The short version of my question: What is $\langle x | p \rangle$ and why does it look like it looks?

Now the long version: Let $|x\rangle$ be the orthonormalized eigenstates and of the spatial operator $\hat x$ with eigenvalues x and $|p\rangle$ be the orthonormalized eigenstates and of the momentum operator $\hat p$ with eigenvalues p. I'm "guessing" that $$ \tag{a}\label{eq:x_on_p} \langle x | p \rangle = \frac 1 {\sqrt{(2 \pi)^d \hbar}} e^{ipx/\hbar}, $$ where $d$ is the number of spatial dimensions, but I haven't found a convincing proof for this. So I tried to find one on my own, which has a problem that I want to share, hoping to find a solution with your help.

As a postulate I'm assuming that $$ \tag{b}\label{eq:x_on_p_operator} \langle x | \hat p = -i \hbar \nabla \langle x |. $$

Using this, I now want to find out how $\langle x | p \rangle$ looks. Starting with the equation $$ \hat p | p \rangle = p | p \rangle,$$ multiplying it with $\langle x|$ from the left and using \eqref{eq:x_on_p_operator} leads to $$ -i \hbar \nabla_x \langle x|p \rangle = p \langle x|p \rangle. $$ This ODE has the solution $$ \tag{c}\label{eq:unnormalized_solution} \langle x|p \rangle = C \exp\left( i p x / \hbar \right) $$ with $C$ being a constant. Now let's assume that $| q \rangle$ are the same eigenstates of $\hat p$ as our $|p\rangle$, but of course with eigenvalues $q$. Then $$ \tag{d}\label{eq:one_equals} 1 = \int \text{d} q \ | q \rangle \langle q | \int \text{d} x \ | x \rangle \langle x | \int \text{d} p \ | p \rangle \langle p | \overset{\eqref{eq:unnormalized_solution}} = |C|^2 \int \text{d} q \int \text{d} p \ | q \rangle \langle p | \int \text{d} x \ e^{i(p-q)x/\hbar} . $$ Now I'm using the fact, that the delta distribution has a representation $$ \delta(p-q) = \frac 1 {(2\pi)^d \hbar} \int \text d x \ e^{i(p-q)x/\hbar}, $$ where $d$ is the number of dimensions of the argument $(p-q)$. Plugging that into \eqref{eq:one_equals} yields $$ |C| = \frac 1 {\sqrt{(2 \pi)^d \hbar}}. $$

This looks very close to what I wanted to proof, but i'm stuck at this point. How can I show that $C$ has no imaginary party? Usually when textbooks on QM come to this point, it's said that the complex phases of the eigenstates don't matter and can therefore chosen that $C$ appearing in \eqref{eq:unnormalized_solution} is real valued. But I think this is not the true story in this case. My thoughts go like this: Take an arbitrary $|x_0\rangle$. Then we can modify all of our $|p\rangle$s so that $C$ in $\langle x_0 | p \rangle$ is real valued. By doing so, the phases of all our $|p\rangle$ are now fixed. Now lets take a different spatial eigenstate $|x_1\rangle$. Of course we can adjust it's phase, that for some specific $|p_0\rangle$ the $C$ in $\langle x_1 | p_0\rangle$ is real, but what we need is $\langle x_1 | p \rangle$ to be real for ALL $|p\rangle$. Summing it up, it seems as the problem of making $C$ real valued can not be solved by modifying the phases of the spatial and momentum eigenstates.

Is there another way to show that $C$ is real?


I think I immediately found an answer after posting this question: The key points are that in our ODE $p$ is a constant and in the solution $C$ is also a constant, which means it does not depend on $x$.

So for our fixed $p$ we modify the phase of $|p\rangle$ that $C$ is real. Then this $C$ is real for all $\langle x|p\rangle$ with that specific $p$ we had in our ODE. Then we can repeat this procedure for all other $p$.

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    $\begingroup$ You can eliminate any p dependence from C if you want, but it's artificial and is equivalent to fixing a gauge (which isn't great). Instead, it's better to take the conjugate of that object and do an analogous development swapping p and x. $\endgroup$ – Emilio Pisanty Sep 2 '17 at 13:09
  • $\begingroup$ Could you explain that in a bit more detail? I don't completely get the point. $\endgroup$ – psicolor Sep 2 '17 at 13:19

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