I've started reading "Quantum Mechanics: A Modern Development" by Leslie E. Ballentine and have some trouble understanding how to handle 3-vector operators (i.e. an operator $\mathbf{A}$ with $\mathbf{A} = (A_{1}, A_{2} , A_{3})$) like the position operator $\mathbf{Q}$ and the momentum operator $\mathbf{P}$ when it comes to multiplication.

General problem

Do I multiplicate those vectors componentwise, i.e.

\begin{equation} \mathbf{Q} \mathbf{P} = \sum_{\alpha = 1}^{3} Q_{\alpha} P_{\alpha} \end{equation}

or do I have to do it another way?

Specific problem

The context in which my problem arises is the following: On pp. 77 and 78 of the book by Ballentine one finds (I added some explanatory notes to the original text):

We assume the position operator for the particle to be $\mathbf{Q} = (Q_{1}, Q_{2}, Q_{3})$, where by definition

\begin{equation} Q_{\alpha} | \mathbf{x} \rangle = x_{\alpha} | \mathbf{x} \rangle \qquad (\alpha = 1, 2, 3)\tag{3.36} \end{equation}

has an unbounded continuous spectrum. [...]

The space displacement $\mathbf{x} \rightarrow \mathbf{x}^{\prime} = \mathbf{x} + \mathbf{a}$ involves a displacement of the localized position eigenvectors,

\begin{equation} | \mathbf{x} \rangle \rightarrow | \mathbf{x} \rangle^{\prime} = e^{−i \mathbf{a} \cdot \mathbf{P}} | \mathbf{x} \rangle = | \mathbf{x} + \mathbf{a} \rangle\tag{3.40} \end{equation}

where $i$ is the imaginary unit and $\mathbf{a}$ is a space displacement. [...]

The displaced observables bear the same relationship to the displaced vectors as the original observables do to the original vectors, as was discussed in Sec. 3.1. In particular,

\begin{equation} \mathbf{Q} \rightarrow \mathbf{Q}^{\prime} = e^{−i \mathbf{a} \cdot \mathbf{P}} \mathbf{Q} e^{i \mathbf{a} \cdot \mathbf{P}}\tag{3.41} \end{equation}


\begin{equation} Q_{\alpha}^{\prime} | \mathbf{x} \rangle^{\prime} = x_{\alpha} | \mathbf{x} \rangle^{\prime} \qquad (\alpha = 1, 2, 3).\tag{3.42} \end{equation}

But since $ | \mathbf{x} \rangle^{\prime} = | \mathbf{x} + \mathbf{a} \rangle$, a comparison of (3.42) with (3.36) implies that

\begin{equation} \mathbf{Q}^{\prime} = \mathbf{Q} − \mathbf{a} I\tag{3.43} \end{equation}

where $I$ is the identity matrix. [...]

Equating terms of first order in $a$ from (3.43) and (3.41), we obtain

\begin{equation} [Q_{\alpha}, a \cdot \mathbf{P}] = i_{\alpha} I,\tag{3.44} \end{equation}

(where $[A, B] = A B - B A$ is the commutator of the operators $A$ and $B$) which can hold for arbitrary directions of $a$ only if

\begin{equation} [Q_{\alpha} , P_{\beta}] = i \delta_{\alpha \beta} I\tag{3.45} \end{equation}

where $\delta_{\alpha \beta}$ is the Kronecker Delta.

I'm not sure whether the result $[Q_{\alpha}, a \cdot \mathbf{P}] = i_{\alpha} I$ is correct or contains typos. Ballentine doesn't say what $i_{\alpha}$ means and usually he adds a "$\cdot$" between two variables only when he wants to indicate an inner product between two vectors, so maybe the $a$ should be the vector $\mathbf{a}$ and not its norm. Furthermore when I do the Taylor expansion of $e^{−i \mathbf{a} \cdot \mathbf{P}} \mathbf{Q} e^{i \mathbf{a} \cdot \mathbf{P}}$ to the first order in $\mathbf{a}$ I get

\begin{equation} e^{−i \mathbf{a} \cdot \mathbf{P}} \mathbf{Q} e^{i \mathbf{a} \cdot \mathbf{P}} \approx \mathbf{Q} + i [\mathbf{Q}, \mathbf{a} \cdot \mathbf{P} ] \end{equation}

so that I come to the result

\begin{equation} [\mathbf{Q}, \mathbf{a} \cdot \mathbf{P}] = i \mathbf{a} I . \end{equation}

I have some problems getting to equation (3.45) from this point. Especially I'm unsure how to get from the general product of the 3-vector operators, $\mathbf{Q} \mathbf{P}$, to the componentwise one, $Q_{\alpha} P_{\beta}$, and whether I can move the vector $\mathbf{a}$ around like a scalar, i.e.

\begin{equation} [\mathbf{Q}, \mathbf{a} \cdot \mathbf{P}] = \mathbf{a} \cdot [\mathbf{Q}, \mathbf{P}] \end{equation}

or not.

I hope I have made clear what my problems are without being unnecessarily wordy.

  • 1
    $\begingroup$ Yes, $[Q_{\alpha}, {\bf a} \cdot {\bf P}] = i_{\alpha} I$ is a typo in Ballentine's book. It should be $[Q_{\alpha}, {\bf a} \cdot {\bf P}] = i a_{\alpha} I$. $\endgroup$
    – Qmechanic
    Apr 4, 2012 at 13:24

1 Answer 1



1) There are "foundamental" ordinary operators without any vector indices, say $\hat A,\hat B, ...$

2) There is notation $\hat A_i$ which stands for $\hat A_1, \hat A_2$ or $\hat A_3$, depending on the index $i$. Each of these is an ordinary operator from 1). Hence, you can encounter operators, which look like $\hat A_i, \hat B_i,...$

2.1) Notation ${\bf \hat A}$ stands for $\sum_i \hat A_i{\bf e}_i$, where ${\bf e}_i$ are basis 3-vectors like those you studied in linear algebra or analytic geometry.

3) Whenever an operator ${\bf \hat A}$ of the group 2) is involved in a vector operation like dot or vector product, the operation involves only vector indices of the operators. For example: ${\bf \hat A}\cdot{\bf \hat B}\equiv\sum_i \hat A_i\hat B_i$. If there are several vector operations in an expression, you should make sure that the indices used in each operation are unique. In fact, the most unambiguous way to work with expressions is to write them in components.

Expression $[{\bf \hat Q},{\bf \hat a}{\bf \hat P}]=i{\bf \hat a}\hat I$ in components is written as $[\hat Q_i,\sum_j \hat a_j \hat P_j]=i\hat a_i \hat I$. Now, as it holds for any ${\bf \hat a}$, you can chose ${\bf \hat a}=(\hat I,\hat 0, \hat 0)$ to obtain:

$$ \sum_j \hat a_j \hat P_j = \hat I \hat P_1 = \hat P_{1}\\ $$ $$ [\hat{Q}_1,\hat{P}_1]=i\hat{I} \hat{I} = i\hat{I}, [\hat Q_{2,3},\hat P_1]=\hat 0 $$ The last line reads off $[\hat{Q}_i,\hat{P}_1]= i\delta_{i1}\hat{I}$. Chosing different ${\bf \hat{a}}$ you can trivially derive the full $[\hat{Q}_i,\hat{P}_j]= i\delta_{ij}\hat{I}$.

In fact, indeed the vector notation is a bit abusive, because one could use the dot symbol for ordinary multiplication of non-vector operators, like this: $\hat A \cdot \hat B$. This is a good reason to work in components at least when doing derivations, as the components are standard operators, and you will never get confused between dot products of vector operators and ordinary products of simple scalar operators.

  • $\begingroup$ Thank you very much. I don't know why I haven't thought of it that way before. $\endgroup$
    – Philipp
    Mar 19, 2012 at 17:20

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