Physics 491: Quantum Mechanics 1 Problem Set #3: Solutions, Lecture notes of Quantum Mechanics

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J 3k ae C6ud.) | 2018 Fall Physics 491: Problem Set 3 Problem 2 Contents 1 More examples in momentum space 1 1.1 Normalization constant . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1 1.2 Momentum space wave function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2 1.3 Position and momentum uncertainties . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3 1.4 Superposition of two Gaussians . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4 1.4.1 Normalization . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4 1.4.2 Momentum space wave function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5 1.4.3 Position and Momentum Uncertainties . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7 1 More examples in momentum space We consider the wave function (x) = Ax exp ⇣ x 2 4 2 ⌘ . It is useful to recall the form of a normalized Gaussian probability distribution and its properties. The distri- bution with mean x 0 and variance 2 is as follows. P Gauss = 1p 2⇡2 exp ✓ (x x 0 )2 22 ◆ (1) The moments of this distribution are as follows. The zeroth moment gives the normalization. Z +1 1 dxx0P Gauss = Z +1 1 dxP Gauss = 1 (2) The first moment gives the mean. Z +1 1 dxx1P Gauss = Z +1 1 dxxP Gauss = x 0 (3) The second moment gives the variance. Z +1 1 dxx2P Gauss = Z +1 1 dxx2P Gauss = 2 + x 2 0 (4) 1.1 Normalization constant We find the normalization by requiring that the integral of the square of the absolute value of the wave function integrated over all space is 1. Z +1 1 dx| (x)|2 = 1 =) |A|2 Z +1 1 dx x exp ✓ x 2 42 ◆ 2 = 1 (5) For our wave function we have the following. Z +1 1 dx| (x)|2 = |A|2 Z +1 1 dx x exp ⇣ x 2 4 2 ⌘ 2 = |A|2 p 2⇡2 1p 2⇡2 Z +1 1 dxx2 exp ⇣ x 2 4 2 ⌘ = |A|2 p 2⇡2 2 (6) 1 1.4 Superposition of two Gaussians Now we consider the wave function (x) = A exp x+ a 2 2 42 ! + exp x+ a 2 2 42 !! (27) This can be written an equal superposition of two normalized Gaussian wavefunctions. (x) = B( + (x) + (x)) (28) Here ± refer to normalized Gaussian wavefunctions centered at ±a 2 . ±(x) = 1 (2⇡2) 1/4 exp ✓ (x⌥ a 2 )2 42 ◆ (29) 1.4.1 Normalization Normalization leads to the following condition. Z +1 1 dx| (x)|2 = 1 =) |B|2 Z +1 1 dx| + (x) + (x)|2 = 1 =) |B|2 Z +1 1 dx| + (x)|2 + |B|2 Z +1 1 dx| (x)|2 + 2|B|2 Z +1 1 dxRe ( + (x) (x)) = 1 (30) The integral with each normalized Gaussian wave functions is 1. Therefore the normalization condition is as follows. 2|B|2 ✓ 1 + Z +1 1 dx + (x) (x) ◆ = 1 (31) The cross term is as follows. + (x) (x) = 1 (2⇡2) 1/4 exp ✓ (x a 2 )2 42 ◆ 1 (2⇡2) 1/4 exp ✓ (x+ a 2 )2 42 ◆ = 1 (2⇡2) 1/2 exp (x2 a 2 4 ) 22 ! (32) The integral of the cross term can be evaluated using the integral of a normalized Gaussian. Z +1 1 dx + (x) (x)) = Z +1 1 dx 1 (2⇡2) 1/2 exp (x2 a 2 4 ) 22 ! = exp ✓ a 2 82 ◆ 1 (2⇡2) 1/2 Z +1 1 dx exp ✓ x 2 22 ◆ = exp ✓ a 2 82 ◆ (33) Therefore, we can find |B| 2|B|2 ✓ 1 + exp ✓ a 2 82 ◆◆ = 1 =) |B| = 1r 2 ⇣ 1 + exp ⇣ a 2 8 2 ⌘⌘ (34) We choose the overall phase to make the wavefunction real. Therefore the normalized wavefunction is the following. (x) = 1r 2 ⇣ 1 + exp ⇣ a 2 8 2 ⌘⌘ 1 (2⇡2) 1/4 exp ✓ (x a 2 )2 42 ◆ + 1 (2⇡2) 1/4 exp ✓ (x+ a 2 )2 42 ◆! (35) Therefore the normalization constant A is the following. A = 1r 2 ⇣ 1 + exp ⇣ a 2 8 2 ⌘⌘ 1 (2⇡2) 1/4 (36) The wave functions are plotted below. When a is large compared , the two Gaussian are distinguishable. 4 1.4.2 Momentum space wave function The position space wavefunction is an equal superposition of + (x) and (x). We can use the linearity of the Fourier transform to write the momentum space wave function as equal superposition of the Fourier transforms ̃ + (p) and ̃(p) ̃(p) = B ⇣ ̃ + (p) + ̃(p) ⌘ (37) The functions + (x) and (x) are displaced versions of the Gaussian wavefunction centered at zero, Gauss (x). + (x) = Gauss x a 2 (x) = Gauss x+ a 2 (38) Therefore, we use the phase shift property of the Fourier transform to find the momentum space wave function. 5 ̃(p) = B ⇣ ̃ + (p) + ̃(p) ⌘ = B ⇣ ̃ Gauss (p) exp ⇣ +i pa 2~ ⌘ + ̃ Gauss (p) exp ⇣ i pa 2~ ⌘⌘ = 2B̃ Gauss (p) cos ⇣ pa 2~ ⌘ (39) Recall that for a Gaussian wavefunction with variance in position and mean position x = 0, the momentum space wavefunction is as follows. ̃ Gauss (p) = ✓ 22 ⇡~2 ◆ 1/4 exp ✓ p 2 2 ~2 ◆ (40) Using this and the expression for B we found earlier, we have the following. ̃(p) = 2 1r 2 ⇣ 1 + exp ⇣ a 2 8 2 ⌘⌘ ✓ 22 ⇡~2 ◆ 1/4 exp ✓ p 2 2 ~2 ◆ cos ⇣ pa 2~ ⌘ =) ̃(p) = 1r 1 + exp ⇣ a 2 8 2 ⌘ ✓ 82 ⇡~2 ◆ 1/4 exp ✓ p 2 2 ~2 ◆ cos ⇣ pa 2~ ⌘ (41) These are plotted below. When a is large compared , the phase di↵erence between the two Gaussians leads to interference fringes. 6