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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. 3.6.4.3 Solution harmdc 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 xaxis 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 3.6.5.1 Solution harmea 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 counterclockwise 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 xdirection has a component along the xaxis 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 xaxis 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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This note was uploaded on 11/13/2011 for the course PHY 4458 taught by Professor Garvin during the Fall '11 term at University of Florida.
 Fall '11
 GARVIN
 mechanics

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