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Fourier transformations:

$$\phi(\vec{k}) = \left( \frac{1}{\sqrt{2 \pi}} \right)^3 \int_{r\text{ space}} \psi(\vec{r}) e^{-i \mathbf{k} \cdot \mathbf{r}} d^3r$$

for momentum space and

$$\psi(\vec{r}) = \left( \frac{1}{\sqrt{2 \pi}} \right)^3 \int_{k\text{ space}} \phi(\vec{k}) e^{i \mathbf{k} \cdot \mathbf{r}} d^3k$$

for position space.

How do we know that $\psi$ is not the Fourier transform of $\phi$ but we suppose that its the other way around ($\psi$ would be proportional to $\exp[-ikr]$ and $\phi$ would be proportional to $\exp[ikr]$)? If there was no difference in the signs, wouldn't there be a problem in the integration from minus inf. to plus inf. if the probability is asymmetric around zero?
What is the physical reason that in the integral for momentum space we have $\exp[-ikr]$? I agree about the exponent for position space which can be explained as follows: its the sum of all definite momentum states of the system, but what about the Fourier of the momentum space? How can we explain the integral (not mathematically)?

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  • $\begingroup$ If anyone has a better title, please feel free to suggest one. $\endgroup$ – TheQuantumMan Aug 27 '15 at 13:51
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    $\begingroup$ Which way round you put the minus sign makes no difference and is purely a matter of convention. All that matters is that you are consistent in where you put it. $\endgroup$ – By Symmetry Aug 27 '15 at 13:54
  • $\begingroup$ But when you integrate, won't it make a difference if the limits of integration are not from minus infinity to plus infinity? $\endgroup$ – TheQuantumMan Aug 27 '15 at 13:55
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    $\begingroup$ ok,thanks a lot. if you want to write it as an answer, I will mark it as the accepted answer. $\endgroup$ – TheQuantumMan Aug 27 '15 at 14:23
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    $\begingroup$ @LandosAdam I understood the question fine, and I used your notation in my answer. I just wanted to let you know the correct notation in the future. $\endgroup$ – Jahan Claes Aug 27 '15 at 22:41
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Let's say $\Phi$ is a delta function, $\Phi(k)=\delta(k-k_0)$. Presumably, you want this to be an eigenstate of the momentum operator with momentum $\hbar k_0$. With the convention you've chosen, we can convert this to a real-space wavefunction (I'm ignoring normalization for convenience):

$$ \Psi(r)= \int dk \delta(k-k_0)e^{ikr}=e^{ik_0 r} $$

We can then find the momentum of the state by applying the momentum operator $-i\hbar \frac{\partial}{\partial r}$ and finding the eigenvalue. We see that this state has momentum $\hbar k_0$, as desired.

Had you defined the Fourier transform with your signs switched, you would find that the state defined by $\Phi(k)=\delta(k-k_0)$ would have momentum $-\hbar k_0$, which would be inconvenient. That's why we define the Fourier transform as above. Without any particular preference as to what we want $\Phi(k)$ to represent, we could have chosen either one as long as we were consistent.

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  • $\begingroup$ If the domain of integration was asymmetrical, wouldn't there be a problem with the integral? $\endgroup$ – TheQuantumMan Aug 27 '15 at 19:27
  • $\begingroup$ Also, what about the physical interpretation of the integral of Φ? Why is it the sum of exp [-ikr]? $\endgroup$ – TheQuantumMan Aug 27 '15 at 19:29
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    $\begingroup$ A state with momentum $k$ in position space is $e^{ik\cdot r}$. When we're computing $\Phi(k)$, we want to find out the amount of that state within $\psi(r)$. This is done by taking the inner product with that state, which adds a complex conjugation. $\endgroup$ – knzhou Aug 27 '15 at 19:42
  • $\begingroup$ @LandosAdam If your domain of r-integration is asymmetrical, that has no effect on your domain of k-integration. And the domain of k-integration can always be made symmetrical. Can you give an example where the range of k was asymmetrical? $\endgroup$ – Jahan Claes Aug 27 '15 at 21:39
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    $\begingroup$ @LandosAdam my point is that such an example doesn't EXIST. A Fourier transform has to integrate over all possible Fourier modes. Now, it might be true that certain modes are irrelevant (they might all be zero, for a certain wavefunction), so we only integrate over a finite range of $k$s. In this case, the range of ks we integrate over will change appropriately if we switch $e^{-ikr}$ with $e^{ikr}$. So in that sense, it could kind of matter. But again, it's just a matter of appropriately shuffling minus signs $\endgroup$ – Jahan Claes Aug 27 '15 at 22:34
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The identification of one transform as the Fourier transform and the other as the inverse transform is a matter of definition. The Fourier transform predates quantum mechanics so the reason for the assignment has nothing to do with QM and everything to do with mathematics history.

In 1807 Fourier submitted a manuscript to the Institut de France containing, among other things, what we now call the Fourier cosine transform and its inverse. These are his transforms:

$$F_c(u)=\frac{2}{\pi}\int_0^\infty f(x)\cos(ux)dx $$ $$f(x)=\int_0^\infty F_c(u)\cos(ux)du. $$

Cauchy's 1827 generalization of Fourier's relations entailed complex-valued functions, and an ineluctable sign asymmetry in form of the transforms. Trying to preserve symmetry does not help. As the article below notes, it may be shown that if the same sign is taken for both the forward and inverse formulae, "one formula is not exactly the inverse of the other one."

It is a long and helpful exercise to verify that $\hat{f}$ and $f$ inhabit dual spaces with a high degree of symmetry. For example, a function contains the same "energy" as its FT (Plancherel). Whether physics would be equally well served had a different convention been chosen is moot, even if we find particular instances that seem to point to the road not taken.

Much of this material can be found in an article the Jan/Feb 2016 issue of IEEE's Pulse, which in turn draws from a sequence of notes by Deakin listed in the references.

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The de Broglie hypothesis essentially says that the states of definite momentum $p$ are of the form $\psi(r) = e^{ipr}$. These are orthogonal w.r.t. the inner product

$$\langle f|g\rangle := \int f(r)\overline g(r)dr$$

(for any scaling these would obviously still be orthogonal). By the postulates of quantum mechanics, they generate the full state space.

States of definite position $r_0$ are of the form $\phi(r) = \delta(r_0 - r)$.

In a finite dimensional vector space $V$ with an inner product and a basis $\psi_p$, every element $v\in V$ is of the form

$$v = \sum_pa_p\psi_p.$$

If the $\psi_p$ are orthonormal, it is immediately clear that

$$a_p = \langle v|\psi_p\rangle.\tag1$$

We could view $a$ as a function of $p$: $a(p) := a_p$, giving the coefficient of $v$ in the basis $\psi_p$.

If we have another orthonormal basis $\phi_r$, in which the index has been suggestively denoted by a different symbol, we also have

$$v = \sum_rb_r\phi_r,$$

and

$$b(r) := b_r = \langle v|\phi_r\rangle.$$

For the bases described before $a$ (the coefficient function) would be the momentum representation of $v$ and $b$ as the position representation. We have $\langle\psi_p|\phi_r\rangle = e^{ipr}$, and your first expression is the infinite dimensional analogue of (1), whereas your second expression is the analogue of

$$b(r) \equiv \langle v|\phi_r\rangle = \sum a_p\langle\psi_p|\phi_r\rangle$$

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