\n", "Right column: The period of the sine wave, (middle curve) which is determined by $k$, is now changed compared to the left-hand figure and now matches the period of the function. The product $f(x)\\sin(kx)$ is now only positive, and integrates to a finite number and so appears as a peak in the transform.\n", "_______" ] }, { "cell_type": "markdown", "metadata": { "deletable": false }, "source": [ "## 6.5 Phase sensitive detection and the lock-in amplifier\n", "\n", "In measuring signals buried in noise, the technique of phase sensitive detection is a very effective way of extracting the data and removing noise. In this method, the input to an experiment is modulated at a fixed frequency and the signal produced by the experiment is measured at this same frequency by a device known as a _lock-in amplifier_. This device illustrates the principle underlying the Fourier transform, although it is not a transform method.\n", "\n", "In using a lock-in amplifier to measure fluorescence the light used to excite the molecules, and so stimulate the fluorescence, is modulated by rotating a slotted disc (chopper) in the exciting light's path. The photomultiplier or photodiode detects the modulated (on - off ) fluorescence signal together with any noise and this signal is passed to the lock-in amplifier. The lock-in also receives a reference signal directly from the chopper and it electronically multiplies this with the fluorescence signal (see fig 20). The schematic of an instrument is shown in Fig.21. \n", "\n", "Multiples (higher harmonics) of the fundamental reference frequency are filtered away, the resulting signal is integrated over many periods of the fundamental frequency, and a DC output signal is produced. As shown in Fig. 20, when the product of reference and signal is integrated, frequencies dissimilar to the reference, $f(x)$ in the figure, will average to something approaching zero.\n", "\n", "If the reference signal is $r = r_0\\sin(\\omega t)$ and the noise free signal $s = s_0\\sin(\\omega t + \\varphi)$ then the output of the lock-in is\n", "\n", "$$\\displaystyle V_s= \\frac{r_0s_0}{T}\\int_0^T \\sin(\\omega t + \\varphi)\\sin(\\omega t)dt$$\n", "\n", "where $T = 2\\pi n/\\omega$ and $n \\gg$ 1 is an integer and $\\varphi$ is the phase (time) delay between the reference and the signal and is due to detectors and the amplifiers and other components in the experiment, but can be changed by the user. Expanding the sines and integrating gives\n", "\n", "$$\\displaystyle V_s= \\frac{r_0s_0}{T}\\int_0^T \\cos(\\varphi)-\\cos(2\\omega t+\\varphi)dt\\\\\n", "=\\frac{r_0s_0}{4\\omega T}\\big(\\sin(\\varphi)+2T\\omega \\cos(\\varphi)-\\sin(2\\omega T+\\varphi) \\big) $$\n", "\n", "The sine at twice the reference frequency is electronically filtered away leaving a signal that is constant because $T$ is the integration time set by the experimentalist and normally ranges from a few milliseconds to a few seconds. The measured signal is\n", "\n", "$$\\displaystyle V_s=\\frac{r_0s_0}{4\\omega T} \\big(\\sin(\\varphi)+2T\\omega \\cos(\\varphi) \\big) \\tag{29}$$\n", "\n", "and as the phase $\\varphi$ can be adjusted by the user, this signal can be maximized when $\\theta=\\tan^{-1}(1/(2T\\omega))$.\n", "\n", "Now consider the situation when noise is present and assume that this has a wide range of frequencies $\\omega_{1,2..}$ and amplitudes $n_{1,2..}$. The signal from an instrument is normally noisy and is represented as\n", "\n", "$$\\displaystyle s_0\\sin(\\omega t+\\varphi)+n_1\\sin(\\omega _1t+\\varphi_1)+n_2\\sin(\\omega _2t+\\varphi_2)+\\cdots$$\n", "\n", "where $s_0, \\omega$, and $\\varphi$ are respectively the amplitude, frequency, and phase (relative to the reference) of the data. \n", "\n", "The first term of the signal arises from the information we wish to measure and produces $V_s$ equation (29). We need only consider one noise term, for all the others behave similarly. Multiplying by the reference at frequency $\\omega$ but ignoring the phase $\\varphi$, as this adds nothing fundamental but makes the equations more complicated, gives the term,\n", "\n", "$$\\displaystyle 2\\sin(\\omega_1 t)\\sin(\\omega t)=\\cos((\\omega-\\omega_1)t)-\\cos((\\omega+\\omega_1)t) $$\n", "\n", "Integrating produces\n", "\n", "$$\\displaystyle \\begin{align} V_n &= \\frac{r_0n_1}{2T} \\int_0^T \\cos((\\omega-\\omega_1)t)-\\cos((\\omega+\\omega_1)t) dt\\\\ &=\\frac{r_0n_1}{4T} \\left( \\frac{ \\sin\\big((\\omega_1-\\omega)T\\big)}{\\omega_1-\\omega} -\\frac{\\sin\\big((\\omega_1+\\omega)T\\big)}{\\omega_1+\\omega} \\right) \\end{align}$$\n", "\n", "The sum frequency term is filtered by the instrument and is removed from the output leaving the term $\\displaystyle \\frac{ \\sin((\\omega_1-\\omega)T)}{\\omega_1-\\omega}$ which is the sinc function, see Fig. 15. Suppose that the frequency $\\omega_1$ represents white noise that contains all frequencies more or less equally. As these frequencies differ from $\\omega$ and the absolute value $| \\omega_1-\\omega |$ becomes larger the sinc function rapidly becomes very small. This means that the reference sine wave picks out just that frequency containing the signal and rejects almost all of the noise. The total signal is $V + V_n$ and although it still contains noise at the reference frequency $\\omega$ it contains very little at other frequencies, and the signal to noise ratio is increased very considerably. Often signals can be extracted from what appears to be completely noisy data.\n", "\n", "As a practical consideration, the reference frequency should always be chosen to be a prime number so that the chance of detecting one of the multiples of electrical mains frequency is reduced. Also, this frequency should be in a region where the inherent noise of the experiment is low and if possible be of a high enough frequency to allow a short time $T$ to be used in the integration step, allowing many separate measurements to be made in a reasonable time.\n", "\n", "![Drawing](fourier-fig21.png)\n", "\n", "Figure 21. Schematic of phase sensitive detection and a lock-in amplifier.\n", "________" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "## 6.6 Parseval or Plancherel theorem.\n", "\n", "This theorem is important because it proves that there is no loss of information when transforming between Fourier transform pairs. This is rather important because otherwise how would it be possible to tell what information has been lost or added? Fortunately, it can be shown that\n", "\n", "$$\\displaystyle \\int_{-\\infty}^{\\infty} g^*(k)g(k)dk = \\int_{-\\infty}^{\\infty} f^*(x)f(x)dx \\tag{30}$$\n", "\n", "where the asterisk denotes the complex conjugate. The Fourier transform of $f(x)$ is $g(k)$, which is integrated over its variable $k$, and similarly $f(x)$ is integrated over its variable $x$. As the total integral taken over all coordinate space $x$ and that over its conjugate variable $k$ is the same, all the information in the original function is retained in the transformation, i.e. it looks as if the transform is a different beast but this is only a disguise as it contains exactly the same information. This, of course, means that if something is done to the transform, then, in effect, the same is done to the function.\n", "\n", "The Plancherel theorem (also called Rayleigh's theorem as it was first used by him in the theory of Black-Body radiation) is effectively the same but usually written as \n", "\n", "$$\\displaystyle \\int_{-\\infty}^{\\infty} |g(k)|^2dk = \\int_{-\\infty}^{\\infty} |f(x)|^2 dx $$\n", "\n", "Graphically it means that the shaded areas are the same. The figure shows the transform of a square wave as in fig 15. The transform is \n", "\n", "$$\\displaystyle g(k)=\\frac{\\sin(ak)}{ak}$$\n", "\n", "where $a = 4$. The function $f(x) = 1 $ in the range $-2\\to 2$ so $\\int f(x)^2 dx= 4$. \n", "\n", "![Drawing](fourier-fig21a.png)\n", "\n", "Figure 21a. Illustrating the Parseval or Plancherel theorem. The shaded areas are the same size, although they do not appear to be. The absolute square of the transform $|g(k)|^2$ extends to $\\pm \\infty$ which makes up the area since the function is always positive a small part of which is shown magnified.\n", "___________\n", "\n", "This theorem is very important in quantum mechanics. Should $f(x)$ represent a wavefunction that varies as a function of distance $x$, which could be the displacement from equilibrium of an harmonic oscillator, then variable $k$ can be interpreted as the momentum (usually given the letter $p$) making $g(k)$ the wavefunction in 'momentum space'. This means that calculations can be formed either in spatial coordinates, i.e. distance or in 'momentum space' depending upon which is the most convenient mathematically. The change in displacement, $\\delta x$, and change in momentum, $\\delta p$, are conjugate pairs of variables and are linked by the Heisenberg uncertainty principle $\\delta x\\delta p \\ge \\hbar/2$.\n", "\n", "If $x$ is time $t$ and $k$ frequency $\\omega$ then Parseval's theorem has a simple physical interpretation. On the right of the equation $f(t)$ is the electric field of some radiation, for example i.r. in a spectrometer or a laser pulse, then $|f(t)|^2$ is proportional to the total radiated power and the integral the total radiated energy. On the left, $\\int|g(\\omega)|^2d\\omega$ is the total amplitude squared of the spectrum, and $g(\\omega)$ the energy per unit frequency interval. Thus Parseval's theorem represents the conservation of energy.\n", "\n", "### **(i) The delta function as an integral**\n", "\n", "The Parseval theorem can be proved using the delta function in its integral form. First we find the delta function. Suppose that a fourier transform is inserted into itself, starting with eqn. 28,\n", "\n", "$$\\displaystyle f(x) = \\frac{1}{\\sqrt{2\\pi}}\\int_{-\\infty}^\\infty g(k)e^{+ikx}dk \\qquad\\tag{28}$$\n", "\n", "then from eqn. 27 substitute $g(k)$ and change the dummy integration variable to $x'$ to avoid confusion, \n", "\n", "$$\\displaystyle f(x) = \\frac{1}{2\\pi} \\int_{-\\infty}^\\infty \\int_{-\\infty}^\\infty f(x')e^{-ikx'}dx'e^{+ikx}dk $$\n", "\n", "and by sleight of hand, interchange order of integration, \n", "\n", "$$\\displaystyle f(x) = \\int_{-\\infty}^\\infty f(x') \\left[\\frac{1}{2\\pi} \\int_{-\\infty}^\\infty e^{-ik(x-x')}dk\\right]dx'$$\n", "\n", "For this equation to be true the expression in square brackets must now be interpreted as a delta function making\n", "\n", "$$\\displaystyle \\delta(x-x')=\\frac{1}{2\\pi} \\int_{-\\infty}^\\infty e^{-ik(x-x')}dk \\tag{28a}$$\n", "\n", "which means that one side is interpreted as having the same effect as the other, $\\delta(x-x')=1$ if $x=x'$ otherwise it is zero and similarly for the integral. Recall that the delta function $\\delta(x-x')$ picks out just one term in any integral at $x=x'$.\n", "\n", "$$\\displaystyle \\int_{-\\infty}^\\infty f(x)\\delta(x-x')dx = f(x') \\tag{28b}$$\n", "\n", "$$\\displaystyle \\int_a^b\\delta(x-x')dx = 1 \\text{ if } a\\lt x\\lt b \\text{ otherwise } 0$$\n", "\n", "\n", "The delta function also has these properties\n", "\n", "$$\\displaystyle \\begin{array}{lll}\\\\ \\hline \\delta(x)&=\\delta(-x)\\\\ x\\delta(x)&=0\\\\ \\delta(ax)&=\\displaystyle\\frac{1}{|a|}\\delta(x)\\\\ \\delta(x^2-a^2)&=\\displaystyle\\frac{1}{|a|}(\\delta(x-a)+\\delta(x+a))\\\\ \\hline\\end{array}$$\n", "\n", "To prove the Parseval theorem we start with $\\int|f(x)|^2dx=\\int f(x)^*f(x)dx $ where * indicated the complex conjugate, substitute for $f(x)$ using eqn 28\n", "\n", "$$\\displaystyle \\begin{align}\\int_{-\\infty}^{\\infty} |f(x)|^2 dx &=\\int_{-\\infty}^{\\infty}dx\\frac{1}{2\\pi}\\int_{-\\infty}^\\infty g^*(k)e^{-ikx}dk\\int_{-\\infty}^\\infty g(k')e^{+ik'x}dk' \\\\ &=\\int_{-\\infty}^{\\infty} g^*(k)dk\\int_{-\\infty}^{\\infty} g(k')dk'\\left[\\frac{1}{2\\pi}\\int_{-\\infty}^\\infty e^{i(k'-k)x} \\right]dx\\\\&=\\int_{-\\infty}^{\\infty} g^*(k)dk\\int_{-\\infty}^{\\infty} g(k')\\delta(k'-k)dk'=\\int_{-\\infty}^{\\infty} g^*(k)g(k)dk=\\int_{-\\infty}^{\\infty} |g(k)|^2dk \\end{align}$$\n", "\n", "### **(ii) The delta function and comb of delta functions.**\n", "\n", "Using a comb of delta functions and convoluting this with a more complex function is a direct way of making a repeated pattern such as an image or positioning molecules in each unit cell in a crystal. This image can then be fourier transformed to imitate the diffraction pattern expected from a given crystal.\n", "\n", "A delta function has the property that it extracts just a single term from an integral, i.e. \n", "\n", "$$\\displaystyle \\int_{-\\infty}^\\infty f(x)\\delta(x-x_0)dx = f(x_0)$$\n", "\n", "and is because $\\delta(0)=1$ and is otherwise zero. This is also its shifting property, the function is moved from $f(x)$ to $f(x_0)$, which is equivalent to multiplying its transform by a phase factor $e^{-ikx_0}$.\n", "\n", "The delta function's transform is\n", "\n", "$$\\displaystyle g(k)=\\frac{1}{\\sqrt{2\\pi}}\\int_{-\\infty}^\\infty \\delta(x)e^{-ikx}dx =\\frac{1}{\\sqrt{2\\pi}} e^{0}=\\frac{1}{\\sqrt{2\\pi}}$$\n", "\n", "and so the inverse transform of $\\delta(x)$ is $1$ and so we can write \n", "\n", "$$\\displaystyle \\delta(x)=\\frac{1}{\\sqrt{2\\pi}}\\int_{-\\infty}^\\infty e^{ikx}dk$$\n", "\n", "and this means that the delta function is the transform of a constant value. The transform of $\\cos(k_0x)$ can now be found when the cosine is put into its exponential form.\n", "\n", "$$\\displaystyle \\begin{align} g(k)&=\\frac{1}{2\\sqrt{2\\pi}}\\int_{-\\infty}^\\infty (e^{ik_0x}-e^{-ik_0x})e^{-ikx} dx\\\\\n", "&=\\frac{1}{2\\sqrt{2\\pi}}\\int_{-\\infty}^\\infty e^{-i(k-k_0)x}-e^{-i(k+k_0)x} dx\\\\\n", "&=\\frac{1}{2} (\\delta (k-k_0)+\\delta(k+k_0)) \\end{align}$$\n", "\n", "which means that the transform has two spikes at wavevector $\\pm k_0$. This expression can be used to describe interference by two thin slits producing Young's fringes. It also shows a single frequency in a transform corresponds to a cosine wave in 'real' space, i.e. an extended wave train containing a single frequency gives a 'spike' or a single line in its spectrum (i.e. its transform). This extends to several waves of different frequencies giving different lines in a spectrum. Of course, the negative frequency has no physical meaning.\n", "\n", "The transform of a shifted delta function $\\delta(x-x_0)$ is\n", "\n", "$$\\displaystyle g(k)=\\frac{1}{\\sqrt{2\\pi}}\\int_{-\\infty}^\\infty \\delta(x-x_0)e^{-ikx} dx =\\frac{1}{\\sqrt{2\\pi}} e^{-ikx_0}$$\n", "\n", "A function that is itself a sequence of delta functions is called 'comb' or 'sha' function,\n", "\n", "$$\\displaystyle f(x)=\\sum_n \\delta(x-x_n)$$\n", "\n", "has a transform\n", "\n", "$$\\displaystyle g_n(k) =\\frac{1}{\\sqrt{2\\pi}}\\sum_ne^{-ikx_n}$$\n", "\n", "When there is a series of equally spaced delta functions \n", "\n", "$$\\displaystyle f(x)=\\sum_{n=-\\infty}^\\infty \\delta(x-na)$$\n", "\n", "the transform is also never ending a series of delta functions. To sum the transform it is necessary to suppose that $n$ is a large number $N$ and do the summation as a geometric series\n", "\n", "$$\\displaystyle \\sqrt{2\\pi}g(x)=\\sum_{n=-N}^N e^{-ikaN}=\\frac{e^{iak(N+1)}-e^{-iakN} }{1-e^{-iak}}$$\n", "\n", "which multiplying top and bottom by $e^{iak/2}$ and using the definition of a sine; $\\sin(x)=(e^{ix}-e^{-ix})/2i$ produces\n", "\n", "$$\\displaystyle g(k)=\\frac{1}{\\sqrt{2\\pi}}\\frac{\\sin\\left((N+\\frac{1}{2})ak\\right)}{\\sin\\left(\\frac{ak}{2}\\right)} \\tag{28c}$$\n", "\n", "The function appears as a series of spikes separated by $2\\pi/a$ when $N$ is large. Figure 21b shows the transform when $N=100, a=2$, and it is clear that the delta function is being approached even with small $N$. In the limit $N\\to \\infty$ the transform of a comb of separation $a$ is a comb of delta functions with separation $2\\pi /a$. \n", "\n", "$$\\displaystyle f(x)=\\sum_{n=-\\infty}^\\infty \\delta(x-na), \\qquad g(k)=\\frac{1}{\\sqrt{2\\pi}}\\sum_{n=-\\infty}^\\infty \\delta(x-2\\pi n/a) \\tag{28d}$$\n", "\n", "![Drawing](fourier-fig21b.png)\n", "\n", "Figure 21b. The transform of a comb of separation $a$ is a comb of separation $2\\pi/a$. In this figure $N=100, a=2$. In the limit $N\\to\\infty$ a comb of $\\delta$ functions result.\n", "________________\n", "\n", "The reason that the transform becomes more like a series of delta functions as $N$ increases is that as $N$ increases the frequency of the sine wave in the numerator increases but that of the denominator remains the same. These waves are only in phase at each $2\\pi/a$ and as the high frequency numerator is oscillating very rapidly positive and negatively compared to $ak/2$ it makes the ratio effectively zero elsewhere. " ] }, { "cell_type": "markdown", "metadata": { "deletable": false }, "source": [ "## 6.7 Uncertainty Principle\n", "\n", "It is known from experiment that when an emission line from an atomic or molecular transition has a very broad frequency spread, then the lifetime $\\tau$ of the state involved is short lived and vice versa. This is called the 'time-energy' uncertainty relationship $\\Delta E\\tau \\ge \\hbar/2$ or equivalently $v\\tau\\ge 1/2$ as $E=hv$. A similar effect is observed when a time varying signal is measured for example a voltage on an oscilloscope. The product of the signal's duration and its bandwidth (its spread in frequency) has a certain minimum value. This is a consequence of the variables, time and frequency being related via a fourier transform. Time and frequency are called _conjugate_ variables. \n", "\n", "To show this can be quite tricky since the variance of the transform has to be calculated and this may not be integrable. The proof is given by Bracewell 'The Fourier Transform and its Applications'. Instead of giving this proof to illustrate the effect some examples of particular cases are described.\n", "\n", "As a measure of the spread in a value its standard deviation can be used, see Chapter 4 Integration eqn. 26. The square of the standard deviation is the variance and is defined as \n", "\n", "$$\\displaystyle \\sigma^2 =\\langle x^2\\rangle - \\langle x\\rangle^2$$\n", "\n", "where the brackets $\\langle \\rangle$ indicate the average value. The average value of a function $p(x)$ is defined as \n", "\n", "$$\\displaystyle \\langle x_p\\rangle= \\frac{\\int xp(x)dx}{\\int p(x)dx}$$\n", "\n", "and the average of the square \n", "\n", "$$\\displaystyle \\langle x_p^2\\rangle= \\frac{\\int x^2p(x)dx}{\\int p(x)dx}$$\n", "\n", "The denominator is the normalisation. If the function is symmetrical about zero the mean is also zero and $\\langle x\\rangle=0$ and can be ignored. All that is necessary is to calculate $\\langle f^2\\rangle\\langle g^2\\rangle$ for a function $f$ and its transform $g$ and take the square root to obtain the standard deviation of the product. \n", "\n", "Often the transform and function may not be good distributions in which case their square is taken as the function to use thus we make the average squared ( also called the Energy Function) as \n", "\n", "$$\\displaystyle \\langle x^2\\rangle= \\frac{\\int x^2f^*f dx}{\\int f^*f dx}$$\n", "\n", "and a similar equation for the transform $g$. As the function and transform is usually complex the square is obtained using the complex conjugate i.e. $p^2 \\to p^*p$.\n", "\n", "This method will be demonstrated below but sometimes the transform integrals become infinity, in this case we choose to take the deviation as the measure from the peak of the transform ($g$) to the first zero. \n", "\n", "### **(i) A Gaussian shaped pulse** \n", "\n", "A wavepacket comprising several waves of varying frequency can have an overall profile that is Gaussian in shape. This can apply, for example, to a laser pulse or a summation of harmonic oscillator wavefunctions. The normalised Gaussian function is \n", "\n", "$$\\displaystyle f(t)= \\frac{e^{-t^2/2\\sigma^2}}{\\sigma \\sqrt{2\\pi} } $$\n", "\n", "where $\\sigma$ is the standard deviation, or width of the Gaussian, see figure 4, Chapter 13 'Data Analysis'. The transform of a Gaussian is also a Gaussian but with a different width. \n", "\n", "$$\\displaystyle g(v)= \\frac{1}{\\sqrt{2\\pi}}\\int_{-\\infty}^\\infty f(t)e^{-ivt} dt = \\frac{1}{\\sqrt{2\\pi}}e^{-i\\sigma^2v^2/2}$$\n", "\n", "As the standard deviation of $f$ is by definition $\\sigma$ and by comparison the standard deviation of $g$ must be $1/\\sigma$ hence $\\sigma_t\\sigma_v=\\sigma/\\sigma=1$.\n", "\n", "Calculating using the method outlined above to find $\\langle t^2\\rangle$ and $\\langle v^2\\rangle$ we expect a different result because we now use the square of the function and transform.\n", "\n", "Taking all integrations between $\\pm\\infty$, the function \n", "\n", "$$\\displaystyle \\int f(t)^2dt= \\frac{1}{2a\\sqrt{\\pi}},\\quad\\text{and}\\quad \\displaystyle \\int t^2f(t)^2dt = \\frac{a}{4\\sqrt{\\pi}}$$\n", "\n", "and their ratio is \n", "\n", "$$\\displaystyle \\Delta t= \\frac{a^2}{2}$$\n", "\n", "The transform has similar integrals\n", "\n", "$$\\displaystyle \\int g(v)^2dv= \\frac{1}{2a\\sqrt{\\pi}}\\quad\\text{and}\\quad \\displaystyle \\int v^2g(v)^2dv = \\frac{1}{4a^3\\sqrt{\\pi}}$$\n", "\n", "and their ratio is \n", "\n", "$$\\displaystyle \\Delta v= \\frac{1}{2a^2}$$ \n", "\n", "The product of uncertainties after remembering that we calculate the square of the values is \n", "\n", "$$\\displaystyle \\Delta t\\Delta v = \\frac{1}{2}$$ \n", "\n", "which is the minimum possible value.\n", "\n", "### **(ii) Excited state decay**\n", "\n", "An excited electronic state can decay by emitting a photon, fluorescence if the transition is allowed, or phosphorescence if from a triplet excited state to a singlet ground state. The time profile can be measured as can the spectral width of the transition and they demonstrate the time - energy/frequency relationship. The excited state has a lifetime $\\tau$ and transition frequency $\\omega_0$. A spatial analogue is momentum broadening as a result of collisions, where the lifetime is replace by the mean free path.\n", "\n", "The field of the emitted photon is \n", "\n", "$$\\displaystyle f(t)=e^{i\\omega_0 t}e^{-t/2\\tau}$$\n", "\n", "which is that of a plane wave of frequency $\\omega_0$ that decays away (or is damped) with a lifetime $\\tau$. The detector measures the 'square' or absolute value squared of this which is \n", "\n", "$$\\displaystyle |f(t)|^2\\equiv f(t)^*f(t) = e^{-t/\\tau}$$ \n", "\n", "The transform is \n", "\n", "$$\\displaystyle g(\\omega)= \\frac{1}{\\sqrt{2\\pi}}\\int_{-\\infty}^\\infty f(t)e^{-i\\omega t} dt = \\sqrt{\\frac{2}{\\pi}}\\left(\\frac{1}{2i (\\omega-\\omega_0)+1/\\tau}\\right)$$\n", "\n", "making the measured spectral profile, which has the shape of a Lorentzian curve,\n", "\n", "$$\\displaystyle g(\\omega)^*g(\\omega)=\\frac{2}{\\pi}\\left( \\frac{1}{4(\\omega-\\omega_0)^2+1/\\tau^2} \\right) $$\n", "\n", "The full width at half maximum of this curve is $1/\\tau$ which makes $\\Delta t\\Delta \\omega = \\tau/\\tau=1$. Figure 54 in the answer to question 7 shows the exponential decay and the Lorentzian curves. As the energy is related to the frequency as $\\Delta E =\\hbar \\Delta \\omega$ then $\\tau\\Delta E = \\hbar$\n", "\n", "The calculation using $\\displaystyle \\int g(\\omega)^2 d\\omega$ as in the previous example will not work here because this integral is infinite.\n", "\n", "### **(iii) Finite wave-train**\n", "\n", "If a plane wave of frequency $\\omega_0$ exists for a short time $\\pm a$ and is zero elsewhere then over this range $\\displaystyle f(t)=e^{i\\omega t};\\quad -a\\le t \\le a$, see figure 19 where $a=10$. The spread of the wave $\\Delta x = a$. The 'square' of $f$ is $f(x)^*f(x)=1$\n", "\n", "The transform has the form of a sinc function \n", "\n", "$$\\displaystyle g(\\omega)=\\int_{-\\infty}^\\infty f(t)e^{-i\\omega t}dt= \\sqrt{\\frac{2}{\\pi}}\\frac{\\sin\\left( a (\\omega-\\omega_0)\\right)}{\\omega-\\omega_0}$$\n", "\n", "The function $g^2$ cannot be normalised easily as it produces another integral (the Sine integral) but it is clear that the zeros of the function are at the same place in $g$ and its square and so we can take the spread $\\Delta \\omega$ to be the value from $\\omega_0$ to the first minimum. The zeros of the sinc function occur at $\\mathrm{sinc}(n\\pi)$ where $n$ is an integer greater than zero, thus the first zero occurs at $\\pm \\pi/a$ from the central frequency, see figure 15.\n", "\n", "If this zero is associated with $\\Delta \\omega$ then $\\Delta \\omega =\\pi/a$ and the product $\\Delta x\\Delta \\omega= \\pi$.\n", "\n", "### **(iv) Heisenberg Uncertainty**\n", "\n", "The relationship $\\Delta \\omega\\Delta x \\ge 1$ as described in the last example above is not an inherent property of quantum mechanics but is a property of Fourier Transforms. The last example shows that it is not possible to form a train of electromagnetic waves for which it is possible to measure, at the same time, the position and wavelength with complete accuracy. \n", "\n", "However, when considering quantum mechanics a particle is given wavelike properties via the de-Broglie relation. Such a material particle (such as an electron or a molecule) of energy $E$ and momentum $\\vec p$ is now associated with a wave of angular frequency $\\omega=2\\pi\\nu$ and wavevector $\\vec k$ i.e. $E=\\hbar\\omega; \\vec p=\\hbar \\vec k$ and leads to $\\Delta x\\Delta p \\ge \\hbar/2$. The fact that the Schroedinger equation is linear and homogeneous means that for particles a superposition principle applies which gives them wavelike properties. The small value of Planck's constant makes the limitations of the uncertainty principle totally negligible for anything macroscopic, i.e. greater than approx micron size. " ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "## 6.8 Summary of some Fourier Transform properties\n", "\n", "$$\\displaystyle \n", "\\begin{array}{ll}\n", "\\hline\n", "&\\text{The transform pair is} & f (x)& \\rightleftharpoons & g(k)\\\\[0.15cm] \n", "\\text{(i)}&\\text{Shift or delay } & f (x - x_0) & \\rightleftharpoons & e^{- ikx_0}g(k)\\\\[0.15cm] \n", "\\text{(ii)}&\\text{frequency shift} & f (x)e^{ik_0x} & \\rightleftharpoons & g(k - k_0) \\\\[0.15cm]\n", "\\text{(ii)}&\\text{Scaling/Similarity} \\quad a \\gt 0 & \\displaystyle f(ax) & \\rightleftharpoons & \\displaystyle \\frac{1}{ |a|}g\\left(\\frac{k}{a}\\right)\\\\[0.15cm]\n", "\\text{(iv)}&\\text{Reversal } & f (-x) & \\rightleftharpoons & g(-k) \\\\[0.15cm]\n", "\\text{(v)}&\\text{Complex Conjugate } & f (x)^* & \\rightleftharpoons & g(-k)^* \\\\[0.15cm]\n", "\\text{(vi)}&\\text{Derivative} & \\displaystyle F\\left(\\frac{d}{dx}f(x)\\right) & \\rightleftharpoons &ik\\,g(k)\\\\[0.15cm]\n", "\\text{(vii)}& n^{th}\\text{ Derivative} & \\displaystyle F\\left(\\frac{d^n}{dx^n}f(x)\\right) & \\rightleftharpoons &(ik)^n\\,g(k)\\\\[0.15cm]\n", "\\text{(viii)}&\\text{Derivative} & \\displaystyle F\\left(\\frac{d}{dt}f(x,t)\\right)& \\rightleftharpoons & \\int \\displaystyle \\frac{d}{dt}g(k,t)e^{-ikx} dx \\\\[0.15cm]\n", "\\text{(ix)}&\\text{Product} & F(xf(x)) & \\rightleftharpoons &\\displaystyle i\\,\\frac{d}{dk}g(k)\\\\[0.15cm]\n", "\\text{(x)}&\\text{linearity} & F\\left(af_a(x) + bf_b(x)\\right)& \\rightleftharpoons & ag_a(k) + bg_b(k)\\\\ \n", "\\hline\n", "\\end{array}\n", "$$\n", "\n", "The capital $F(\\cdots)$ means 'do the transform'. The derivation of these results is given below.\n", "\n", "### **(i), (ii) Shift, delay.**\n", "\n", "$\\qquad$Let $s=x-x_0$ then $ds=dx$ and provided $f$ exists at $x_0$,\n", "\n", "$$\\displaystyle F(f(x- x_0)) = \\int_{-\\infty}^{\\infty} f(x- x_0)e^{-ik x}dx = \\int_{-\\infty}^{\\infty} f(s)e^{-ik(s+ x_0)}ds = e^{-ik x_0} g(k)$$\n", "\n", "### **(iii) Scaling/Similarity.**\n", "\n", "$\\qquad$Make the substitution $s=ax,\\; a>0$\n", "\n", "$$\\displaystyle F(f(ax))=\\int_{-\\infty}^{\\infty}f(ax)e^{-ikx}dx=\\frac{1}{|a|}\\int_{-\\infty}^{\\infty}f(s)e^{-iks/a}ds=\\frac{1}{|a|}g\\left(\\frac{k}{a}\\right)$$\n", "\n", "### **(iv) Reversal.** \n", "\n", "$\\qquad$Start with scaling (as above (iii) ) and let $a=-1$ giving \n", "\n", "$$\\displaystyle F(f(-x)) = g(-k)$$\n", "\n", "### **(v) Complex conjugate.**\n", "\n", "$$\\displaystyle g(-k)= \\int_{-\\infty}^\\infty f(x)e^{ikx}dx$$\n", "\n", "$\\qquad$taking the complex conjugate of both sides gives\n", "\n", "$$\\displaystyle g(-k)^*= \\int_{-\\infty}^\\infty f(x)^*e^{-ikx}dx=f(x)^*$$\n", "\n", "### **(vi)(vii) Fourier derivatives.** \n", "\n", "$\\qquad$These are important when solving differential equations. First derivative\n", "\n", "$$\\displaystyle \\begin{align} \n", "F\\left(\\frac{df}{dx}\\right) = \\int_{-\\infty}^\\infty \\frac{df}{dx}e^{-ikx}\\, dx\n", "\\end{align}$$\n", "\n", "$\\qquad$Integrate by parts using $\\int udv=uv-\\int vdu$ where $u=e^{-ikx}$ and $dv=df/dx$ gives\n", "\n", "$$\\begin{align} F\\left(\\frac{df}{dx}\\right)&= \\left |f(x)e^{-ik x}\\right |^\\infty_{-\\infty} + ik \\int_{-\\infty}^\\infty f(x)e^{-ik x} \\,dx\\\\ & =0+ ik F\\left( f(x)\\right) \\\\ & = ik g(k) \\end{align}$$\n", "\n", "$\\qquad$By induction the $n^{th}$ derivative's transform is \n", "\n", "$$\\displaystyle F\\left(\\frac{d^nf}{dx^n}\\right)=(ik)^n g(k)$$\n", "\n", "### **(viii) Product.** \n", "\n", "$\\qquad$The product $xf(x)$ can be useful in solving differential equations.\n", "\n", "$$\\displaystyle \\begin{align}F\\left(xf(x)\\right)&=\\int_{-\\infty}^{\\infty} xf(x)e^{-ikx}dx =\\int_{-\\infty}^{\\infty} f(x)\\frac{d}{dk}\\left(ie^{-ikx}\\right) dx\\\\&=i\\frac{d}{dk}\\int_{-\\infty}^{\\infty} f(x)e^{-ikx} dx=i\\frac{d}{dk}g(k)\\end{align}$$\n", "\n", "### **(i,ii) Example using the Shift Theorem**\n", "\n", "When we measure a spectrum we expect to see a number of lines depending on the particular molecules being excited. These lines each have their own width and energy, i.e wavelength or frequency. However, if the light entering the spectrometer has a little of it delayed, such as by a beam splitter, and which then enters the spectrometer extra lines can be observed due to the phase-change introduced by the delay. The extra lines are caused by beating between the two light fields. The experimental detail is shown in the diagram.\n", "\n", "![Drawing](fourier-fig21c.png)\n", "\n", "figure 21c. Left The arrangement whereby two beams enter the spectrometer. Right The spectrum as it should be (left) and (right) after contamination with a little ($5$%) of the delayed beam. The ripples are separated in frequency by $2\\pi/\\Delta$.\n", "____________\n", "\n", "If the spectrum has the field $E(t)$ the detector measures the absolute value squared of the fourier transform $F$ of the total field, which in this case is the sum of the direct and reflected parts, i.e. a 'square law detector' measures\n", "\n", "$$\\displaystyle W(\\omega)=\\bigg| F\\left(E(t)+cE(t-\\Delta)\\right)\\bigg|^2$$\n", "\n", "where $\\Delta$ is the time delay and $c$ the amount in the delayed beam. This is single a part of the autocorrelation of the field $E$ taken at one delay $\\Delta$. Using the Shift Theorem produces\n", "\n", "$$\\displaystyle \\begin{align}W(\\omega)&=\\bigg|E(\\omega)(1+ce^{-i\\omega\\Delta})\\bigg|^2\\\\\n", "&= |E(\\omega)|^2(1+ce^{-i\\omega\\Delta})(1+ce^{+i\\omega\\Delta})\\\\\n", "&= |E(\\omega)|^2(1+c^2+2c\\cos(\\omega \\Delta))\\end{align}$$\n", "\n", "where the identity $2\\cos(\\theta)=e^{+i\\theta}+e^{-i\\theta}$ was used. This expression shows us that the spectrum $E(\\omega)$ now has oscillations/ripples that might look like new transitions but are an artefact of allowing two beams into the spectrometer. The frequency separation between peaks of the ripples is $2\\pi/\\Delta$. The way round this problem is to use a beam-splitter that is anti-reflection coated on its far side, or use a far thicker one and use a knife edge to remove the weaker beam.\n", "\n", "The spectrum is shown in fig. 21c. The fringes are very clear even when $c$ is small, for example the figure shows the spectrum when $c=0.05$ or $5\\text{%}$ of the more intense beam. The oscillation at the peak is $\\approx \\pm 10\\text{%}$ of the peak height, rather large for such a small amount of extra light.\n", "\n", "## 6.9 Table of useful Fourier transforms\n", "The capital $F(\\cdots)$ means 'do the transform'.\n", "\n", "$$\\displaystyle \\begin{array}{ll}\n", "\\hline\n", "\\text{(i) Gaussian} &f(x)=e^{-ax^2} & F(f(x)) & \\rightleftharpoons &\\displaystyle \\frac{1}{\\sqrt{2a}}e^{-k^2/(4a)}\\\\[0.15cm]\n", "\\text{(ii) cosine }&f(x)=\\cos(2\\pi k_0 x) & F(f(x)) & \\rightleftharpoons &\\frac{1}{2}(\\delta(k-k_0)+\\delta(k+k_0)) \\\\[0.15cm]\n", "\\text{(iii) sine }&f(x)=\\sin(2\\pi k_0 x) & F(f(x)) & \\rightleftharpoons &\\frac{1}{2}i(\\delta(k+k_0)-\\delta(k-k_0)) \\\\[0.15cm]\n", "\\text{(iv) exponential }&f(x)=e^{-2\\pi k_0 |x|} & F(f(x)) & \\rightleftharpoons &\\displaystyle \\frac{1}{\\pi}\\frac{k_0}{k^2+k_0^2} \\\\[0.15cm]\n", "\\text{(v) sinc }&f(x)=\\sin(x)/x & F(f(x)) & \\rightleftharpoons & \\text{top hat, }\\\\\n", "& & & &1 \\text{ if } 1/2\\le x\\le 1/2\\; \\text{ else } 0 \\text{ (fig }15) \\\\[0.15cm]\n", "\\text{(vi) complex exponential }&f(x)=e^{ik_0x} & F(f(x)) & \\rightleftharpoons & \\delta(k-k_0) \\\\[0.15cm]\n", "\\text{(vii) a constant} & f(x)=1& F(f(x))=& \\rightleftharpoons & \\delta(k) \\\\[0.15cm]\n", "\\hline\\end{array}$$" ] }, { "cell_type": "code", "execution_count": null, "metadata": {}, "outputs": [], "source": [] } ], "metadata": { "kernelspec": { "display_name": "Python 3 (ipykernel)", "language": "python", "name": "python3" }, "language_info": { "codemirror_mode": { "name": "ipython", "version": 3 }, "file_extension": ".py", "mimetype": "text/x-python", "name": "python", "nbconvert_exporter": "python", "pygments_lexer": "ipython3", "version": "3.9.6" } }, "nbformat": 4, "nbformat_minor": 2 }