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I'm just beginning Shankar's 2nd edition Quantum Mechanics and having some trouble with notation. He defines his vectors as "$\left|V\right>$" . And with a scalar multiplier as "$a\left|V\right>$" . When he gets into inner products a couple of pages later he does this: $$\left<V\right|\Bigl(a\left|W\right>+b\left|Z\right>\Bigr) \equiv \left<V|aW+bZ\right>$$

My question is whether $aW$ and $bZ$ has meaning? Or, is this just a definition of a new notation to simplify $\left<V\right|\Bigl(a\left|W\right>+b\left|Z\right>\Bigr)$? If so, does this mean that aW shouldn't be used outside of this specific case?

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Hi Mitchell, if you hit edit, you can look at what I did to format everything using $\LaTeX$. $\equiv$ is "\equiv" in $\LaTeX$, it's some kind of equivalence. There can, however, be different kinds of equivalence. – Peter Morgan Mar 6 '12 at 16:11
Peter - Thank you so much! – Mitchell Kaplan Mar 6 '12 at 18:32
Here's another related question: – David Z Mar 6 '12 at 19:18
David - thank you. It does seem to be the same thing, only a little more complicated. Since I'm just beginning reading this it's helpful to be able trust the notation. I'm sure that once I'm more familiar with it shortcuts (or even sloppiness) in notation will be easier to get through. – Mitchell Kaplan Mar 6 '12 at 20:02

It is sloppy formulation. Either he introduces silently the abbreviation $|aW+bZ\rangle$ for $a|W\rangle+b|Z\rangle$, or he meant to write $a⟨V|W⟩+b⟨V|Z⟩$ in place of $⟨V|aW+bZ⟩$. As you didn't provide more context, it is difficult to say more.

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Thanks Arnold. I don't have the book with me at the moment so I can't look it up, but I think I remember that the next expression was $a⟨V|W⟩+b⟨V|Z⟩$. Based on your answer and what it looks like he's developed I'm guessing it's a shortcut. I'll check the book this evening. Thanks again. – Mitchell Kaplan Mar 6 '12 at 19:59

Bra/Ket notation can be used in different ways. (1) It takes the $W$ in $\left|W\right>$ to be a vector itself, so that the brackets just indicate whether we are dealing with the vector or covector form, $\left|W\right>$ or $\left<W\right|$. In this form, $aW$ makes sense, it is a scalar multiple of the vector $W$, and so does $aW+bZ$.

(2) Bra/Ket notation sometimes takes $\left|W\right>$ to be an eigenvector of some operator, say $\hat A$, with eigenvalue $W$, so that $\hat A\left|W\right>=W\left|W\right>$. In this case, $W$ is a number, and $\left|W\right>$ is the eigenvector of $\hat A$ with the eigenvalue $W$. One has to keep track of which operator one is working with, which is sometimes done by using arrows, distinguishing between $\left|\leftarrow\right>,\left|\rightarrow\right>$ and $\left|\uparrow\right>,\left|\downarrow\right>$, say, instead of writing $\left|+1\right>,\left|-1\right>$ for both, but not infrequently it's obvious to experts which operator is intended but not necessarily so obvious to others. [The vacuum vector in quantum field theory is often denoted $\left|0\right>$, for example, because it is the zero-eigenvalue eigenstate of every annihilation operator, but it is not an eigenstate of any creation operator. EDIT: note that the vacuum vector $\left|0\right>$ is not the zero vector.] If this is how the notation is being used here, then it's problematic because in general $\hat A(a\left|W\right>+b\left|Z\right>)\not=(aW+bZ)(a\left|W\right>+b\left|Z\right>)$. This is so egregious that I cannot imagine this is the way that Shankar is using the notation.

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Peter, Thanks very much for the information. This is near the beginning of his book, so I'm pretty sure it's just a convenient bit of notation. But I was concerned that I was missing something. I'll go over it again this evening after having read everyone's helpful comments. Thanks again. – Mitchell Kaplan Mar 6 '12 at 22:37

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