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Quantum Engi Q and A_Part_12

Quantum Engi Q and A_Part_12 - 34 CHAPTER 3 BASIC IDEAS OF...

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Unformatted text preview: 34 CHAPTER 3. BASIC IDEAS OF QUANTUM MECHANICS The other eigenfunctions do not necessarily have their maximum magnitude at the origin: for example, the shown states ψ100 and ψ010 are zero at the origin. For large negative values of its argument, an exponential becomes very small very quickly. So if the distance from the origin is large compared to ℓ, the wave function will be negligible, and it will be zero in the limit of infinite distance. For example, if the distance from the origin is just 10 times ℓ, the exponential above is already as small as 0.000,000,000,002 which is clearly negligible. As far as the value of the other eigenfunctions at large distance from the origin is concerned, note from table 3.1 that all eigenfunctions take the generic form ψnx ny nz = polynomial in x polynomial in y polynomial in z . ex2 /2ℓ2 ey2 /2ℓ2 ez2 /2ℓ2 For the distance from the origin to become large, at least one of x, y , or z must become large, and then the blow up of the corresponding exponential in the bottom makes the eigenfunctions become zero. (Whatever the polynomials in the top do is irrelevant, since an exponential includes, according to its Taylor series, always powers higher than can be found in any given polynomial, hence is much larger than any given polynomial at large values of its argument.) It may be noted that the eigenfunctions do extend farther from the nominal position when the energy increases. The polynomials get nastier when the energy increases, but far enough away they must eventually always lose from the exponentials. Solution harmd-c Question: Write down the explicit expression for the eigenstate ψ213 using table 3.1, then verify that it looks like figure 3.4 when looking along the z -axis, with the x-axis horizontal and the y -axis vertical. Answer: The generic expression for the eigenfunctions is ψnx ny nz = hnx (x)hny (y )hnz (z ) and substituting nx = 2, ny = 1 and nz = 3, you get ψ213 = h2 (x)h1 (y )h3 (z ). Now substitute for those functions from table 3.1: ψ213 [2(x/ℓ)2 − 1][2y/ℓ][2(z/ℓ)3 − 3(z/ℓ)] −x2 /2ℓ2 −y2 /2ℓ2 −z2 /2ℓ2 = e e e √ 2 3 (πℓ2 )3/4 where the constant ℓ is as given in table 3.1. 3.6. THE HARMONIC OSCILLATOR 35 √ The first polynomial within square brackets in the expression above is zero at x = ℓ/ 2 √ and x = −ℓ/ 2, producing the two vertical white lines along which there is zero probability of finding the particle. Similarly, the second polynomial within square brackets is zero at y = 0, producing the horizontal white line. Hence looking along the z -direction, you see the distribution: yT √ −ℓ/ 2 √ ℓ/ 2 E x Figure 3.4: Energy eigenfunction ψ213 . Seen from above, you would see four rows of three patches, as the third polynomial between brackets produces zero probability of finding the particle at z = − 3/2ℓ, z = 0, and z = 3/2ℓ, splitting the distribution into four in the z -direction. This example illustrates that there is one more set of patches in a given direction each time the corresponding quantum number increases by one unit. 3.6.5 Degeneracy Solution harme-a Question: Just to check that this book is not lying, (you cannot be too careful), write √ down the analytical expression for ψ100 and ψ010 using table 3.1, then (ψ100 + ψ010 ) / 2 and √ (ψ010 − ψ100 ) / 2. Verify that the latter two are the functions ψ100 and ψ010 in a coordinate 36 CHAPTER 3. BASIC IDEAS OF QUANTUM MECHANICS system (¯, y , z ) that is rotated 45 degrees counter-clockwise around the z -axis compared to x¯ the original (x, y, z ) coordinate system. Answer: Take the rotated coordinates to be x and y as shown: ¯ ¯ yT s d d yd ¯ d d d d d d d  ¯ x E x d d d d d A vector displacement of magnitude x in the x-direction has a component along the x-axis of ¯ √ ◦ magnitude x cos 45 , equivalent to x/ 2. Similarly, a vector displacement of magnitude y in √ the y -direction has a component along the x-axis of magnitude y cos 45◦ , equivalent to y/ 2. ¯ So in general, for any point (x, y ), x+y x= √ . ¯ 2 Similarly you get y−x y= √ . ¯ 2 Turning now to the eigenfunctions, taking the generic expression ψnx ny nz = hnx (x)hny (y )hnz (z ) and substituting nx = 1, ny = nz = 0, you get ψ100 = h1 (x)h0 (y )h0 (z ). Now substitute for h0 and h1 from table 3.1: √ 2x/ℓ −x2 /2ℓ2 −y2 /2ℓ2 −z2 /2ℓ2 e e e ψ100 = (πℓ2 )3/4 where the constant ℓ is as given in table 3.1. You can multiply out the exponentials: √ 2x/ℓ −(x2 +y2 +z2 )/2ℓ2 e . ψ100 = (πℓ2 )3/4 ...
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