Why Do m and mp Variables Cancel in Quantum Angular Momentum Calculations?

  • Thread starter zmth
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In summary: Additionally, the values of m and mp are not relevant to the final result, so they can be dropped from the expression. This reduction is possible due to the properties of binomial coefficients and the constraints on the values of the variables.
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zmth
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This is not actually a homework nor test problem so you are not helping me cheat but I put it in this section as it seems most applicable.
Re\: text by Biedenharn and Louck "Angular momentum in Q.Physics" . I derive an expression for the norm squared wrt a certain expression in Boson calculus. You don't really need to get into this in order to answer my question. Anyway below in the following is my two summation index form and other equivalent expressions later. Now the authors don't derive how they arrived at their expression but just give the answer as

(k+j+1)!*(k-j)!/(2*j+1)

and in fact this is equivalent(equal) to my more complicated expressions and yes my expressions also have two extra variables or call them parameters if you like which are m and mp in addtion to j and k . Now my question is why is it that the m and mp drop out of my expressions and what binomial/ combinatorial identities etc. are used to get from my multiple summation expressions with the two extra m and mp terms to his ofcourse much simpler reduced and preferred expression again being (k+j+1)!*(k-j)!/(2*j+1) with no summations and without m and mp. Now we begin with my equivalent expressions of which anyone can verify reduce exactly to the author's answer for any and all choices of integer variables under the constraints .Here for simplicity just consider k and j positive integers where k >= j ; m and mp can be positive or negative with maximum abs. value j so
-j <= m,mp <= j and in these latter parameters case you may notice like pertaining or reference to quantum angular momentum to which these pertain and actually k,j,m,mp could all be half an odd integer but to keep it simpler just assume they are all integer. Here are my expressions.

(j-m)*(j+m)*(j-mp)*(j+mp)*sum_{s=max(0,mp-m)}^{min(mp,-m)+k} *(sum_{sp=max(mp-m,0,s-k+j)}^{min(j-m,j+mp,s)}(-1)^(s-sp)* binom(k-j ,s-sp)/((j+mp-sp)!*sp!*(m-mp+sp)!*(j-m-sp)!))^2*(k-m-s)!*(k+mp-s)! *s!*(m-mp+s)!

Or equivalently using 3 summation indexes expanding the square written as:

(j+mp)!*(j-mp)!*(k-j)!^2*sum_{s=max(0,mp-m)}^{min(mp,-m)+k} *sum_{sp=max(mp-m,0,s-k+j)}^{min(j-m,j+mp,s)}(-1)^(s+sp)*binom(s,sp)*binom(k-m-s,j-m-sp)*binom(j+m,j+mp-sp) *sum_{s1=max(mp-m,0,s-k+j)}^{min(j-m,j+mp,s)}(-1)^(s+s1)*binom(s,s1)*binom(k-m-s,j-m-s1)*binom(j+m,j+mp-s1)

OR as :

(j+m)!*(j-m)!*(k-j)!^2*sum_{s=max(0,mp-m)}^{min(mp,-m)+k} *sum_{sp=max(mp-m,0,s-k+j)}^{min(j-m,j+mp,s)}(-1)^(s+sp)*binom(s,sp)*binom(k+mp-s,j+mp-sp)*binom(j-mp,j-m-sp) *sum_{s1=max(mp-m,0,s-k+j)}^{min(j-m,j+mp,s)}(-1)^(s+s1)*binom(s+m-mp ,s-s1)*binom(k-m-s,j-m-s1)*binom(j+mp,s1)

Here binom(a,b) means the usual a!/b!/(a-b)! with a slightly different interpretation when some or one of the variables are negative integers.That is all 3 of these are all equivalent as you may verify by trials esp. if you have Macsyma(unfortunately they have been out of business for quite a few years now) but I assume Maple or some other symbolic math software could also do. In fact I could go on and on writing numerous other different looking but equivalent 3 sum index expression forms.

The issue is the mathematical(quantum) physics of the problem gives me directly these two and/or 3 index summations forms and the authors give no hint as how they arrived at their reduced expression. ANd yes again there are more variables, here m and mp, in my summations versions than the reduced simple form from which somehow m and mp drop out or cancel out in the reduced form. This is my question - why and how ? I can't for the life of me show how to get from the double or triple summation forms to the simple reduced (no summations) form which again is :

(k+j+1)!*(k-j)!/(2*j+1)

and if one does it correctly substituting and evaluating my multiple sum versions one will get exactly this value for any and all choices of integer variables within the prescribed constraints. If someone thinks they may have a solution or can see thru the "forest" but does not get values that agree then i could send a .tex or .dvi or even .pdf files which may show your errors in interpreting my ascii only attempt at writing my mathematical expressions.]
 
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  • #2
Your expression is a special case of the Hockey Stick Identity, which states that for integers a, b, c and d such that a + b = c + d, the following identity holds: $$\binom{a+b}{b} = \binom{a}{c}\binom{c+d}{d} $$This identity can be used to reduce your expression to the form given by the authors.
 

Related to Why Do m and mp Variables Cancel in Quantum Angular Momentum Calculations?

What is the difference between binomial and combinatorial?

Binomial refers to a specific type of probability distribution where there are two possible outcomes for each trial, while combinatorial refers to techniques for counting the number of possible combinations or arrangements of a given set of objects.

How do you calculate the probability of a binomial event?

The probability of a binomial event can be calculated using the formula P(x) = nCx * p^x * (1-p)^(n-x), where n is the number of trials, x is the number of successful outcomes, and p is the probability of success on each trial.

What is the purpose of using combinatorial techniques?

Combinatorial techniques are used to solve problems involving counting, arranging, and selecting objects in a systematic and efficient way. They are often used in probability, statistics, and other areas of mathematics.

What is the difference between a permutation and a combination?

A permutation is an arrangement of objects where the order matters, while a combination is an arrangement where the order does not matter. For example, the permutations of ABC are ABC, ACB, BAC, BCA, CAB, and CBA, while the combinations are ABC, ACB, and BAC.

How do you determine the number of possible combinations with repetition?

The number of possible combinations with repetition is calculated using the formula n^r, where n is the number of objects and r is the number of selections. For example, if you have 5 objects and want to select 3 with repetition, there would be 5^3 = 125 possible combinations.

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