# Heisenberg relation

Given that $A(k)=\frac{N}{k^2+\alpha^2}$, show that $\Delta k \Delta x >1$.

Considering the above example, according to my textbook, it is written that I must square the above function and determine when does the square fall to 1/3 of it's peak value. What does that mean, practically speaking?

This should enable me to determine a value for $\Delta k$. Similarly we proceed in order to determine $\Delta x$ but by squaring $\psi(x,0)$, the wave function and seeing where does it drop off to $1/3$ for it's peak value.

• I have no idea about your last paragraph: what I should think you would need to do is find the inverse Fourier transform of $A(k)$ (being $\propto e^{-\alpha |t|}$) and then find the widths of the "pulses" $A(k)$ and of the pulse $e^{-\alpha |t|}$. I get $\Delta k \Delta x =1/\sqrt{2}$ if I use the rms value as my "spread" measure, whereas I get $\Delta k \Delta x =1/2$ for Gaussian pulses (which saturate the inequality) – Selene Routley Jan 1 '14 at 12:41
• @WetSavannaAnimalakaRodVance Could you show me how you do that in detail please ? – amir Jan 1 '14 at 12:50
• Asking for step by step calculations that show you how to solve the problem is exactly what you're not supposed to do, but otherwise this is a fine thing to ask. I've edited your question accordingly. – David Z Jan 2 '14 at 2:41
• I've added some more discussion of your "Full Width Third Maximum" spread measurement. Your question is asking to compute "spreads" with an unconventional spread measure, which is fine in this case. It seems the question is trying to give you an intuitive feel for the idea that as a function gets "pointier" and "narrower", its Fourier transform gets "broader" and contrariwise. However, as I discuss, such measures of spread are not appropriate for a general discussion of the Heisenberg Uncertainty Principle. – Selene Routley Jan 2 '14 at 3:18
• PS: Sorry that I was a bit slow on the uptake: I was a bit bewildered by the last paragraph, but then I should have known better because FWHM, FWTM and like spreads are often used in microscopy, which I have worked with quite a bit in the past. – Selene Routley Jan 2 '14 at 3:20

I think your question is simply illustrating the following fundamental property of the Fourier transform: if $f:\mathbb{R}\to\mathbb{R}$ is a function and $F:\mathbb{R}\to\mathbb{R}$ its Fourier transform, then the product of the root mean square spreads of both functions is bounded as follows. Without loss of generality, assume that $f(x)$ is real and $\int_{-\infty}^\infty x\,f(x)\,\mathrm{d}\,x = \int_{-\infty}^\infty k\,F(k)\,\mathrm{d}\,k = 0$ (i.e. the function and its Fourier transform have means of nought), then:

$$\sqrt{\frac{\int_{-\infty}^\infty x^2\,|f(x)|^2\,\mathrm{d}\,x}{\int_{-\infty}^\infty |f(x)|^2\,\mathrm{d}\,x}}\;\sqrt{\frac{\int_{-\infty}^\infty k^2\,|F(k)|^2\,\mathrm{d}\,k}{\int_{-\infty}^\infty |F(k)|^2\,\mathrm{d}\,k}} \geq \frac{1}{2}\qquad(1)$$

Now, I'm using the unitary definition of the Fourier transform here:

$$F(k)=\frac{1}{\sqrt{2\pi}}\int\limits_{-\infty}^\infty e^{-i\,k\,x}\,f(x)\,{\rm d}x\qquad(2)$$

So, if I plug $A(k)$ into Mathematica, I find its inverse FT is (proportional to) $e^{-\alpha\,|t|}$. The actual relationship is:

$$f(x) = e^{-\alpha\,|t|};\qquad F(k) \frac{\sqrt{\frac{2}{\pi }} \alpha}{\alpha^2+k^2}\qquad(3)$$

but the proportionality constants don't matter, they all cancel out in (1). So if I plug the functions in (3) back into (1) and simplify (with Mathematica in my case) I get $\Delta x \Delta k = 1/\sqrt{2}$ where:

$$\Delta x = \sqrt{\frac{\int_{-\infty}^\infty x^2\,|f(x)|^2\,\mathrm{d}\,x}{\int_{-\infty}^\infty |f(x)|^2\,\mathrm{d}\,x}} = \frac{1}{2\,\alpha}\qquad(4)$$

and

$$\Delta k = \sqrt{\frac{\int_{-\infty}^\infty k^2\,|F(x)|^2\,\mathrm{d}\,k}{\int_{-\infty}^\infty |F(k)|^2\,\mathrm{d}\,k}} = \alpha\qquad(5)$$

If I check this with a Gaussian pulse $f(x) = e^{-a\,x^2}$ I get:

$$F(k) = \frac{e^{-\frac{k^2}{4 a}}}{\sqrt{2} \sqrt{a}}\qquad(6)$$

then:

$$\Delta x = \frac{1}{2\,\sqrt{a}};\qquad \Delta k = \sqrt{a}$$

and the Gaussian saturates the inequality (1).

This is, of course, important for the Heisenberg Uncertainty Principle because momentum co-ordinates and position co-ordinates (more generally, eigen-co-ordinates corresponding to any observables $\hat{X}$, $\hat{P}$ which fulfill the canonical commutation relationship $[\hat{X}, \hat{P}]=i\,\hbar\,{\rm id}$) are needfully related by a Fourier transform (with a scaling of the momentum co-ordinates by $\hbar$ thrown in after the Fourier transform). For more detail, see my answer here.

Footnotes:

Notice that, if you define a family of pulses $f_\alpha(x) = f(\alpha x)$ from a "prototype" $f(x)$, then the Fourier transforms of the family are $F(k\,/\,\alpha)\,/\,\alpha$, so the product of uncertainties in (1) is the same for all family members: a broad pulse has a narrow FT.

Also, to prove (1) for the class of Tempered Distributions (see also here) we note that proving (1) is equivalent to the problem:

Minimise $\int_\mathbb{R} k^2 |F(k)|^2 \,{\rm d} k$ subject to:

1. $\int_\mathbb{R} x^2 |f(x)|^2\, {\rm d} x = const$ (find smallest wavenumber domain spread for a constant position domain spread);

2. $\int_\mathbb{R} x |f(x)|^2\, {\rm d} x = 0$ (mean of nought in "position co-ordinates");

3. $\int_\mathbb{R} k |F(k)|^2\, {\rm d} k = 0$ (mean of nought in "wavenumber co-ordinates");

4. $\int_\mathbb{R} |f(x)|^2\, {\rm d} x = 1$ (constant norm functions. Note that we do not need to assume $\int_\mathbb{R} |F(k)|^2\, {\rm d} k = 1$ as this follows, by the Plancherel / Parseval theorems from $\int_\mathbb{R} |f(x)|^2\, {\rm d} x = 1$;

Since the product of Fourier transforms is the Fourier transform of the convolution of the relevant tempered distributions, we can rewrite $k^2 |F(k)|^2$ as:

$$\frac{1}{2\pi} \int_\mathbb{R} e^{-i\,k\,x} (f * f^*)(x) {\rm d} x$$

where $f * f^*$ is the convolution of $f$ with its complex conjugate. Then we integrate $k^2 |F(k)|^2$ over the whole real line, switch order of integration, taking heed that multiplication of the FT by $i\,k$ is equivalent to taking the FT of the derivative $f^\prime(x)$and so, in the distributional sense:

$$\int_\mathbb{R} k^2 |F(k)|^2 \,{\rm d} k = \sqrt{\frac{2}{\pi}} \int_\mathbb{R} \delta(x)\,{\rm d}_x^2\left( (f * f^*)(x)\right)\,{\rm d} x = -\sqrt{\frac{2}{\pi}}\int_\mathbb{R} f^*(x) f^{\prime\prime}(x) {\rm d} x$$

Likewise:

$$\int_\mathbb{R} k |F(k)|^2 \,{\rm d} k = -i\sqrt{\frac{2}{\pi}}\int_\mathbb{R} f^*(x) f^{\prime}(x) {\rm d} x$$

Thus if we bring to bear standard calculus of variation techniques, calculating the variation of our integral to be minimised with the constraints accounted for by Lagrange multipliers we find:

$$2{\rm Re}\left(\int_\mathbb{R}\delta f^*(x)\left( -\sqrt{\frac{2}{\pi}}f^{\prime\prime}(x) +\lambda_1 f^\prime(x)+(\lambda_2 x^2 + \lambda_3 x + \lambda_4)\,f(x)\right){\rm d} x\right) = 0$$

where $\delta f^*(x)$ is an arbitrary variation function. Hence, functions which our integral is extremal for fulfill:

$$-\sqrt{\frac{2}{\pi}}f^{\prime\prime}(x) +\lambda_1 f^\prime(x)+(\lambda_2 x^2 + \lambda_3 x + \lambda_4)\,f(x)=0$$

a differential equation which defines a general Gaussian pulse $\exp(-a\,x^2 + b\,x + c)$ where $a,\,b,\,c\in\mathbb{C}$ and ${\rm Re}(a) > 0$. When we put such a general Gaussian pulse into the left hand side of (1), we find that its extremum is $1/2$.

Second footnote:

I now understand what your question's getting at. When we discuss the HUP, "spreads" of momentum and position domain functions and the "uncertainty" product, we wontedly use the root-second-moment (standard deviation) measure of spread. This is because this is a "good" "overall" measurement of spread. The measurement your question calls for is the "Full Width Third Maximum" (FMWTM) measure of spread (more often people use the "Full Width Half Maximum" (FWHM) - the pulse's width measured between the points where its intensity $|\psi|^2$ falls to half its peak value). For the functions you have, this spread metric is "well representative": there are no "wiggly bits" on your functions. The FWHM for $A(k)$ is $2 \sqrt{\sqrt{3}-1} \alpha$, whilst for its inverse Fourier transform it is $\log3/\alpha$. The uncertainty product, by the FWTM metrics, is then $4 \sqrt{\sqrt{3}-1} \log 3$ which works out to about 1.9.

For monotonically decreasing functions, the FWHM, FWTM and like measures of spread give an intuitive, reliably idea of a function's spread. However, functions can have "arbitrarily thin wiggly bits" which make these measures unreliable. Think, for example, about the function $\exp(-1000\,x^2) + \exp(-x^2)/4$. You have a very thin pointy bit $\exp(-1000\,x^2)$ in the middle, yet most of the function's square norm is contributed by the $\exp(-x^2)/4$ part. The FWTM is of the order of 0.001. But most of the function's Fourier transform comes from the $\exp(-x^2)/4$ bit. In the Fourier space, the FWHM will be of the order of 1. So we get an uncertainty product of 0.001. You can use this kind of example to show that, with measures like the FWHM, FWTM, you can achieve arbitrarily small uncertainty products. These measures do not always give a good intuitive idea of a functions overall behaviour, and, as we can see above, they do not capture the idea of the Heisenberg Uncertainty Principle.