Heisenberg's inequality for Fourier transform

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1 Heisenberg's inequality for Fourier transform Riccardo Pascuzzo Abstract In this paper, we prove the Heisenberg's inequality using the Fourier transform. Then we show that the equality holds for the Gaussian and the strict inequality holds for the function e t. Contents 1 Fourier transform 1 Heisenberg's inequality 3 3 Examples Gaussian function Exponential function A Appendix 6 A.1 Gaussian function A. Exponential function Fourier transform Denition 1. Let f L (R) and t R. dened by The Fourier transform of f(t) is F f(t); ω f(ω) A and the inverse Fourier transform by where AB 1 π. f(t) B e ±iωt f(t) dt ω R (1) e iωt f(ω) dω t R () Remark 1.1. Since there are dierent denitions of Fourier transform, in order to include most of them, in (1) and () we have used the symbol ± and the generic constants A and B, that can be chosen in three ways: 1. A 1 and B 1 π 1

2 1 FOURIER TRANSFORM. A B 1 π 3. A 1 π and B 1. As we will point out in the sequel, each choice of A and B is suitably adopted in order to simplify some formulas. We recall some properties of the Fourier transform that will be useful to prove the Heisenberg's inequality. Proposition 1.1. If f L (R) then f L (R). Theorem 1.1 (Convolution). Let f, g L (R). We dene the convolution of f and g as Then 1. (f g)(t) L (R) (f g)(t). F (f g)(t); ω 1 A f(ω)ĝ(ω). f(τ)g(t τ)dτ t R Remark 1.. In this case we should choose A 1, so the above equation become F (f g)(t); ω f(ω)ĝ(ω). The previous theorem is necessary to prove the next fundamental equality: Theorem 1. (Parseval). Let f L (R) and t R. Then A called Parseval's formula. f(t) dt B f(ω) dω (3) Remark 1.3. In this case we should choose A B 1 π, so that (3) become f(t) dt f(ω) dω. The above equation can be read as a preservation of the L norm: recalling that f L f f(t) dt we obtain that f f. Remark 1.4. As a consequence of the Parseval's formula we nd that ( A + ) f(t) dt A ( B + ) B B A f(ω) dω B A ( f(ω) dω) (4) Theorem 1.3 (Time dierentiation). Let f L (R) and t R. Then n N d n f F dt n ; ω ( iω) n f(ω) (5)

3 HEISENBERG'S INEQUALITY 3 Heisenberg's inequality The uncertainty principle is partly a description of a characteristic feature of quantum mechanical system and partly a statement about the limitations of one's ability to perform measurements on a system without disturbing it. When translated into the language of quantum mechanics, it says that the values of a pair of canonically conjugate observables such as position and momentum cannot both be precisely determined in any quantum state. On the mathematical side, when one asks for a precise quantitative formulation of the uncertainty principle, the most common response is the following inequality, usually called Heisenberg's inequality: Theorem.1 (Heisenberg's inequality). If f, tf(t) and ω f(ω) L (R) then ( ) ( ) t f(t) dt ω f(ω) dω A ( f(t) dt) (6) 4B B 4A ( f(ω) dω) (7) Proof. First we observe that, since f, tf(t) and ω f(ω) L (R), then (6) is well dened. Because f L (R), from Proposition 1.1 also f L (R) and this means that (7) is well dened too. From (4) it follows immediately (7). Let's now prove (6). Since (f )(ω) ω f(ω) from (5), the niteness of ω f implies that f is absolutely continuous and f L (R). This allows us to dene I(λ) λtf(t) + f (t) dt λ R. (8) In fact, because tf(t) and f (t) L (R), also tf(t) + f (t) L (R) and this means that I(λ) <, so (8) is well dened. Since f(t) f(t)f(t) and λtf(t) + f (t) λtf(t) + f (t) λt f(t) + f (t) λ t f(t) + λt f(t)f (t) + f (t)f(t) + f (t) then we obtain I(λ) λ t f(t) dt + λ and integrating by parts I(λ) λ t f(t) t dt + lim λ a f(t) a + t f(t)f (t) + f (t)f(t) dt a λ + f(t) dt + f (t) dt f (t) dt.

4 3 EXAMPLES 4 Since I(λ) < and all the integrals in the above expression are convergent, then the second addend must be zero, for otherwise f(t) would be comparable to t 1 for large t and f would not be in L (R). Therefore I(λ) λ t f(t) dt λ f(t) dt + f (t) dt. Since f L (R), we can apply (3) to f (t) and, using (5) with n 1, we nd that so we obtain f (t) dt B A B A I(λ) λ t f(t) dt λ (f )(ω) dt B A ω f(ω) dω, f(t) dt + B A iω f(ω) dt ω f(ω) dω that is a quadratic equation in λ. Since I(λ) for any value of λ, its discriminant must be nonpositive: ( ) t f(t) dt 4 B ( ) ( ) t f(t) dt ω f(ω) dω A and nally this leads to (6). Remark.1. We can rewrite the Heisenberg's inequality in terms of the L norm: 3 Examples tf(t) ω f(ω) A 4B f 4 B 4A f 4. In this section we'll verify (6) and (7) for two special function: the Gaussian and e t. 3.1 Gaussian function Let f(t) e α t the Gaussian function; one can prove 1 that f L (R) and f(ω) A. 1 See Appendix A.1. α e ω

5 3 EXAMPLES 5 One can prove that the integrals that appear in (6) are in this case I 1 I I 3 I 4 Moreover, using B 1 πa, we obtain f(t) dt α dω A π 3 t f(t) dt α 3 (α) ω f(ω) dω A π 3 α therefore I 3 I 4 A 4B (I 1) A π 4B α A π π 3 (α) A π 3 α A π A 4B (I 1) so we have proved that (6) is an equality for the Gaussian. Furthermore, since B 4A (I ) B 4A A4 π 3 α A π A 4B (I 1) we have shown that also (7) holds for the Gaussian. 3. Exponential function Let f(t) e t an exponential function; one can prove 3 that f L (R) and f(ω) A 1 + ω. The integrals that appear in (6) are in this case 4 See Appendix A.1. 3 See Appendix A.. 4 See Appendix A.. I 1 I I 3 I 4 f(t) dt 1 dω πa t f(t) dt 1 ω f(ω) dω πa.

6 A APPENDIX 6 Then we nd that I 3 I 4 1 πa πa (9) A 4B (I 1) A 4B πa πa 4 B 4A (I ) B 4A (πa ) π A 3 B π A 3 πa πa (1) (11) so we have proved that for the function e t the left hand side in (6) is exactly twice the right hand side and this means that in this case (6) is a strict inequality. Finally, because we have shown then (7) holds for the exponential. A 4B (I 1) πa B 4A (I ) A Appendix In this appendix we give full proof of the results presented in the two previous examples. A.1 Gaussian function Let f(t) e α t the Gaussian function, with α >. Lemma A.1. The Gauss integral is e α t dt α (1) Proposition A.1. The following properties are valid: f L (R) and f(ω) A. (13) α e ω Proof. First we have to show that f(t) dt <, i.e. f L (R): f(t) dt α <, e α t dt where in the last equality we have applied (1). e α t dt

7 A APPENDIX 7 Since f L (R), we can verify (13). Let ω R, then f(ω) A A A Ae ω A α e ω e ±iωt e α t dt A e α t iω iω α t+( α) +α( α) iω dt iω α(t e α) ω dt e αt dt e α(t iω α t) dt Proposition A.. I 1 I I 3 I 4 f(t) dt α dω A π 3 t f(t) dt α (14) (15) (α) 3 (16) ω f(ω) dω A π 3 α (17) Proof. We have already proved (14) in Proposition A.1. Let's show (15): I dω π A α e ω dω A π α A α π πα A π 3 α. e ω α dω

8 A APPENDIX 8 Let's show (16): I 3 t f(t) dt t e αt dt 1 Finally we prove (17): ( t te αt) dt 1 te αt + ( 1 ) ( ) π α 3 (α). I 4 ω f(ω) dω π ω A A π α A π α (α) 3 A π 3 α. α e ω ω e ω α dω e αt dt dω A. Exponential function Let f(t) e t an exponential function. Proposition A.3. The function f L (R) and for ω R f(ω) A 1 + ω (18) Proof. First we have to show that f(t) dt <, i.e. f L (R): f(t) dt e t dt e t dt <. e t e t dt +

9 A APPENDIX 9 Since f L (R), we can verify (18). Let ω R, then f(ω) A A A e ±iωt e t dt e ±iωt+t dt + A e t(±iω+1) dt + A A e t(±iω+1) ±iω A 1 A ±iω + 1 A 1 A ±iω 1 ±iω 1 (±iω + 1) ω 1 A ω 1 A 1 + ω e ±iωt t dt e t( iω+1) ±iω 1 e t( iω+1) dt + Proposition A.4. I 1 I I 3 I 4 f(t) dt 1 (19) dω πa () t f(t) dt 1 Proof. We have already proved (19) in Proposition A.3. Let's show rst (1). I 3 t f(t) dt + te t e t 1 (1) ω f(ω) dω πa. () t e t dt t e t dt t e t + + e t dt + te t dt We prove () using a consequence of the residue theorem.

10 A APPENDIX 1 Proposition A.5. Assume f is an analytic function on C with a nite number of poles. Considering only the poles z 1, z,..., z n that are in the upper half plane, then n f(x)dx πi Resf; z j (3) where Resf; z j is the residue of f in z j. Recall that, if z j is a pole of order m for the function f, then 1 Resf; z j lim x z j (m 1)! j1 d m 1 dx m 1 (x z j) m f(x) (4) Dening g(ω) 4A (1 + ω ) 4A (ω + i) (ω i), we observe that there are two poles z 1 i and z i of order. For Theorem A.5 we can consider only z 1 C +. Let's compute Resg; i, using (4): d Resg; i lim ω i dω (ω 4A i) (ω + i) (ω i) 4A lim ω i d dω 1 (ω + i) 4A lim ω i (ω + i) 3 4A (i) 3 A i 3 ia so we have proved that Resg; i ia. Using this result and (3) we obtain I dω A 1 + ω dω 4A (1 + ω ) dω πi Resg; i πa i πa Finally we prove () imitating the proof of (). Dening h(ω) ω f(ω) 4A ω (ω + i) (ω i), we observe that there are two poles z 1 i and z i of order. For Theorem

11 REFERENCES 11 A.5 we can consider only z 1 C +. Let's compute Resh; i: Resh; i lim ω i 4A lim ω i d (ω i) 4A ω dω d dω (ω + i) (ω i) ω (ω + i) 4A lim ω i ω(ω + i) ω (ω + i) (ω + i) 4 4A ω(ω + i) ω + i ω lim ω i (ω + i) 4 4A iω lim ω i (ω + i) 3 4A i (i) 3 4iA ia 4 so we have proved that Resh; i ia. Using this result and (3) we obtain I 4 and this conclude our proof. dω ω A 1 + ω dω ω 4A (1 + ω ) dω πi Resh; i πa i πa References 1 Folland Gerald B. and Sitaram Alladi, The Uncertainty Principle: A Mathematical Survey, The Journal of Fourier Analysis and Application, Volume 3, Number 3 (1997), Ciaurri O. and Varona J.L., An Uncertainty Inequality for Fourier-Dunkl Series, J. Comput. Appl. Math. 33 (1), Papoulis A., The Fourier Integral and its Applications, McGraw-Hill Companies (196). 4 Course notes of mathematical physics by Professor Francesco Mainardi.