This preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: 20 0perat0rs 20—1 Operations and operators All the things we have done so far in quantum mechanics could be handled
w1th ordinary algebra, although we dld from time to time show you some special
ways of writing quantummechanical quantities and equations. We would like
now to talk some more about some interesting and useful mathematical ways of
describing quantummechanical things. There are many ways of approaching the
subject of quantum mechanics, and most books use a different approach from the
one we have taken. As you go on to read other books you might not see right
away the connections of what you will ﬁnd in them to what we have been doing.
Although we Will also be able to get a few useful results, the main purpose of this
chapter is to tell you about some of the different ways of writing the same physics.
Knowmg them you should be able to understand better what other people are
saying. When people were ﬁrst working out classical mechanics they always wrote
all the equations in terms of x, y, and zcomponents. Then someone came along
and pointed out that all of the writing could be made much simpler by inventing
the vector notation. It’s true that when you come down to ﬁguring something
out you often have to convert the vectors back to their components. But it’s
generally much easier to see what’s going on when you work with vectors and also
easier to do many of the calculations. In quantum mechanics we were able to
write many things in a simpler way by using the idea of the “state vector.” The
state vector Ii/z) has, of course, nothing to do with geometric vectors in three
dimensions but is an abstract symbol that stands for a physical state, identiﬁed
by the “label,” or “name,” 1/2. The idea is useful because the laws of quantum
mechanics can be written as algebraic equations in terms of these symbols. For
instance, Our fundamental law that any state can be made up from a linear com
bination of base states is written as Iw=2om, mo where the C, are a set of ordinary (complex) numbers—the amplitudes CI = (i I up)
—while I l), I 2), I 3), and so on, stand for the base states in some base, or repre
sentation. If you take some physical state and do something to it—like rotating it, or
like waiting for the time At—you get a different state. We say, “performing
an operation on a state produces a new state." We can express the same idea by
an equation: I ¢> = A I it). (20.2) An operation on a state produces another state. The operator xi stands for some
particular operation. When this operation is performed on any state, say It), it
produces some other state I qs). What does Eq. (20.2) mean? We deﬁne it this way. If you multiply the
equation by (i I and expand W) according to Eq. (20.1), you get mw=2mimmw. we (The states I j) are from the same set as I 1).) This is now just an algebraic equation.
The numbers (1’ I ¢) give the amount of each base state you will ﬁnd in I¢), and
it is given in terms of a linear superposition of the amplitudes (j I 1/1) that you ﬁnd 20—1 20—1 Operations and operators
20—2 Average energies 20—3 The average energy of an
atom 20—4 The position operator
20—5 The momentum operator
20—6 Angular momentum 20—7 The change of averages with time  ill) in each base state. The numbers (i l AA [ j) are just the coeﬂicients which tell how much of (j 1 III) goes into each sum. The operator 151‘ is described numerically
by the set of numbers, or “matrix,” An E (i l A” U). (20.4) So Eq. (20.2) is a highclass way of writing Eq. (20.3). Actually it is a little
more than that; something more is implied. In Eq. (20.2) we do not make any
reference to a set of base states. Equation (20.3) is an image of Eq. (20.2) in
terms of some set of base states. But, as you know, you may use any set you wish.
And this idea is implied in Eq. (20.2). The operator way of writing avoids making
any particular choice. Of course, when you want to get deﬁnite you have to choose
some set. When you make your choice, you use Eq. (20.3). So the operator
equation (20.2) is a more abstract way of writing the algebraic equation (20.3).
It’s similar to the diﬂ'erence between writing c = a X b
instead of
CI = aubz — azby,
c1, = asz — 01b1,, C; = aIbJ, — aybz. The ﬁrst way is much handier. When you want results, however, you will eventually
have to give the components with respect to some set of axes. Similarly, if you
want to be able to say what you really mean by A”, you will have to be ready to
give the matrix AU in terms of some set of base states. So long as you have in
mind some set A“, Eq. (20.2) means just the same as Eq. (20.3). (You should
remember also that once you know a matrix for one particular set of base states
you can always calculate the corresponding matrix that goes with any other base.
You can transform the matrix from one “representation” to another.) The operator equation in (20.2) also allows a new way of thinking. If we
imagine some operator AA, we can use it With any state I (b) to create a new state
.«l  50). Sometimes a “state” we get this way may be very peculiar—it may not
represent any physical situation we are likely to encounter in nature. (For instance,
we may get a state that is not normalized to represent one electron.) In other
words, we may at times get f‘states” that are mathematically artiﬁcial. Such
artiﬁcial “states" may still be useful, perhaps as the midpoint of some calculation. We have already shown you many examples of quantummechanical op
erators. We have had the rotation operator Ry(6) which takes a state ip) and
produces a new state, which is the old state as seen in a rotated coordinate system.
We have had the parity (or inversion) operator P, which makes a new state by
reversing all coordinates. We have had the operators 6",, ﬁg, and 6', for spin one
half particles. The operator J; was deﬁned in Chapter 17 in terms of the rotation operator
for a small angle 6. R.(e) = 1 + i; e L. (20.5) This just means, of course, that
Rz(e) I l) = M + g e J‘. i it). (20.6) In this example, jz ] zp) is li/ie times the state you get ifyou rotate I (L) by the small
angle 6 and then subtract the original state. It represents a “state” which is the
diﬁ'erence of two states. One more example. We had an operator Iii—called the momentum operator
(xcomponent) deﬁned in an equation like (20.6). If DAL) is the operator which 20—2 displaces a state along x by the distance L, then [3, is deﬁned by 1m) = 1 + g, 6a, (20.7) where 6 is a small displacement. Displacmg the state I 1%) along x by a small dis
tance 6 gives a new state l W). We are saying that this new state is the old state
plus a small new piece ,1 air). The operators we are talking about work on a state vector like  11/), which is
an abstract description of a phySical situation. They are quite different from
algebraic operators which work on mathematical functions. For instance, d/dx
is an “operator” that works on f(x) by changing it to a new function f’(x) =
df/dx. Another example is the algebraic operator V2. You can see why the same
word is used in both cases, but you should keep in mind that the two kinds of
operators are different. A quantummechanical operator xi does not work on an
algebraic function, but on a state vector like lW Both kinds of operators are
used in quantum mechanics and often in Similar kinds of equations, as you Will
see a little later. When you are ﬁrst learning the subject it is well to keep the
distinction always in mind. Later on, when you are more familiar With the subject,
you will ﬁnd that it is less important to keep any sharp distinction between the
two kinds of operators. You will, indeed, ﬁnd that most books generally use the
same notation for both! We’ll go on now and look at some useful things you can do with operators.
But ﬁrst, one special remark. Suppose we have an operator xi whose matrix in
some base is A” E (i  A“ lj). The amplitude that the state xi  1/!) 1s also in some
other state [ ¢>) is (¢>  xi l tp). Is there some meaning to the complex conjugate of
this amplitude? You should be able to show that <¢l1l¢>* = <¢ l 2* I «a. (20.8) where AA (read “A dagger”) is an operator whose matrix elements are Al. = (A..)*. (20.9)
To get the i, j element of A’r you go to the j, 1‘ element of a? (the indexes are reversed)
and take its complex conjugate. The amplitude that the state A“r  ¢) is in l i/x) is
the complex conjugate of the amplitude that xi [1/1) is in l 4)). The operator all is
called the “Hermitian adjoint" of xi. Many important operators of quantum
mechanics have the special property that when you take the Hermitian adjoint,
you get the same operator back. If B is such an operator, then at: a, and it is called a “selfadjoint” or “Hermitian," operator. 20—2 Average energies So far we have reminded you mainly of what you already know. Now we
would like to discuss a new question. How would you ﬁnd the average energy of
a system—say, an atom? If an atom is in a particular state of deﬁnite energy and
you measure the energy, you will ﬁnd a certain energy E. If you keep repeating
the measurement on each one of a whole series of atoms which are all selected to
be in the same state, all the measurements will give E, and the “average” of your
measurements will, of course, be just E. Now, however, what happens if you make the measurement on some state
N) which is not a stationary state? Since the system does not have a deﬁnite
energy, one measurement would give one energy, the same measurement on another
atom in the same state would give a different energy, and so on. What would you
get for the average of a whole series of energy measurements? 2073 We can answer the question by proyecting the state I it) onto the set of states
of deﬁnite energy. To remind you that this is a special base set, we’ll call the states
I 1),). Each of the states I 17,) has a deﬁnite energy E,. In this representation, W) = E Q] m). (2010) When you make an energy measurement and get some number 15,, you have found
that the system was in the state m But you may get a different number for each
measurement Sometimes you will get E1, sometimes E2, sometimes E3, and so
on. The probability that you observe the energy E1 is Just the probability of ﬁnding
the system in the state  771), which is, of course, just the absolute square of the
amplitude C1 = (771  tb). The probability of ﬁnding each of the possible energies
E, is P, = IC, [2. (20.11) How are these probabilities related to the mean value of a whole sequence
of energy measurements? Let’s imagine that we get a series of measurements like
this: E1, E7, E11, E9, E1, E10, E7, E2, E3, E9, E5, E4, and so on. We contlnue
for, say, a thousand measurements. When we are ﬁnished we add all the energies
and divide by one thousand. That’s what we mean by the average. There’s also
a shortcut to adding all the numbers. You can count up how many times you get
E1, say that is N1, and then count up the number of times you get E2, call that
N 2, and so on. The sum of all the energies is certainly just NlEl + N2E2 + N353 +   = 2 ME. The average energy is this sum divided by the total number of measurements which
iSJuSt the sum of all the N,’s, which we can call N; 21. NIE?
E,V ~ ——N—— (20.12)
We are almost there. What we mean by the probability of something happen
ing 18 just the number of times we expect it to happen divided by the total number
of tries. The ratio N ,/ N should—for large N—be very near to P,, the probability
of ﬁnding the state  11,), although it Will not be exactly P, because of the statistical
ﬂuctuations. Let’s write the predicted (or “expected”) average energy as (E),,,.; then we can say that
<E>av = Z PIE“ (20.13) The same arguments apply for any measurement. The average value of a measured
quantity A should be equal to <A)av = 2 P114“ where A, are the various possible values of the observed quantity, and P, is the
probability of getting that value.
Let’s go back to our quantummechanical state  1/1). lt’s average energy is (E),,v = Z [c.le, = Z CTC,E,. (20.14)
Now watch this trickery! First, we write the sum as 1/ Next we treat the lefthand 0p l as a common “factor.“ We can take this factor
out of the sum, and write it as 0 (2 l 170120.104. 1 204 This expression has the form (It I ¢>,
where I ¢> is some “cookedup” state deﬁned by
¢> = Z  m>Et<m I it). (20.16) It is, in other words, the state you get if you take each base state I 1],) in the amount E1071 I 1”) Now remember what we mean by the states I 17,). They are supposed to be
the stationary states—by which we mean that for each one, HIM) 2 Elm) Since E, is Just a number, the righthand side is the same as  n,)E,, and the sum
in Eq. (20.16) is the same as 2 RI 771><77zI ¢) Now 1' appears only in the famous combination that contracts to unity, so
2 HI meI‘J’) = HZ I m><ml¢> = HM. Magic! Equation (20.16) is the same as l¢> = H  1/) (20.17) The average energy of the state I IL) can be written very prettily as <5)... = (w I 1W). (20.18) To get the average energy you operate on lip) with H, and then multiply by (‘P I.
A simple result. Our new formula for the average energy is not only pretty. It is also useful,
because now we don’t need to say anything about any particular set of base
states. We don’t even have to know all of the possible energy levels. When we go
to calculate, we’ll need to describe our state in terms of some set of base states,
but if we know the Hamiltonian matrix 11,, for that set we can get the average
energy. Equation (19.18) says that for any set of base states Ii), the average
energy can be calculated from <E>nv = 2 (WI i><i I H I no I a. (2019) where the amplitudes (i I H I j) are just the elements of the matrix H”.
Let's check this result for the special case that the states I i) are the deﬁnite
energy states. For them, H Ij) = E, Ij), so (1 I H Ij) = E, 5,, and (15>... = Z (t I i>E.6.,<jI w) = 2 EM I WW), U
which is right. Equation (20.19) can, incidentally, be extended to other physical measure
ments which you can express as an operator. For instance, 1:, is the operator of
the zcomponent of the angular momentum L. The average of the zcomponent
for the state I 4/) is (Lz>av = (1” I132 I1» One way to prove it is to think of some situation in which the energy is proportional
to the angular momentum. Then all the arguments go through in the same way. 20—5 In summary, if a physical observable A is related to a suitable quantum
mechanical operator ff, the average value of A for the state I ¢> is given by (Am = («I l A It). (20.20)
By this we mean that
Aav = (w l ¢), (20.21)
with A
44>) = A ¢> (20.22) 203 The average energy of an atom Suppose we want the average energy of an atom in a state described by a
wave function Mr); How do we ﬁnd it? Let’s ﬁrst think of a onedimensional
situation with a state  (0) deﬁned by the amplitude (x 1 up) = ﬁx). We are asking
for the special case of Eq. (20.19) applied to the coordinate representation. Follow
ing our usual procedure, we replace the states  i) and [j) by l x) and I x’), and
change the sums to integrals. We get (5)“ = [/0  x)(x 1 H] x’)(x’ 1 (0) dx dx’. (2023)
This integral can, if we wish, be written in the following way:
[GP l X><x I d>> dx, (20.24)
with
(x l 4,) = f (x 1 m x’)(x’ l 1P) dx’. (20.25) The integral over x’ in (20.25) is the same one we had in Chapter l6~see Eq.
(16.50) and Eq. (l6.52)—and is equal to ’12 2
~ m 2d)? (00 + V0020). We can therefore write (x l ¢> = {— g7; 3‘1};— + V(x)]¢(x). (20.26) Remember that (1; l x) = (x  l//>* = ¢*(x); usmg this equality, the average
energy in Eq. (20.23) can be written as 2 2
<E>av = /¢*(X) {— 2h": 5; + V] 1/06) dx. (20.27) Given a wave function 11/(x), you can get the average energy by doing this integral.
You can begin to see how we can go back and forth from the statevector ideas
to the wavefunction ideas.
The quantity in the braces of Eq. (20.27) is an algebraic operator.I We will
write it as 50
h2 d2 3": 7539+" With this notation Eq. (20.23) becomes
<E>nv = f¢*(x)3‘cip(x)dx. (20.28)
The algebraic operator 3‘0 deﬁned here is, of course, not identical to the quantummechanical operator 17. The new operator works on a function of
position up(x) = (x I t/I) to give a new function of x, ¢(x) = (x 1 ¢); while H IThe “operator” V(x) means “multiply by V(x).”
20—6 operates on a state vector  ll!) to give another state vector i 4:), without implying
the coordinate representation or any particular representation at all. Nor is 50
strictly the same as H even in the coordinate representation. If we choose to
work in the coordinate representation, we would interpret H in terms of a matrix
(xl 17 l x’) which depends somehow on the two "indices" x and x’; that is, we
expect—according to Eq. (20.25)rthat (x  (to) is related to all the amplitudes
(x I to) by an integration. On the other hand, we ﬁnd that 5C is a differential op
erator. We have already worked out in Section 16—5 the connection between
(xl [all x’) and the algebraic operator CR3. We should make one qualiﬁcation on our results. We have been assuming
that the amplitude $(x) = (x lip) is normalized. By this we mean that the scale
has been chosen so that /I¢(X)l2dx=1; so the probability of ﬁnding the electron somewhere is unity. If you should choose
to work with a ﬁx) which is not normalized you should write (E)av = ———— (20.29) It‘s the same thing. Notice the Similarity in form between Eq. (20.28) and Eq. (20.18). These
two ways of writing the same result appear often when you work with the xrepre
sentation. You can go from the ﬁrst form to the second with any}? which is a
local operator, where a local operator is one which in the integral [<x I A  x'><x'  to) we can be written as 5% ll/(X), where (i is a diﬁerential algebraic operator. There are,
however, operators for which this is not true. For them you must work with the basic equations in (20.21) and (20.22).
You can easily extend the derivation to three dimensions. The result is thatI (E).w = fip(r)5c.p(r)dei, (20.30)
with
 212 9
3c = — ‘27: v + V(r), (20.31) and With the understanding that
[I ‘0 ldeol = 1. (20.32) The same equations can be extended to systems with several electrons in a fairly
obvious way, but we won’t bother to write down the results. With Eq. (20.30) we can calculate the average energy of an atomic state
even without knowing its energy levels. All we need is the wave function. It‘s
an important law. We’ll tell you about one interesting application. Suppose you
want to know the groundstate energy of some system~—say the helium atom, but
it’s too hard to solve Schrodinger’s equation for the wave function, because there
are too many variables. Suppose, however, that you take a guess at the wave
function—pick any function you like—~and calculate the average energy. That is,
you use Eq. (20.29)—generalized to three dimensions—to ﬁnd what the average
energy would be if the atom were really in the state described by this wave function.
This energy Will certainly be higher than the groundstate energy which is the lowest IWe write d Vol for the element of volume. It is. of course. Just dx dy dz. and the
integral goes from — ac to + w in all three coordinates. 20—7 P(X) X Fig. 20—1. A curve of probability
density representing a localized particle. possible energy the atom can have.1t Now pick another function and calculate its
average energy. If it is lower than your ﬁrst choice you are getting closer to the
true groundstate energy. If you keep on trying all sorts of artiﬁcial states you
will be able to get lower and lower energies, which come closer and closer to the
groundstate energy. If you are clever, you will try some functions which have a
few adjustable parameters. When you calculate the energy it will be expressed
in terms of these...
View
Full Document
 Spring '09
 dfdedf

Click to edit the document details