nick.25 I see, I get why they want to introduce the shorthand notation but it seems weird to introduce it so early when you just want to understand the concepts clearly $\endgroup$ Jul 9, 2021 at 10:17

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