The Born interpretation states that for a particle with a wave function $\Psi(x)$, the total probability of finding that particle at some point in space is equal to $\int_{-\infty}^{\infty}\Psi(x)^*\Psi(x)dx = 1$.

Suppose we have a state $\rvert\psi\rangle$ in the Hilbert space $\mathcal{H} = \mathcal{L}^2(\mathbb{R})$. The position operator here is $\hat{x}$ with eigenvalues of $x$. In order to calculate probabilities, the state must be normalized. To check if $\rvert\psi\rangle$ is normalized, we calculate its norm: $$\langle\psi\rvert\psi\rangle = \langle\psi\rvert\hat{I}\rvert\psi\rangle = \langle\psi\rvert\left(\int_{-\infty}^{\infty}dx\rvert x\rangle\langle x\rvert\right) \rvert\psi\rangle = \int_{-\infty}^{\infty}dx\langle\psi\rvert x\rangle\langle x\rvert\psi\rangle = \int_{-\infty}^{\infty}\psi(x)^*\psi(x)dx.$$

Now suppose we have a state $\rvert\phi\rangle$ in the Hilbert space $\mathcal{H} = \mathcal{L}^2(\mathbb{R}^n)$. The position operator here is, if I understand correctly, $\hat{\textbf{r}}_n$ with vector eigenvalues of $\textbf{r}_n$ (per this answer: https://physics.stackexchange.com/a/126763/117677). Now again, we want to make sure that $\rvert\phi\rangle$ is normalized. Its norm (generalizing from the previous case) is given by $$\langle\phi\rvert\phi\rangle = \langle\phi\rvert\hat{I}\rvert\phi\rangle = \langle\phi\rvert\left(\int_{-\infty}^{\infty}d\textbf{r}_n\rvert \textbf{r}_n\rangle\langle \textbf{r}_n\rvert\right) \rvert\phi\rangle.$$ This doesn't make much sense to me, as we have the differential of a vector, $d\textbf{r}_n$, and the ket $\rvert\textbf{r}_n\rangle$, which is like double labeling a vector.

So how do you calculate the norm of a state in more than one dimension? Did I generalize incorrectly, or am I just missing some key intuition?

  • 3
    $\begingroup$ Why are you writing $\mathrm{d}\mathbf{r}_n$, or rather: Why are you using the position kets at all. If you have a wavefunction $\phi(\vec x)$, you just write $\int \phi(\vec x)\phi^\ast(\vec x)\mathrm{d}^nx$, i.e. the usual n-dimensional integral over $\mathbb{R}^n$. $\endgroup$
    – ACuriousMind
    Commented Jun 3, 2016 at 22:21

1 Answer 1


Were you bothered by having the differential $dx$ and the ket $|x\rangle$ in the 1D example? If not, what makes this different?

Let's expand out your formula a little more. $$ \begin{array}{rcl} \langle \phi | \phi \rangle &=& \int d^3\vec{r}\langle\phi|\vec{r}\rangle\langle \vec{r}|\phi\rangle \\ &=& \int d^3 \vec{r} \phi^*(\vec{r})\phi(\vec{r})\\ &=& \int d^3 \vec{r} |\phi(\vec{r})|^2 \end{array} $$

So this is clearly just the integral over all space of the probability density, as desired.


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