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I am reading a paper on the Wigner function and during a calculation they used $$\langle x|p\rangle = \frac{1}{\sqrt{2\pi\hbar}}e^{\frac{ixp}{\hbar}}$$ and referenced Griffiths "Introduction to Quantum Mechanics" p. 103.

I looked it up and there they want to find the eigenfunctions and eigenvalues of the momentum operator.

$$ -i\hbar \frac{d}{dx} f_p(x) = p f_p(x) \\ \Rightarrow f_p(x) = A e^{\frac{ixp}{\hbar}}. $$

By restricting the eigenvalues to real numbers and setting $A = \frac{1}{\sqrt{2\pi\hbar}}$ they get orthonormality

$$ \int_{-\infty}^\infty f^*_{p^{'}}(x) f_p(x) dx = |A|^2 \int_{-\infty}^\infty e^{\frac{i(p-p')x}{\hbar}} dx = |A|^2 2\pi \hbar \delta(p-p') = \delta(p-p'). $$

Therefore

$$f_p(x) = \frac{1}{\sqrt{2\pi\hbar}} e^{\frac{ixp}{\hbar}}.$$

How does this connect to the above inner product of $x$ and $p$ and how can I obtain the inner product's solution from this?

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  • $\begingroup$ Is this not just the statement that to change basis from position to momentum space one uses a fourier transform? $\endgroup$
    – jacob1729
    May 23, 2018 at 16:10
  • $\begingroup$ For a discrete basis, orthonormality reads $\langle e_i, e_j\rangle= \delta_{ij}$. So for a continuous basis $f_p$ as above one naturaly requires orthonormality in the form $\langle f_p, f_q\rangle=\delta(p-q)$. Notice that $A$ has been chosen to satisfy this. $\endgroup$
    – Gold
    May 23, 2018 at 16:31
  • $\begingroup$ Possible duplicate: physics.stackexchange.com/q/41880/2451 $\endgroup$
    – Qmechanic
    May 23, 2018 at 17:08

2 Answers 2

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This is essentially a notational issue. In Dirac notation, a pure state is represented by a vector (ket) in Hilbert space $|\Psi\rangle$ which contains all the information that one can have about that state. The eigenvectors of momentum are then denoted as $|p\rangle$. This notation is representationally independent - i.e. it does not assume you are in the position representation or the momentum representation. The ket $|p\rangle$ represents the state and says nothing about in which representation it's in. In order to get $|p\rangle$ in the position representation we take the inner product of it with position eigenvector $|x\rangle$. So the position representation of $|p\rangle$ is denoted in Dirac notation by $\langle x |p\rangle$. This position representation of the momentum eigenvector is exactly that Griffiths calculated on page 103.

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To speak about this precisely, one has to built up the setting a bit more carefully. It is also a bit of an abuse to use the term Hilbert space since this functions are not formally a basis for the space, given that they are not normalizable. So I would suggest, to gain some insight on your question, to consider instead of $x\in(-\infty,\infty)$, take $x\in[-1,1]$. Along with the condition of $f$ being square-integrable. Eventually this is called $\mathcal{H}=\mathcal{L}^2([-1,1])$ which is a proper Hilbert space. It turns out that the Fourier transform is a unitary automorphism of this space, so that you can think of $x$ and $p$ as two bases (see https://en.wikipedia.org/wiki/Fourier_transform#On_L2). One can then realize the dual space $\mathcal{H}^*$ is itself and treat $\mathcal{H}$ in let's say $p$ coordinates and $\mathcal{H}^*$ in $x$ coordinates while the relation being the automorphism. A bit Functional analysis will serve the purpose of explaining all this in detail. (The formalism is always deeper in comparison to how the concepts are discovered)

The physical(ist) approach (I will ignore factors of $2\pi$, so forgive me for that :) ) is looking to solve the eigenfunction equation of the momentum operator (obviously real values, since that is what one can measure in the lab) and build a complete base out of it. Once you have solved for the eigenfunctions as above, you should be able to express any state, even $| x\rangle$ in this new base. So what you do is project into every basis element $|p\rangle$. $$|x\rangle = \int dp |p\rangle \langle p|x\rangle$$ However you have implicitly imposed the Fourier relation when you were solving the eigenfunction equation, since $$-i\hslash \frac{d}{dx}f_p(x) = p f_p(x)$$ is turned into through a Fourier transform (including the $\hslash$ in the kernel), which $$p'\tilde{f}_p(p') = p \tilde{f}_p(p')$$ which is solved by picking $p=p'$ (so we can drop the subindex $p$ on the Fourier version while adding the subscript to the $x$ position to which it is associated, as written below), which is basically telling you any function of just $p$ can be converted into an eigenfunction if you transform it via a Fourier transform back to $x$. Up to this point you could pick any set of those functions (depending just on $p$) to be your new basis. So that if you wanted to write this functions in terms of $x$ then you employ the inverse Fourier transform to get: $$f_p(x) = \int dp\; e^{ipx/\hslash}\tilde{f}(p)$$ And where as pointed out before, $\tilde{f}(p)=\langle p|f\rangle$ is the projection of $f$ into the basis labeled by $p$. Having imposed that the relation between the coordinates is a fourier transform and demanding orthonormality, you can determine suitable $f$'s.

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