# Is a bra-ket always a scalar?

I am reading a derivation of the representation of the position operator $$\hat{\mathbf{r}}$$ in momentum space that goes like this:

$$\langle\mathbf{p}|\hat{\mathbf{r}}|\psi\rangle =\int d^{3}x\langle\mathbf{p}|\mathbf{r}\rangle\langle\mathbf{r}|\hat{\mathbf{r}}|\psi\rangle =\int d^{3}xe^{-i\mathbf{p}\cdot\mathbf{r}/\hbar}\mathbf{r}\langle\mathbf{r}|\psi\rangle =\int\mathbf{r}\psi(\mathbf{r})e^{-i\mathbf{p}\cdot\mathbf{r}/\hbar}d^{3}x =i\hbar\nabla_{\mathbf{p}}\int\psi(\mathbf{r})e^{-i\mathbf{p}\cdot\mathbf{r}/\hbar}d^{3}x =i\hbar\nabla_{\mathbf{p}}\bar{\psi}(\mathbf{p})$$

And I have trouble understanding the step $$\langle\mathbf{r}|\hat{\mathbf{r}}|\psi\rangle=\mathbf{r}\langle\mathbf{r}|\psi\rangle$$, since I'm used to thinking of the inner product on the left as a scalar, and the right-hand side seems to be a vector.

This similarly leads me to wonder how $$\langle\mathbf{p}|\hat{\mathbf{r}}|\psi\rangle = i\hbar\nabla_{\mathbf{p}}\bar{\psi}(\mathbf{p})$$ for the same reason, since the left seems to be a scalar and the right (due to $$\nabla_{\mathbf{p}}$$) is a vector.

Am I missing something simple?

• The analogy between vectors and brakets is the following: bras are row vectors, kets are column vectors and operators and matrices. In this case, $\hat{\mathbf{r}}$ is a "matrix vector", i.e. a collection of three matrices. Multiplying a bra, an operator and a ket will give a scalar. In this case, you are calculating the average of three operators ($\hat{x}$, $\hat{y}$, $\hat{z}$) which are usually represented as a vector (collection of three scalars). Feb 22 '19 at 0:55

There are two different notions of vector at play here. The kets are vectors in the Hilbert space, and bra-kets are inner products in this vector space. The operator $$\hat{\mathbf{r}}$$ is a vector in a three-dimensional euclidean space. The matrix elements of such a vector operator are scalars with respect to the Hilbert space, but vectors in the euclidean space.