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I'm reading Griffiths and he has this section where he states that $|\hat{Q}f\rangle$ is mathematical nonsense and that really we should write $\hat{Q}|f\rangle$, where the latter makes more sense to me, as it states (to my knowledge) that we have some operator acting on a vector ($Ax$ in "standard" linear algebra), but I cannot express what $|\hat{Q}f\rangle$ would be. By playing around a little I came to think of $|\hat{Q}f\rangle$ as $(Ax)$, meaning a new vector on which the operator has acted upon whilst $\hat{Q}|f\rangle$ is an expression where the operator "has yet to act". But of course, this is not very logical since as Griffith states, $|\hat{Q}f\rangle$ is mathematical nonsense (although I would claim that a great deal of what makes this notation good and intuitive is the distribution of operators inside the bra and ket.)

Furthermore, he goes on to justify that the expression $\langle\hat{Q}f|$ is valid, which to me seems increadibly unreasonable if $|\hat{Q}f\rangle$ is not valid, as I thought the entire point is that distributing an operator inside the bra and kets is essentially nonsensical although everyone uses it all the time. I played the thought around with to justify $\langle \hat{Q}f|$ is that strictly speaking, this is not really a vector, but something more closely resembling an operator ($\langle f|$ is a linear functional and $\hat{Q}$ an operator. I'm not sure of the terminology, but it is certainly not a vector), and thus we distinguish it from a vector by distributing the operator inside of the bra.

After doing further research like on this thread: Difficulties with bra-ket notation where it is claimed that $\langle \psi|A|\psi\rangle =\langle\psi|A\psi\rangle$, which again feels weird after having read Griffiths.

Lastly I came across this MIT lecture https://ocw.mit.edu/courses/8-05-quantum-physics-ii-fall-2013/resources/lecture-9-diracs-bra-and-ket-notation/ where at $58:30$ the lecturer defines $\langle\psi|A|\psi\rangle=\langle\psi,A\psi\rangle$ where the right hand side is in "inner product notation" and the right hand side is supposedly in "bra-ket" notation. This appears logical to me, and the natural conclusion is that when we write $\langle \psi|A \psi \rangle$, we are actually using "inner product notation" as opposed to bra ket notation where we exchanged the "$|$" with ",".

My question boils down to (I) How can we justify $\langle\hat{Q}f|$ as being "not mathematically nonsensical", (II) Should we really think of the bra-ket notation and inner product notation as separate things to justify the distribution of operators? (III) In general, how can we justify the distribution of operators inside of bra and kets if this is mathematical nonsense (but is so widely used!)?

Apologies for the thread being all over the place, just wanted to show my thought/research process.

Thank you!

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  • $\begingroup$ Notation can be good or bad, but not "mathematical nonsense" as it only has the meaning we give it to it. So either the concepts encapsulated in the notation is nonsense or not. $\endgroup$ Commented Dec 1, 2022 at 9:12
  • $\begingroup$ That being said, I think the Dirac notation is more confusing than it helps. Yes, it simply means an inner product and the usual convention is that $\langle \psi |A|\phi\rangle := \langle \psi, A\phi\rangle$. You can of course write $|A\psi\rangle$ and $\langle A\psi|$ -but again: It has the meaning you (or the textbook) gives to it. So to before you ask whether or not it makes sense to do that, you should explain what these symbols mean. $\endgroup$ Commented Dec 1, 2022 at 9:15
  • $\begingroup$ @TobiasFünke don't you mean $\langle \psi |A| \phi \rangle=\langle \psi ,A \phi \rangle $? $\endgroup$
    – William
    Commented Dec 1, 2022 at 9:20
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    $\begingroup$ Where is the difference to what I wrote? $\endgroup$ Commented Dec 1, 2022 at 9:21
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    $\begingroup$ Schuller has a nice discussion of Dirac's bra-ket (~15min), connecting this notational convenience to the actual mathematical objects: inner product of two vectors $\langle \cdot, \cdot \rangle$, the Hilbert space $\mathcal{H}$, and its dual space $\mathcal{H}^*$. $\endgroup$
    – Ben H
    Commented Dec 1, 2022 at 14:00

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You seem to be referring to Griffiths 3rd edition, p. 153. In there, Griffiths explains that he writes $\langle \hat{Q} f |$ due to the lack of a better notation.

(I) How can we justify $\langle \hat{Q} f |$ as being "not mathematically nonsensical"

I believe what Griffits has in mind is that $\langle \hat{Q} f |$ is not nonsensical because it is just notation. Nevertheless, in my opinion, the same could be said about $| \hat{Q} f \rangle$.

(II) Should we really think of the bra-ket notation and inner product notation as separate things to justify the distribution of operators?

They are separate things, although closely related. Notice that $\langle g, \hat{Q}, f \rangle$ is non-sensical in inner-product notation. In fact, it is even ambiguous, since it doesn't make clear whether you are acting to the left or right. This is, in fact, a limitation of Dirac notation: it makes some expressions ambiguous because it becomes unclear whether the operator is acting to the left or right. This is usually amended by taking the convention that the operator always acts to the right, and when it acts to the left you write it inside the bra, for example.

The two notations are different. They have different advantages. Namely, Dirac notation is wonderful for dealing with Hermitian operators and doing calculations quickly, but it makes some other concepts more confusing. I'd say your question is an example of how it can lead to confusion in some situations (notice that if the operator was surely Hermitian, there would be no confusion at all because $\langle \hat{Q} f | = \langle f | \hat{Q}$ would never be ambiguous.

(III) In general, how can we justify the distribution of operators inside of bra and kets if this is mathematical nonsense (but is so widely used!)?

You define the meaning of putting the operator inside the bra/ket in the way you prefer. Griffiths's point is a bit pedantic. Technically, $\hat{Q}$ does act on the vector rather than its name, but it is natural to define and understand that whenever we write $|\hat{Q} f \rangle$ we actually mean $\hat{Q} |f \rangle$. This is exactly what he does when we write $\langle \hat{Q} f |$ to mean $(\hat{Q} | f \rangle)^{\dagger}$.

This is similar to how you could complain that $3_{\mathbb{R}}$, the real number, is not $3_{\mathbb{N}}$, the natural number. After all, $3_{\mathbb{N}}$ is an element of $\mathbb{N}$, but $3_{\mathbb{R}}$ is a Dedekind cut of rationals, which are equivalence classes of ordered pairs of integers, which are equivalence classes of ordered pairs of natural numbers. Technically, they are different, but there really isn't that much of a reason to distinguish them in $99.99%$ of the situations.

Griffiths probably raises this point because it is important to point it out when you're learning so you actually understand the meaning of the expressions, but I guess most physicists ignore this most of the time.

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    $\begingroup$ I always like to point to this post by Terry Tao on MathOverflow when I'm talking about notation: mathoverflow.net/a/366118 $\endgroup$ Commented Dec 1, 2022 at 9:17
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    $\begingroup$ ". This is usually amended by taking the convention that the operator always acts to the left, and when it acts to the left you write it inside the bra, for example." - did you mean right (instead of the first left)? $\endgroup$ Commented Dec 1, 2022 at 9:24
  • $\begingroup$ @TobiasFünke Yes, thank you for pointing it out =) $\endgroup$ Commented Dec 1, 2022 at 9:27
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As far as I'm concerned:

$$\hat p |p\rangle = p|p\rangle $$

Where $|p\rangle$ is a momentum eigenstate of the $\hat p$ operator. The eigenvalues of $\hat p$ is the number $p$. It is the same number used to label the eigenstate. So for an eigenvalue equation, one has:

$$ [\rm operator][ket] = [number][ket] $$

Where :

$$ [{\rm operator}] \rightarrow \hat{{\rm symbol}} $$

That is, a symbol for a physical quantity, with a hat over it. Like momentum $p$ has an operator $\hat p$.

The key is as follows:

$$ [{\rm ket}] \rightarrow |{\rm label}\rangle$$

That is, a basis element of the orthonormal Hilbert space is a label in the ket construction. The label is specifically a number, like $3.3/\hbar\ {\rm cm^{-1}}$, or just $p$, if it indeterminant. If you don't like units in your labels, predeclare them (like we do with spin). Then the state:

$$|3.3\rangle$$

has a momentum operator eigenvalue:

$$\hat p|3.3\rangle = 3.3\hbar\ {\rm cm}^{-1}|3.3\rangle$$

So both the ket label and the eigenvalue are numbers, but only the eigenvalue might need units.

ASFIK:

$$|\hat p p\rangle -->|\hat p 3.3\rangle = 3.3|\hat p > --> [\rm confused emoji] $$

Since the label is just a number. (It an also be a character like '3.3', or an arrow, even an emoji....as a label it doesn't know what to do with an operator. It needs to label a ket, which can play with operators.

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