# Dirac Formalism/Notation Question

I'm reading the 3rd edition of Sakurai and Napolitano's Modern Quantum Mechanics, which (probably rightly) relegates wave mechanics to an appendix. Instead, it carefully develops Dirac's formalism, and uses it to expose quantum mechanics. Because notation is at the heart of this exposition, I want to be sure I get it exactly right.

The authors note that the usual wave function corresponding to a state $$\left|\alpha\right>$$ in the position basis is given by $$\psi_{\alpha}(x') = \langle x' | \alpha\rangle$$, and that, given two kets $$\left|\alpha\right>$$ and $$\left|\beta\right>$$, the expectation value of some operator depending on position $$f(x)$$ is given by

$$\begin{equation*} \left<\beta\right|f(x)\left|\alpha\right> = \int{\mathrm{d}x'\,\langle \beta |x'\rangle f(x')\langle x' | \alpha\rangle} = \int{\mathrm{d}x'\,\psi^{\ast}_{\beta}(x')f(x')\psi_{\alpha}(x')}. \end{equation*}$$

Later, the familiar problem of the infinite square potential well is treated as an exercise, and the usual wave functions are

$$\begin{equation*} \psi_{n}(x') = \langle x | n\rangle = \sqrt{\frac{2}{a}}\sin{\left(\frac{n\pi}{a}x'\right)}. \end{equation*}$$

Now suppose I want to calculate $$\left$$. I know the way to calculate it is

$$\begin{equation*} \left = \int{\mathrm{d}x'\psi^{\ast}_{n}(x')x'\psi_{n}(x')}, \end{equation*}$$

but by the notation developed earlier, we should be able to write

$$\begin{equation*} \left = \left<\beta\right|x\left|\alpha\right>. \end{equation*}$$

Here (at long last) is my question: in this case, what are the kets $$\left|\alpha\right>$$ and $$\left|\beta\right>$$? Do I have a tragic misunderstanding of the notation?

Edit: the authors denote the inner product of $$\left|\alpha\right>$$ and $$\left|\beta\right>$$ as $$\langle \alpha | \beta\rangle$$. Therefore, in writing $$\langle x' | n\rangle$$, they seem to suggest the existence of two kets $$\left|x'\right>$$ and $$\left|n\right>$$. Maybe my confusion is here.

Here is an excerpt from the text: • Well, $\alpha$ and $\beta$ are labels, just as $\psi$ or $\psi_n$. I don't understand the question. You can similarly ask: What is $v\in \mathbb R^3$? Answer: It is a vector, which we denote/label by $v$?! So $|\alpha\rangle$ and $|\beta\rangle$ are "what you want them to be". Jan 12 at 11:02
• @TobiasFünke, you're right, of course--$\alpha$ and $\beta$ are labels by which one refers to the two vectors $\left|\alpha\right>$ and $\left|\beta\right>$ in Hilbert space (though the example in $\mathbb{R}^{3}$ is appreciated). So I guess my question is, what do I "want them to be" when calculating the expectation value of $x$ for the wave functions mentioned? Jan 12 at 11:08
• I still don't understand, given any two vectors in the Hilbert space, which me might label as above, you can compute $\langle \alpha|X|\beta\rangle$ (mathematical rigor aside). So again: Where is the problem in choosing $\alpha=\beta=\psi_n$, if you want to compute the expectation value? However, note that in your last equation the notation is misleading. As indicated, the LHS is independent of $\alpha$ and $\beta$; but I think this is a quite common abuse of notation. Jan 12 at 11:10
• If as you say we pick $\left|\psi_{n}\right>$ for the kets in question, it seems like that implies $\left|\psi_{n}\right>$ is the correct notation for the wave function and not $\langle x'|n\rangle$ (which seems to suggest the existence of two vectors, $\left|x'\right>$ and $\left|n\right>$). (The authors define the inner product between $\left|\alpha\right>$ and $\left|\beta\right>$ as $\langle \alpha|\beta\rangle$.) Is $\langle x'|n\rangle$ just used for convenience and not supposed to suggest an inner product? Jan 12 at 11:13
• It is all notation and it has the meaning you give it; if you feel better, take $\alpha=\beta=n$ - in any case, what you mean is, in this example, the vector $\sim \sin \ldots$. If that does not help then sorry, I really don't understand your confusion. Perhaps someone else can help you with that. As a last point it seems that you are confused by the difference between a wave function and an abstract vector in an abstract Hilbert space, is that correct? If so, it might be worth to browse through the questions here; as far as I remember, this was asked a few times. Jan 12 at 11:16

An expectation value is defined for a single state and an operator. For state $$|\alpha\rangle$$, the expectation value of the position operator $$\hat x$$ is given by $$\langle \hat x\rangle _\alpha \equiv \langle \alpha| \hat x |\alpha\rangle.$$

You are not calculating an expectation value if two different states are involved. If two different states are involved as in $$\langle \alpha|\hat x|\beta\rangle$$, with $$|\alpha\rangle \neq |\beta\rangle$$, it is common to call it a transition element, coupling element or off diagonal matrix element, depending on context, since it is basically a probability amplitude for a transition between the two involved states, mediated by the operator.

The state is often suppressed when it is clear which system and set of states is considered due to context. The expectation value is then abbreviated as $$\langle \hat x\rangle$$, leaving out the explicit state we are referring to. But when you expand this you have to use the same state for the bra and ket.

I would also like to add that your expansion is not the most general form. It is correct for operators that admit a diagonal representation in position representation but generally you should insert two resolutions of the identity,

$$\langle \alpha|\hat O |\alpha \rangle = \int\int dx dx' \langle \alpha|x\rangle \langle x|\hat O|x'\rangle \langle x'|\alpha\rangle = \int\int dx dx' \bar \alpha(x')O(x',x)\alpha(x)$$ Most common operators are diagonal in position representation or local and you have $$O(x',x)= O(x)\delta(x-x')$$ which enables you to perform one integral. This simplifies the expression to the common form $$\langle \alpha|\hat O |\alpha \rangle = \int\int dx dx' \bar \alpha(x')O(x',x)\alpha(x)=\int dx \bar \alpha(x) O(x) \alpha(x)$$

Remark due to comment:

It is valid to evaluate the operator first and then calculate the inner product. For example, $$\hat f|\beta \rangle =|\gamma\rangle$$, such that $$\langle \alpha| \hat f|\beta \rangle = \langle \alpha|\gamma\rangle = \int dx \langle \alpha|x\rangle \langle x|\gamma\rangle = \int dx \bar \alpha(x)\gamma(x)$$ Then one resolution of identity suffices.

EDIT:

I see now where your confusion stems from. The notation used in the book is rather unlucky in my opinion since does not make the distinction between the position operator and its eigenvalues and its representation explicit. Lets take equation 1.246. $$\langle \beta| f(x)|\alpha \rangle = \int dx' \beta(x')f(x') \alpha(x')$$ where I use $$\langle x|\beta \rangle = \beta(x)$$ instead of $$\psi_\beta(x)$$.

$$f(x)$$ on the left-hand side from your excerpt is a function of the position operator, so I would prefer to denote it as $$f(\hat x)$$, for example $$f(\hat x)= \hat x^2$$. When you switch to the position representation, a function of the position operator simply goes to a function of $$x$$, you can proof this by Taylor expansion. I.e. $$f(\hat x)\rightarrow f(x)$$. I would have written the expression as $$\langle \beta |f(\hat x)|\alpha \rangle = \int dx' \beta(x')f(x') \alpha(x')$$

I would recommend taking a look at a second book that is more verbose.

• Thank you for your thorough answer! You're right of course that my expansion was incomplete, but I was thinking in terms of the position operator in the position basis. So I take $\langle \alpha |f(x) | \beta \rangle$ to mean the inner product between the vectors $\left|\alpha\right>$ and $f(x)\left|\beta\right>$. Is this the wrong way to look at it? Is $\left<\alpha\right|f(x)\left|\beta \right>$ just convenient notation? Jan 12 at 12:00
• @kandb You would need to define $f(x)$. It is not clear to me what that is supposed to be as you seem to mix a quantity in position representation with a basis free representation, when you write $\langle \alpha| f(x)|\beta \rangle$. Jan 12 at 12:05
• Sorry, you're right. Take $f(x) = x$ to be the position operator. Jan 12 at 12:08
• @kandb I have extended my answer to addresses your comment. Treating it as an inner product of two states once the operators action on one state has been evaluated, is equivalent. The position operator is $\hat x$. It is important to make the distinction between an operator and the representation with respect to a basis. Only when you are working within the position representation, you may use the function $x$. I think some of your confusion stems from a clear understanding of this distinction. Jan 12 at 12:21
• @kandb I hope the following equation helps to make my point clearer, $\hat x = \int dx \ x |x\rangle \langle x| \neq x$. Jan 12 at 12:34

You can just apply the steps in the excerpt to the expectation value $$\langle n |\hat x|n\rangle$$. If we insert the identity, $$𝟙=\int\mathrm d x|x\rangle\langle x|,$$ two times we get \begin{align} \langle n |\hat x|n\rangle&=\langle n |\left(\int\mathrm d x'|x'\rangle\langle x'|\right)\hat x\left(\int\mathrm d x''|x''\rangle\langle x''|\right)|n\rangle\\ &=\int \mathrm d x'\mathrm d x''\,\langle n|x'\rangle\langle x'|{\hat x|x''\rangle}\langle x''|n\rangle\\ &=\int \mathrm d x'\mathrm d x''\,\psi^*_n(x')\,\langle x'|{\hat x|x''\rangle}\psi_n(x'')\\ &=\int \mathrm d x'\mathrm d x''\,\psi^*_n(x')\,\langle x'|{x''|x''\rangle}\psi_n(x'')&&\text{apply }\hat x\text{ to }|x''\rangle\\ &=\int \mathrm d x'\mathrm d x''\,\psi^*_n(x')\,x''\langle x'|{x''\rangle}\psi_n(x'')\\ &=\int \mathrm d x'\mathrm d x''\,\psi^*_n(x')\,x''\delta(x'-x'')\psi_n(x'')&&\langle x'|x''\rangle=\delta(x'-x'')\\ &=\int \mathrm d x'\,\psi^*_n(x')\,x'\psi_n(x')&&\text{integrate over }x'' \end{align}