Trouble understanding bra-ket notation
John J. Sasso Jr.
(or bracket) notation
<a| |a> <a|X|b>
I am not an expert at quantum mechanics (only know the short, basic stuff),
but I have had real analysis, advanced calc, etc. Could anyone explain
to me what this notation means, how it is used, and whatever else may
help me out? I have looked in several QM books on the notation, but I
still have trouble understanding it. Any help on this matter would be
greatly appreciated.
-- John
Allen Knutson
Yeah, it bothered me initially too.
Basically, the deal is that a state of a QM system is given by a 1-D subspace
of the Hilbert space associated with your system. (Okay, this is only a
pure state, I should include density matrices, and so on - JJS doesn't have
to know this yet.) Because subspaces aren't as convenient as vectors,
people pick vectors in them, and to partially eliminate the ambiguity
(which vector to choose) say "The vector should be a unit vector, and,
well, the phase exp(i\theta) doesn't matter either."
The kets |b> are these vectors. You can think of them as column vectors.
Associated to any space V is the dual space V^* = Map(V,C) of linear maps
from V to the complexes. These are the bras <a|. You can think of them as
row vectors. Of course, if you multiply a row vector by a column vector,
you get a 1x1 matrix, a number. Also, you can multiply a row vector on the
right by a matrix and get another row vector, or a column vector on the
left to get another column vector, which is why one sees expressions like
H|b> = |c>. Even better, you can multiply a column vector by a row
vector to get a matrix, which is why one sees expressions like O = |b><a|.
Of course, Hilbert spaces come equipped with an inner product, which is
exactly a way of associating a bra with each ket, i.e. a map V -> V^*.
(Careful again, this map isn't linear but conjugate-linear, or
"anti-linear", a term I have always hated.) This is why it is kosher
to speak of <a| and |a> at the same time, and think that they mean
something.
One last thing: a fine freedom this notation gives you is that of naming
your vectors whatever you want - if it's in between | and >, you know it's
got to be a vector. (I was really frightened by this at first, but one
gets used to it even quicker than Einstein Summation.) One standard trick
is to name an eigenvector with eigenvalue n "|n>", leading to the equation
H |n> = n |n>.
A good book, though a little demanding, and not really relevant to this
stuff directly, is Mackey's "The Mathematical Foundations of QM".
Allen K.
Jon Bell
To start off, think of it as an abstract shorthand for the various integrals
you encounter for expectation values etc.:
*
<a|a> <--> integral of psi psi , i.e. the normalization integral,
a a
which is usually 1, since we usually assume we have normalized wave functions;
<a|X|a> <--> similar to above, but with the position operator x
sandwiched in between the psi's, i.e. the expectation value of position;
<a|P|a> <--> similar to above, but with the momentum operator -i d
---- --
hbar dx
sandwiched in between, i.e. the expectation value of momentum. (WARNING!!
I may have the sign wrong... I'm notorious among my students for this.)
All this is in terms of the "position representation" that practically
everybody starts off studying QM in, that is, functions of x and t which
are solutions to Schroedinger's equation. As you may be aware, any QM state
can also be described in the "momentum representation" which uses the
momentum probability amplitude phi(p,t) instead of the position probability
amplitude psi(x,t). The integrals given above have momentum space analogs
which use the _same_ bra-ket representation:
*
<a|a> <--> integral of phi phi (notice that this integral is
a a
guaranteed to equal 1 if psi is normalized!)
<a|X|a> <--> similar to above, but with the position operator i d
---- --
hbar dp
sandwiched in between; this too is guaranteed to equal the corresponding
integral involving psi;
<a|P|a> <--> similar, but with the momentum operator p sandwiched in
between; this is guaranteed to equal the corresponding psi-integral.
So, to sum up this long-winded example, the bra-ket notation is a
"representation-independent" way of writing expectation values, etc.
For any given system, there are many different ways of representing states,
depending on what physical variables you deem important (or convenient),
but the relationships between the states do not depend on the representation,
and the bra-ket notation is supposed to emphasize this.
In this point of view, it's hard to think of a "bra" (<a|) or a "ket" (|a>)
as having an independent existence, since they would each represent "half
an integral". Nevertheless, it _is_ possible to think of them as
representing "abstract" quantum-mechanical states, which are completely
independent of any representation, and which live in a "Hilbert space"
which has well-defined mathematical properties. This is what that other guy
was going on about... :-)
Jon Bell / Dept. of Physics & C.S. / Presbyterian College / Clinton SC USA
Jim Carr
> (or bracket) notation
>
> <a| |a> <a|X|b>
>
> I am not an expert at quantum mechanics (only know the short, basic stuff),
> but I have had real analysis, advanced calc, etc. Could anyone explain
Ah, that explains your problem. The basic versions of quantum always
work in a particular representation, usually coordinate space, leaving
you totally ignorant of how it really works.
> to me what this notation means, how it is used, and whatever else may
> help me out? I have looked in several QM books on the notation, but I
> still have trouble understanding it. Any help on this matter would be
> greatly appreciated.
The entity |a> is a vector that lives in an abstract mathematical space
known as a Hilbert space. If you have not encountered this in your
mathematics, it is not big deal. It is just like the vector spaces you
know about ( there is a dual space with vectors <a|, normalized vectors
satisfy <a|a>=1, implying that there is some way of evaluating this
contraction defined although it may not be explicit yet, etc ).
In QM, the vector |a> describes some state of a physical system. It does
not imply any particular representation of that state of the system. If
you want to work in coordinate space, for example, you must project the
vector |a> onto that space. It looks like psi(r) = |r><r|a>. You should
view that step as projecting onto a new basis that has a convenient
representation. (This leads to a very clear way of thinking about the
procedure for fourier transforming to work in momentum space; it is
just a change of basis that is executed by doing an integral.) The
point is that you can carry out various manipulations with these vectors
|a> without worrying about what representation you are in or carrying
around lots of integrals and coordinates. You worry about those details
at the end, when it is time to *evaluate* the expression.
Similarly, the V you wrote above is an operator that acts on vectors in
the abstract space. It too can be expressed in various representations,
such as V(r,r') = |r><r|V|r'><r'|, but such details do not matter in
this abstract way of working. Rather, V acting on |a> gives some new
vector, call it |c>, which would be equal to v|a> (where v is just a
number) if |a> happens to be an eigenstate of V.
You evaluate a transition from state |a> to a state |b> mediated by an
operator V by evaluating the matrix element <b|V|a>. You evaluate it
in some representation, of course, but we do not worry about that here.
The reason we work this way is that the state of a system could be
rather awkward to write down, particularly when we are working with
many-body systems. Consider a nucleus with 8 protons and 8 neutrons,
each with a spin vector, and you begin to get the idea.
The standard way of introducing a newbie to this way of thinking is to
use the example of polarized light. This is done in most grad texts,
sometimes near the middle of the book after a review has been made of
all the coordinate space methods.
--
J. A. Carr | "The New Frontier of which I
j...@gw.scri.fsu.edu | speak is not a set of promises
Florida State University B-186 | -- it is a set of challenges."
Supercomputer Computations Research Institute | John F. Kennedy (15 July 60)
Jon Bell
>you want to work in coordinate space, for example, you must project the
>vector |a> onto that space. It looks like psi(r) = |r><r|a>. You should
>view that step as projecting onto a new basis that has a convenient
I thought that in order to get an actual number (such as the position
probability amplitude psi(r)) you had to "close the brackets." Isn't
(integral over all r of) |r><r|a> still an abstract state in Hilbert
space, being a sum of abstract states?
The way I make the connection with psi(r) is to start with (say) the
normalization condition and insert a projection operator:
<a|a> = 1
integral over all r of <a|r><r|a> = 1
which leads to the correspondence psi*(r) = <a|r> and psi(r) = <r|a>.
John C. Baez
Mathematically speaking, both bras and kets are vectors in a kind of
vector space called a Hilbert space. This is a vector space over the
complex numbers equippend with an inner product satisfying various
axioms. So <a| and b> are just vectors and <a|b> denotes their inner
product. The "operator" X is just a linear transformation of the
Hilbert space, and <a|X|b> is just a weird way of writing the inner
product of a and Xb.
Benjamin Weiner
> ... I have looked in several QM books on the notation, but I
> still have trouble understanding it. Any help on this matter would be
> greatly appreciated.
Marvin Chester's book "A Primer of Quantum Mechanics" is an introductory
(advanced undergrad) level book which uses the bra-ket notation; most
books at this level don't. Chester spends a lot of time explaining the
notation, and although his writing style can be irritating (I just now
realized that he must have sat at the metaphorical feet of J. A.
Wheeler), the book is very good for helping you get comfortable with
the notation. I think it helps the transition to graduate-level QM
texts immensely.
Allen Knutson
>Mathematically speaking, both bras and kets are vectors in a kind of
>vector space called a Hilbert space. This is a vector space over the
>complex numbers equippend with an inner product satisfying various
>axioms. So <a| and b> are just vectors and <a|b> denotes their inner
>product. The "operator" X is just a linear transformation of the
>Hilbert space, and <a|X|b> is just a weird way of writing the inner
>product of a and Xb.
John, I think thinking this way is a great hindrance to a person trying
to understand the practice of manipulating bras and kets. What's the
difference between <a| and |a>? Why do you see expressions like X|a>
but never |a>X? And so on. If you naively think of kets (bras) as
column (row) vectors, you can readily follow more complicated expressions,
like |a><a|. (See my earlier post for a more extensive discussion.)
Allen K.
Caveat: I think it's a hindrance because it certainly was for me for a
while.
Jim Carr
>In article <11...@sun13.scri.fsu.edu> j...@ds8.scri.fsu.edu (Jim Carr) writes:
>
>>you want to work in coordinate space, for example, you must project the
>>vector |a> onto that space. It looks like psi(r) = |r><r|a>. You should
>>view that step as projecting onto a new basis that has a convenient
>
>I thought that in order to get an actual number (such as the position
>probability amplitude psi(r)) you had to "close the brackets." Isn't
>(integral over all r of) |r><r|a> still an abstract state in Hilbert
>space, being a sum of abstract states?
Yeah, I screwed up. Been doing too much work lately with density
matrices were the thing *is* still an abstract state in Hilbert space...
I said project out, which means operate with <r|, but did not do it.
Same thing goes for what I said about V....
> ...
>which leads to the correspondence psi*(r) = <a|r> and psi(r) = <r|a>.
Yes. And this also makes clear that you get psi(k) = <k|a> from
psi(r) by inserting {integral over} |k><k| = 1 and noting that you
get a product of <r|k> (the exp{ik.r} term) and psi(k).
John C. Baez
Well, I certainly don't think we have anything other than a pedagogical
argument on our hands here; I was afraid that references to dual
vector spaces -- he only way to make a good mathematical
distinction between "bras" and "kets" -- would be too mathematically
sophisticated, and I didn't think of explaining it in terms of column
and row vectors. (Perhaps this is because until I learned about dual
vector spaces, I found the distinction between column and row vectors to
be ridiculous - as if a list of numbers gave a damn about which way you
wrote it. Only until I learned about natural vs unnatural isomorphisms
did it click.) Hopefully the original questioner will find our
different answers complementary rather than contradictory.
Jim Carr
>In article <1jd41c...@gap.caltech.edu> all...@ugcs.caltech.edu (Allen Knutson) writes:
>>jb...@riesz.mit.edu (John C. Baez) writes:
>>
>>>Hilbert space, and <a|X|b> is just a weird way of writing the inner
>>>product of a and Xb.
>>
>>John, I think thinking this way is a great hindrance to a person trying
>>to understand the practice of manipulating bras and kets. What's the
>>difference between <a| and |a>? Why do you see expressions like X|a>
>>but never |a>X? And so on. If you naively think of kets (bras) as
>>column (row) vectors, you can readily follow more complicated expressions,
>>like |a><a|. (See my earlier post for a more extensive discussion.)
>> Allen K.
I did not see this posted, so I will comment here. You do, of course,
see expressions like |a>X when the operator X is not supposed to act
on |a>. I made some remarks about the nature of operators, but implicit
in what many of us have said is that operators usually act in a particular
direction, and the default direction is to the right.
Now when you are working in a coordinate space representation, this is
only an important distinction when there are gradients in your operator,
since that is the only thing in that representation that has a direction,
but it can be important.
Also, the sequential nature of these steps may not seem important when
you think of <a|X|b> as shorthand for an integral, but sometimes (as in
nuclear shell model calculations where the kets might be in an m-scheme
representation where they are 100,000 element vectors) one actually has
an operator that one applies to the vector |b> and then dots <a| with
the result. There are lots of ways to skin this particular cat, and
the bra-ket notation of Dirac is used because of its power in glossing
over implementation-type details for the key concepts.
Sometimes |a> represents a real column vector and sometimes it represents
a function -- but it actually represents both.
> ... (Perhaps this is because until I learned about dual
>vector spaces, I found the distinction between column and row vectors to
>be ridiculous - as if a list of numbers gave a damn about which way you
>wrote it.
Me too. Sometimes the physicist in me wants to say that they really
are the same space, but the mathematical distinction that arises in
Abstract Algebra is an important one.
Andrew Stanford Klingler
OK, maybe I can make this even more confused. One's outlook varies with
background and interests. For some physical intuition about what the
quantities mean, I find it helpful to think of the inner product. When
you hear "The probability is the square of the amplitude" you're thinking
<a|a> = |a|^2 is the probability or density. In this case, the similarity
between |a> and <a| makes things easier to see. The row-column vector
image helps keep indices and complicated expressions clear, but when
I actually get around to writing down brackets, I'm usually thinking
of projecting onto a basis via an inner product. So the usage (if not
the stand-alone meaning) of |a><a| is perfectly clear. Incidentally, the
expression X|a> is written in that order just because operators act on
functions from the left. X can then be construed as an inf-dim linear
operator or matrix. I have seen |a>X but only with indices attached to
keep things clear. The expression <a|X|a> is a case of deliberate
notational symmetry; it can be construed as X acting on |a> normally
or its adjoint acting on <a|. I think the inner product "attitude" may
be more helpful in, e.g., solving first year QM problems.
ask@ucscb
Michael Weiss
Whitehead said somewhere that progress in mathematics depends not on
thinking about what you're doing, but in crafting notation that makes you
to do the right thing without thinking.
Dirac's bra-ket notation fills Whitehead's prescription to the letter.
Mathematicians seem to prefer a more thought-ful notation, with
correspondences spelled out. Dirac writes <v|T for the dual of T operating
on the dual of |v>; glance in a mathematics text and you are apt to find
T* L_v, or something similar. Again, Dirac writes simply <v|T|w> for
either side of the equation <v,Tw> = <T*v,w>, thereby saving a step in many
computations.
On the topic of mathematicians' vs. physicists' notation, does anyone know
why most mathematicians will write an integral as shown below on the left
(unless they omit the dummy variable x entirely), whereas physicists prefer
the form on the right?
/ /
| f(x) dx | dx f(x)
/ /
With the left-hand form, the integral sign and the dx act as delimiters.
With the right hand form, the combination of integral sign with dx stands
for a linear operator. Neither interpretation seems to favor one side of
the cultural divide.
Emory F. Bunn
>On the topic of mathematicians' vs. physicists' notation, does anyone know
>why most mathematicians will write an integral as shown below on the left
>(unless they omit the dummy variable x entirely), whereas physicists prefer
>the form on the right?
>
> / /
> | f(x) dx | dx f(x)
> / /
>
This is an interesting question. I sometimes write one and sometimes the
other. In general, I write \int f(x) dx if I'm just doing a single integral
(\int is an integral sign, for those who don't TeX), but if it's a multiple
integral, then I write
\int dx \int dy f(x,y)
I think that the reason I usually put the dx at the end is that that's
the way it was in my first calculus textbook, and the reason I put the
dx next to the integral sign in multiple integrals is that it makes it
instantly clear which integral sign (and hence which limits of integration)
go with which variable.
-Ted
Timothy Kimball
:
: I think that the reason I usually put the dx at the end is that that's
: the way it was in my first calculus textbook...
One of my first calculus teachers tried to get us to
drop the dx completely, on the grounds that the integral symbol
served as an operator on a function,
and the x in dx was just a dummy index of integration.
No one listened to him, though... :^)
--
/* tdk -- Opinions are mine, not my employer's. */
Jon Bell
>One of my first calculus teachers tried to get us to
>drop the dx completely, on the grounds that the integral symbol
>served as an operator on a function,
>and the x in dx was just a dummy index of integration.
>No one listened to him, though... :^)
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
Good! :-)
To me, the "dx", or something equivalent, is essential in setting up most
"physical" integrals. It indicates how you are "slicing" something into
"pieces" whose "thickness" goes to zero as you take the limit and go from
a sum to an integral.
I also think the dy/dx notation for derivatives is much more "physical"
than the mathematicians' f'(x).
In all fairness to the mathematicians, I can see where they're coming from.
We physicists are mainly interested in the physical quantities denoted
by the "y" and the "x" (or whatever variables we're using). The mathema-
ticians are interested mainly in the _relationships_ between the variables,
completely independently of what those variables represent.
Matt McIrvin
>On the topic of mathematicians' vs. physicists' notation, does anyone know
>why most mathematicians will write an integral as shown below on the left
>(unless they omit the dummy variable x entirely), whereas physicists prefer
>the form on the right?
> / /
> | f(x) dx | dx f(x)
> / /
My guess:
Some mathematicians like to view an integral as integrating a differential
form; instead of grouping the dx with the integral sign as an "integral
operator," they think of f(x) dx (they'd be more likely to
just write f dx, actually) as a single object, a "1-form", or "cotangent
vector" -valued function. You integrate this thing over a one-
dimensional manifold to get a number. Each point in the manifold has
a linear space associated with it called the cotangent space; the
basis of this space consists of the single cotangent vector dx.
So then the ordering f dx corresponds to the usual way of writing
vector-valued things, where you write the coefficient before the
basis vector. To represent the integral you just stick an integral
sign in front of the 1-form.
Physicists, on the other hand, view this integration as primarily
something you do to a number-valued function, so they put the measure
over next to the integral sign to get all the integration machinery
in one place. (Personally I use either according to mood.)
--
Matt McIrvin
Benjamin Weiner
>quantities mean, I find it helpful to think of the inner product. When
>you hear "The probability is the square of the amplitude" you're thinking
> <a|a> = |a|^2 is the probability or density. ...
Actually, <a|a> = |a|^2 is the squared norm (length) of the vector |a>.
In order to get a probability amplitude, you need to put the vector
together with (another) state vector. I am sorry to pick this nit,
but this was something that S*rf*tti gave us much grief by
misunderstanding. The rest of this is directed to the original poster:
For example,
<psi|x> , which is what is written psi(x) in intro modern physics texts,
is the probability amplitude of finding a particle in state psi at
position x. The probability density is <psi|x><x|psi>, i.e.
|<psi|x>|^2
The probability of finding the particle between x and x + dx is
P(x,x+dx) = |<psi|x>|^2 dx = <psi|x> <x|psi> dx
A properly normalized state |psi> has norm 1, so <psi|psi> = 1.
The particle should be somewhere between -infinity and +infinity,
i.e. it is certain (probability = 1) that it is _somewhere_, so the
total probability should be 1:
+inf
/
P(-inf,+inf) = | <psi|x> <x|psi> dx = 1
/
-inf
/
You see that I inserted a | |x><x| dx into the middle of <psi|psi>
/
(where the integral is understood to be over all possible values of x,
in this case -inf to +inf) without changing the value. This is known
as expanding in the basis "x". I could equally well have expanded in
the basis "p", where p is the momentum of the particle. In some cases
there are discrete bases, such as photon polarization, where vertical
and horizontal form a complete basis - in this case you sum over the
basis rather than integrating. Which basis you use is a matter of
convenience - whatever helps you get the problem done.
Matt McIrvin
>One of my first calculus teachers tried to get us to
>drop the dx completely, on the grounds that the integral symbol
>served as an operator on a function,
>and the x in dx was just a dummy index of integration.
>No one listened to him, though... :^)
Wow! I wonder if he ever tried to change the system of coordinates
in a multiple integral.
--
Matt McIrvin
SCOTT I CHASE
>On the topic of mathematicians' vs. physicists' notation, does anyone know
>why most mathematicians will write an integral as shown below on the left
>(unless they omit the dummy variable x entirely), whereas physicists prefer
>the form on the right?
>
> / /
> | f(x) dx | dx f(x)
> / /
>
I think that it's just a matter of convenience. Physicists often write
long messy integrals with no simple solution. In these situations, it is
nice to know what kind of integral you have, just by looking at the
beginning of the expression. Is it a multivariable integral? What are
the coordinates, i.e., does it say dxdydz or r^2drd(theta)d(phi)? If you
separate the integral sign from the volume element by a long expression, it
is harder to see at a glance what you have in front of you.
-Scott
--------------------
Scott I. Chase "It is not a simple life to be a single cell,
SIC...@CSA2.LBL.GOV although I have no right to say so, having
been a single cell so long ago myself that I
have no memory at all of that stage of my
life." - Lewis Thomas
Keith Ramsay
|why most mathematicians will write an integral as shown below on the left
|(unless they omit the dummy variable x entirely), whereas physicists prefer
|the form on the right?
|
| / /
| | f(x) dx | dx f(x)
| / /
This is just a guess (from a mathematician), but the notation
\integral f(x) dg(x) sometimes is used, by mathematicians at least:
one function integrated relative to another one. The right-hand-side
notation has a possible ambiguity between dx.f(x) and d(xf(x)). If you
are confused, then, you could interpret the right hand side so that it
evaluates trivially to xf(x)+C.
Keith Ramsay "Being a computer means not having
ram...@math.ubc.ca to say you're sorry."
Jim Carr
>
>On the topic of mathematicians' vs. physicists' notation, does anyone know
>why most mathematicians will write an integral as shown below on the left
>(unless they omit the dummy variable x entirely), whereas physicists prefer
>the form on the right?
>
> / /
> | f(x) dx | dx f(x)
> / /
Well, I got my BS in math and my PhD in physics, so no wonder I feel
schizophrenic when I work with integrals.
If all I had was a simple case such as above, I would use the first form,
probably a result of my original training. Where the distinction really
matters is in *real* equations (where there might be 3 integrals, some
nested, and it requires a full page to write the thing down). In that
case I choose on the following basis:
In cases where the limits on integration are simple things like 0 to
infinity but I have nested integrals, I will keep the dx at the end as
a place keeper, in place of a set of parentheses (which I think look ugly
around an integrand, especially when the integrand fits on one line),
to denote what is covered by the integral.
If there is any possibility that I will be changing variables or making
some other transformation, I like keeping the info on the range of
integration and the integration variable together, for the reasons
Emory notes below.
In article <1jn1jv$k...@agate.berkeley.edu> ted@physics1 (Emory F. Bunn) writes:
>
> .... the reason I put the
>dx next to the integral sign in multiple integrals is that it makes it
>instantly clear which integral sign (and hence which limits of integration)
>go with which variable.
This reduces errors when making such manipulations. It also puts the
differential in a place that reminds me to put in the Jacobian.
Not that any of this matters much from a fundamental standpoint: just
do whatever makes it possible to get correct answers efficiently. For
example, I tend to cluster the integral signs together only if that is
how I will program the nested loops. Structuring the algebra similar
to how I will code it up helps eliminate transcription errors.
John C. Baez
On the topic of mathematicians' vs. physicists' notation, does anyone know
why most mathematicians will write an integral as shown below on the left
(unless they omit the dummy variable x entirely), whereas physicists prefer
the form on the right?
/ /
| f(x) dx | dx f(x)
/ /
With the left-hand form, the integral sign and the dx act as delimiters.
With the right hand form, the combination of integral sign with dx stands
for a linear operator. Neither interpretation seems to favor one side of
the cultural divide.
-----
I have noticed this too and have always thought it was because
physicists write more nasty multiple integrals. For these, it is nice
to be able to see the dx in front so you can tell right away what
you are integrating with respect to, before plunging into the thick of
the integrand itself.
Matt Austern
> Feynman presents a nice discussion of Dirac Notation in Volume III of his
> famous lectures. I do not recommend this for learning QM though. Also,
> Baym's book introduces Dirac notation in a nice way but at a somewhat higher
> level.
Actually, I think that the best place to learn about Dirac notation is
Dirac! Specifically: _The Principles of Quantum Mechanics_, 4th
edition. This is a truly wonderful book: a succinct exposition of the
theoretical and mathematical structure of quantum mechanics. It's a
little bit abstract, so you probably shouldn't read it as your first
QM book, but if you already understand a little bit about the
experimental aspects of quantum mechanics, and the physical motivation
for this type of theory, Dirac's book is the ideal place to see how
things really fit together.
--
Matthew Austern Just keep yelling until you attract a
(510) 644-2618 crowd, then a constituency, a movement, a
aus...@lbl.bitnet faction, an army! If you don't have any
ma...@physics.berkeley.edu solutions, become a part of the problem!
Matthew P Wiener
>Well, I got my BS in math and my PhD in physics, so no wonder I feel
>schizophrenic when I work with integrals.
How do you handle spherical coordinates?
--
-Matthew P Wiener (wee...@sagi.wistar.upenn.edu)
Stephen Spackman
|In article <1993Jan21.2...@stsci.edu>, kimball@stsci (Timothy Kimball) writes:
|>One of my first calculus teachers tried to get us to drop the dx
|>completely, on the grounds that the integral symbol served as an
|>operator on a function, and the x in dx was just a dummy index of
|>integration.
|
|Mathematicians dealing with pure function symbols--ie, f, not f(x)--
|will write Int(f) when it's convenient. Which usually means 1-D only.
Ergo, the best notation ought to be Int(\x.f x) or Int(x+>f x) [where \
is a lower-case lambda and +> is a barred arrow]. If the variable you
want to integrate over isn't the function's first argument, you can
always transpose it....
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stephen p spackman +49 681 302 5288(o) 5282(sec) ste...@acm.org
dfki / stuhlsatzenhausweg 3 / d-w-6600 saarbruecken 11 / germany
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