This is a topic I encountered a number of years ago while at LBNL, when I came across the following statement [1]:

Transcription

1 Equivalent signal to noise in diffraction vs. bright-field experiments Charles Sindelar January 2010 SUMMARY This is a topic I encountered a number of years ago while at LBNL, when I came across the following statement [1]: Thus, for the same exposure, holography should be equal to normal phase contrast in performance, and diffraction methods inferior because of the loss of the information on the phases of the Fourier components of the image [emphasis added]. I found this statement problematic, because it seemed to me that if one had perfect experimental apparati, one should in principle obtain the same amount of "information", and therefore the same quality image, no matter which way the electrons were used to acquire information. After going through the derivation, I have now concluded that there was a logical flaw in the argument upon which the statement from [1] is based. In fact, the diffraction experiment will result in two separate measurements of a given Fourier amplitude, each with the same signal to noise ratio as the single estimate of the same Fourier amplitude which results from an imaging experiment using the identical dose. Thus, there is a trade-off: one can either obtain both the amplitude and the phase (imaging), or one can obtain two independent measurements of the amplitude-- each with the same signal to noise ratio as the single measurement in the imaging experiment. Thus, if one collects diffraction information and then retrieves the image, for example through iterative phase retrieval methods, it appears likely that the resulting image will be of comparable quality to the image collected directly. INTRODUCTION While ref. [1] does not contain the full derivation of the "inferior" statement, a citation is made to [2] where a partial argument may be found. Essentially it

2 goes like this: Suppose a measurement made in the diffraction plane where one electron on average is detected at a diffraction spot, and N unscattered electrons are detected at F_0, the unscattered beam. This would give a signal to noise ratio, in the diffraction spot, of sqrt(1 electron count) = 1. Now suppose the equivalent bright field observation is made with N electrons. According to the Fourier transform of the image density function, this will result in approximately 2 * sqrt(n) electrons contributing to the image component from the diffraction spot, vs. approximately N electrons contributing to a uniform noisy background; this background will thus have a noise level of approximately sqrt(n). Therefore, the signal to noise ratio will be 2*sqrt(N)/sqrt(N) = 2. Thus, goes the argument, the SNR for a given Fourier component from the brightfield measurement is actually twice that for the analogous diffraction measurement; but this is accounted for, according to ref. [2], by the presence of two independent measurements of the diffraction spot (Friedel pairs): The factor of 2 arises because, for thin untilted crystals, an electron diffraction pattern has two Friedel-related diffraction spots F(g) and F(-g) of almost identical amplitude, which interfere with the direct beam F(0) to produce an image in which contributions from Friedel-related Fourier components are indistinguishable [2]. As I will show, this identification of the factor 2 is misleading, and leads to the erroneous conclusion. In fact, the factor of 2 comes from squaring of the image wave function. Furthermore, this same factor of 2 also arises in the diffraction measurement, for a similar reason: the diffraction pattern is a measure of the squared fourier components, and when these are square rooted to obtain the Fourier amplitudes, the SNR increases by a factor of 2 for the diffraction measurements. Thus, each diffraction measurement of a given Fourier amplitude will have the same SNR as that estimated from the bright field image. Therefore, one gets more information about the amplitude from the diffraction measurement, at the price of getting none of the phase information.

3 DERIVATION To begin, ref [2] cites ref [3], where the following derivation for electon optics is given. In the weak phase object approximation, the electron wave function in the diffraction plane is given by: s =F 0 s F g e i /2 s g F * g e i /2 s g where F(0) is the square root of the intensity of the unscattered beam, I(0), and F(g) is the Fourier component for a single spatial frequency g. The wave function of the image is then described by FT 1 e i s s where FT is the Fourier transform, γ(s) is the wave distortion due to spherical aberration and defocus. Squaring this wave function gives the image function: I x = F F 0 F g cos[2 g x phase F g ]sin g whose Fourier transform is: I s = F 0 2 s 2 F 0 F g e i phase F g sin g s g 2 F 0 F g e i phase F g sin g s g So, taking the values of I at 0 and g, you get the following ratio: I g I 0 g =2 F sin g F 0 The important thing that this formula says is that, for a well chosen defocus (so sin(gamma) is close to 1), if one takes the Fourier transform of a measured bright-field image, the ratio between the desired Fourier amplitude I(g) and the unscattered beam amplitude I(0) will be twice the ratio of the corresponding Fourier amplitudes obtained from the diffraction measurement.

4 What about diffraction measurements? Here we note the important fact that these measure F^2, the squared magnitude of the Fourier amplitudes. Thus, one takes the square root of the observations to get the F's. Now, what will the signal to noise ratio be in the diffraction and bright field images? SNR in the diffraction case Suppose N electrons are directed at the sample, and the sample is such that a single one of these electrons, on average, would land on a given diffraction spot F(g). In the diffraction measurement one would observe N unscattered electrons, resulting in an estimate of F_0 as sqrt(n); and 1 scattered electron at position g (on average) giving an estimate for F(g) ^2 = 1. Due to counting statistics, therefore, the diffraction measurement of F(g) ^2 will have a noise of ~sqrt(1) = 1. Of course, to get the F's one now must take the square root. Crucially, this step increases the signal to noise ratio by a factor of 2, as follows: Suppose x = a^b, with perfectly known b, and the error in a (variance) is s_a. Then standard statistical theory gives the following result for s_x, the error (variance) in x (see: ) : s_x / x = b * s_a / a Thus, the ratio of the new signal variance to the signal, is scaled from the original ratio of signal variance to signal, by a factor of b. In our case, b is 0.5. This implies that the new signal-to-noise ratio (x/x_s) will be 1/0.5 = twice the original signal to noise ratio, before the square root. Thus, taking the square root of a normally distributed signal with average value 1 (as in our example, one measured electron on average) and variance 1, gives a new normally distributed signal with variance ½. SNR in the bright field case

5 From above, we have that I g I 0 g =2 F sin g F 0 and for the N electron case we therefore get F(g)/F(0) = 1/sqrt(N) = sqrt(n). This means that in the bright field image, of every N electrons, 2*sqrt(N)*sin(gamma(g)) ~ 2*sqrt(N) electrons will contribute to the signal of interest F(g), and the remainder (if F(g) were the only signal) would be unscattered. Assuming sqrt(n) << N, we can conclude that the image will have a uniform noisy background of average value N (from counting statistics). Therefore the noise in this background is sqrt(n). Since we just derived that the signal strength of F(g) will be 2*sqrt(N), we therefore conclude that the signal to noise ratio of the bright field image will be 2*sqrt(N)/sqrt(N) = 2. Summary What I just derived is a more detailed version of the argument that can be compiled by combining the statements of Ref's 1-3. The main difference between my derivation and the conclusion of Ref. 1 is that I find the signal to noise ratio of the diffraction experiment, for the same toy example as in Ref. [2] having one electron contributing to F(g), to be 2. This SNR of 2 is the same as what I find for the bright field version of the toy example. In contrast, Ref's 1-3 appear to work from the assumption that the diffraction signal to noise ratio would be just 1 in this toy example. Note that the argument of Ref's 1-3 is presented rather sketchily; the visual argument given in Fig. 1 of Ref. [2] presents the SNR of the bright field experiment, for the toy problem of one electron per F(g), as being 1; only in the text is the factor of 2 correction mentioned (equation 1 in Ref. [2]) that would bring this SNR to 2. Furthermore, these authors' apparent assumption that the signal-to-noise ratio of the diffraction measurement should be 1 is never stated explicitly. In any case, the chain of logic as presented here would give identical SNR's for amplitude measurements, for either the bright field or the diffraction experiment; thus, since Friedel pairs are observed in typical diffraction experiments, the

6 amplitude can be estimated more accurately than in the bright field experiment. REFERENCES 1. Henderson R (1995) The potential and limitations of neutrons, electrons and X- rays for atomic resolution microscopy of unstained biological molecules. Q Rev Biophys 28(2): HENDERSON, R. (1992). Image contrast in high resolution electron microscopy of biological macromolecules : TMC in ice. Ultramicroscopy 46, R. Henderson and R.M. Glaeser. Quantitative analysis of image contrast in electron micrographs of beam-sensitive crystals. Ultramicroscopy 16 (1985)