A Quantum Mechanical Game of Craps: Teaching the Superposition

Publication Date (Web): January 1, 2001. Cite this:J. Chem. Educ. 78, 1, XXX-XXX ... Journal of Chemical Education. Neto. 1984 61 (12), p 1044. Abstra...
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In the Classroom

A Quantum Mechanical Game of Craps: Teaching the Superposition Principle Using a Familiar Classical Analog to a Quantum Mechanical System Patrick E. Fleming Department of Chemistry and Biochemistry, Arizona State University, Tempe, AZ 85287; [email protected]

Many students struggle with concepts in quantum chemistry. The physical interpretation of the wave function causes particular grief. The Born interpretation that the square of the wave function gives a probability density is useful. It also provides the criterion necessary for normalizing a wave function. Unfortunately, though, it is confusing. Scientists tend to prefer models that infer a cause-andeffect relationship rather than a probabilistic description of observation. Nonetheless, quantum mechanical systems may have several quantum states that can be observed at any one time. The quantity that can be observed is given by the eigenvalue to some quantum mechanical operator acting on the wave function. Each observable value has some associated probability that can be extracted from the wave function for the system. The difficulty is that classical systems can exist in only one given state. When a coin is tossed, it is assumed to land as either “heads” or “tails”. The quantum mechanical coin, however, would be said to be both heads and tails with equal probability for both. This is the foundation of the famous “Schrödinger’s cat” analogy (1). Common sense based on classical Newtonian physics precludes the possibility of the coin’s being both heads and tails. It must be one or the other. This was the root of Albert Einstein’s objection to quantum mechanics. In a letter to Max Born in 1926 (2), Einstein wrote discussing his famous objection to quantum mechanics. Quantum mechanics is very impressive. But an inner voice tells me that it is not yet the real thing. The theory produces a good deal but hardly brings us closer to the secret of the Old One. I am at all events convinced that He does not play dice.

The merits of this objection are still the subject of active research. None the less, the dice themselves are interesting. Dice have the useful property that the measurement defined as reading the number of spots on the top face produces a result that is naturally quantized. Thus dice can be very handy devices in illustrating quantum mechanical properties of wave functions and operators (3). Let’s define a set of wave functions and an operator pertaining to the roll of one die. We don’t know the exact form of the wave function or the operator, but we can define them in terms some of their more important properties. First, the wave functions must be orthonormal so that  ψ m | ψ n  = δnm where δnm is the Kronecker delta, which has the property

δnm =



1 0

n =m n ≠m

The “roll” operator will operate on the wave function and

return an eigenvalue that is the number of spots showing on the top face of the die after it has been observed. There are six possible values that can be measured when a die is rolled, since the die can land with 1, 2, 3, 4, 5, or 6 spots showing. These values are the eigenvalues obtained when the operator operates on the function ^ R | ψ n  = n| ψ n  so n can take any integral value from 1 through 6. Further, since n corresponds to the physically observable measurement that is made on the system, it must be real valued. (Imagine how difficult it would be to play dice if the dice sometimes came up with imaginary values !) The superposition principle suggests that as in any other quantum mechanical system, any linear combination of these basis functions will satisfy the requirements of our system. Thus the general function

Ψ=

6

Σ c n | ψn  n =1

will also satisfy the requirements of a wave function describing the die. The values of the coefficients cn can be determined through normalization (  Ψ | Ψ  = 1).

Ψ|Ψ =

Σ 6

n =1

c n ψn |





6

Σ c m | ψm  =

m=1

6

6

Σ Σ c n c m ψn | ψm

n =1 m=1

Using the orthonormality of the wave functions, the substitution  ψ n | ψ m = δnm can be made:

Ψ|Ψ =

6

6

Σ Σ c nc mδnm

n =1 m=1

The Kronecker delta causes the argument of the double sum to be zero except when m = n. This picks out a single term for each possible value of n. Ψ|Ψ =

6

Σ c n2 n =1

Also, all the coefficients must be equal because the probability of observing each value upon rolling the die is the same as for any other value.

Ψ|Ψ = 6c n2 = 1 and

cn = 1 6 for all n. Further, the probability of observing any one value

JChemEd.chem.wisc.edu • Vol. 78 No. 1 January 2001 • Journal of Chemical Education

57

In the Classroom

is given by the square of the coefficient c n :

Again, this is a fairly intuitive result. The same conclusion can be reached by using the original constraints on the coefficients cn and setting the sum of the probabilities equal to unity.

c n2 = 1 6

This is a perfectly reasonable result because the probability of observing any particular roll on one die is 1/6, since there are six equally probable outcomes. For any quantum mechanical system whose wave function can be expressed as a linear combination of convenient basis wave functions, the probability of observing an eigenvalue associated with a given basis function is always given by the absolute value squared of the Fourier coefficient for that basis function.

Σ Pn = Σ

c n2 = 1

The expectation value for the roll of a single die is also easily calculated. The expectation value will predict the value for the average number that comes up over the course of an infinite number of rolls of the die. The expectation value is given by

R = Ψ R Ψ Making the substitution from above for the unloaded die,

P n = c n2 To further illustrate this point, consider a “loaded die” constructed such that it is twice as likely to throw a 1, 2, or 3 as a 4, 5, or 6. In this case, the Fourier coefficients must satisfy the requirement

Ψ=

6

Σ n =1

1 ψ | n 6

It is seen that 6

c n2 =

 2ββ

2

2

for n = 1, 2, 3 for n = 4, 5, 6

β2 is used here simply for convenience. The wave function

will be given by 3

6

n =1

n=4

Ψ = Σ 2 β|ψn  + Σ β|ψn 

Ψ|Ψ =

3

Σ

n =1 3

= 2β2 Σ

3

Σ δnm + n =1 m=1

n=4

m=1

m=4

3

6

6

3

6

6

Σ n = 1 m=4

Σ

ψn |ψm + β2 Σ

n =4 m=1

3

6

ψn |ψm + 2 β2 Σ

2 β2 Σ

= 2β2 Σ

3

3

Σ

n =4 m=4

3

6

2 β2 Σ

Σ δnm + n = 1 m=4

ψn |ψm +

Ψ|Ψ = 2β

3

6

6

6

Σ Σ δnm + β nΣ=4 m=4 Σ δnm

n =1 m=1

6

n =1

n =4

= 2β2 Σ 1 + β2 Σ 1 = 6β2 + 3β2 Equating this to unity, it is easy to find the value of β: 9β2 = 1 ;

β = 1 ⁄3

and the Fourier coefficients are given by

58

6

6

6

= 1 Σ Σ n ψm|ψn 6 m=1n =1

6

6

Σ Σ n δ nm m=1n =1

The Kronecker delta picks out only the terms for which n = m and the sum simply becomes 6

1 R = Σ m 6 m=1 which is easy to evaluate.

2

3

cn = 2 3 cn = 1 3

6

Σ ψm|  nΣ=1 n|ψn  m=1 6

Σ δnm + β2nΣ=4 m=4 Σ δnm n =4 m=1

3

1 R = 6

R =

At this point, if your students don’t revolt, they will see that the second and third terms vanish, since n and m can never be the same. Thus, the whole integral simplifies to 2

6

Replacing the matrix element using  ψ m| ψ n = δ nm ,

ψn |ψm

3

6

2 β2 Σ

6

Σ ψm|  nΣ=1 R |ψn m=1

which simplifies to

6

2 βψn | + Σ βψn |  Σ 2 β|ψm + Σ β|ψm 

Σ n =1 m=1

Σ m=1

1 R = 6

The value of β is determined by normalization.

6

1 ψ | R 1 |ψ  m  Σ n n =1 6 6 – Pulling out the factors of 1/√ 6 and moving the operator inside of the sum (which can be done because all quantum mechanical operators are linear), R =

for n = 1, 2, 3 for n = 4, 5, 6

21 = 3.5 R = 6 This is actually an intuitive result. Everything has come out as expected. Even without knowing the form of the wave functions ψn or the “roll” operator, it is still possible to apply all the rules of quantum mechanics to “predict” the behavior of the die. It is pedagogically useful here to point out that the expectation value is not one of the eigenvalues. Just as in the particle-in-a-box problem for a particle in the n = 2 state, the position of the particle will never be measured at the center of the box because of the node in the wave function. Nonetheless, the expectation value for the position is the center of the box.

Journal of Chemical Education • Vol. 78 No. 1 January 2001 • JChemEd.chem.wisc.edu

In the Classroom

Using the above methodology, it is possible to calculate the expectation value for the roll of two dice. The form of the operator will be a sum of two one-die roll operators. ^= R ^A + R ^B R ^A operates only on die A and R ^B operates only on die where R B. This will work quite well so long as the wave function is separable and can be expressed as the product of two “onedie” wave functions: ΨAB = ψA ψB If the one-die wave functions are defined as above as a linear combination of the eigenfunctions of the one-die roll operator (using the unloaded dice for fairness and simplicity), the expectation value can be calculated as ΨAB R ΨAB = 6

Σ  n=1

6

6

1 ψ R +R 1 ψ B,m  A B Σ A ,k k=1 6 6

Σ  m=1

1 ψ A ,n  6

6

Σ l=1

1 ψ B,l 6



– Pulling out the factors of 1/√ 6 and allowing the operators to operate, ΨAB R ΨAB = 1 36 = 1 36

6

6

6

6

6

6

6

Σ m=1 ΣΣ Σ ψA ,nψB,m  k=1  n=1 l=1 6

6

6

l ψB,l  Σ kψA ,k Σ Σ ψA ,n  m=1 Σ ψB,m k=1 n=1 l=1

6

6

k + l ψA ,k ψB ,l



= 1 Σ Σ Σ Σ k + l ψA ,n ψB,m ψA ,k ψB ,l 36 n=1 m=1 k=1 l=1 6

6



6

6

= 1 Σ Σ Σ Σ k +l 36 n=1 m=1 k=1 l=1 6

6



ψA ,nψA ,k ψB,mψB,l 6

6

= 1 Σ Σ Σ Σ k + l δnk δml 36 n=1 m=1 k=1 l=1

Expanded, this expression yields a very large number of terms. But as before, the Kronecker delta will annihilate all of those terms for which n ≠ k and m ≠ l. The result is 6

6

ΨAB R ΨAB = 1 Σ Σ k + l = 252 = 7 36 k =1 l =1 36 which is again the intuitive result. So even if the math is a bit more cumbersome than the single-die case, the result is just as expected. The treatment above is similar to the presentation of a multidimensional problem such as the particle in a rectangular box or a polyatomic harmonic oscillator such as the water molecule. This model also exhibits properties similar to energy degeneracy (3), since, for example, there are four unique ways to roll a five. This discussion may not have done much to convince the late Einstein of the merits of quantum mechanics, but he probably would not dispute the results shown here. In fact, the die analogy can be expanded to explain another of Einstein’s objections to the model. This relates to the collapse of a wave function upon measurement of the system. Imagine a quantum-mechanical dice game. For convenience, consider only one die used in the game. A shooter

will roll the die and a dealer will “measure” the system by calling the number of spots showing on the top face. After the shot but before the measurement, a correct wave function for the die would be 6

Ψ = 1 ψ1 + 1 ψ2 + 1 ψ3 + 1 ψ4 + 1 ψ5 + 1 ψ6 = 1 Σ ψn 6 6 6 6 6 6 6 n=1

as discussed before. However, once the die is measured the wave function collapses. For the sake of discussion assume the die shows five spots. The wave function instantaneously becomes Ψ = ψ5 There has been an instantaneous collapse of the wave function! No longer are all measurements possible with equal probability. Considering what happens to the wave function describing the bottom face of the die is even more troubling. Before measurement, a correct wave function would be given by

Ψb = 1 6

6

Σ ψ7–n n=1

where n, as before, gives the number of spots on the top side. Thus, this wave function is coupled to that describing the top face. This is true because the total number of spots on any pair of opposite-facing sides of the die is always seven. The measurement of the top face of the die will cause a collapse not only of the wave function Ψ describing the top of the cube, but also simultaneously it will cause the collapse of the wave function Ψb describing the bottom! Instantaneously, Ψb becomes Ψb = ψ 2 The measurement of one system has affected the wave function of another system, causing a simultaneous collapse of both wave functions. This effect is what Einstein referred to as “spooky action at a distance”. In their famous gedankenexperiment (4), Einstein, Podolsky, and Rosen (EPR) discussed how a system of two particles that interact and then move apart must respond to measurements made on one of the particles. Consider a particle that breaks into two fragments each with spin s = 1⁄2. A π meson is an example of such a particle in that it decays to form an electron and a positron. The orientations of the spins must be correlated to assure conservation of angular momentum. Given that the spin of the ith particle has equal probability of having a component ms = + 1⁄2 (designated αi) as ms = ᎑ 1⁄2 (designated βi), the spin wave function for the system is given by the familiar singlet spin wave function

Ψs = 1 α1 β 2 – β 1 α2 2 This wave function has important properties. First, it is equally likely that either particle has the z component of its spin given by ms = + 1⁄2 as by ms = ᎑ 1⁄2 . Second, the total z component of the spin is 0. This is required by the conservation of angular momentum because the total spin angular momentum must be 0 as it was for the parent π meson. Measurement of the orientation of the spin for one fragment causes an instantaneous change in the wave function describing the spin of the counter fragment as well as for the total system. The counter fragment goes from being both spin up (α) and spin down (β) with equal probability to being one

JChemEd.chem.wisc.edu • Vol. 78 No. 1 January 2001 • Journal of Chemical Education

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In the Classroom

or the other with unit probability. This happens without any measurements being made on the counter fragment! Thus the system must somehow communicate across a distance at a speed faster than the speed of light, violating Einstein’s theory of relativity. Coming to terms with this paradox is an area of active research (5). Unfortunately, simple illustrations using classical analogs to quantum mechanical systems will not provide the answers. The construction of a die makes the treatment of the probabilities governing opposite sides intrinsically different from treating the quantum mechanical properties of correlated particles. For example, a measurement made on one of the correlated particles in the EPR thought experiment requires instantaneous communication over a distance, causing the collapse of the wave function on the partner particle. In the case of a die, opposite sides are in constant contact so no such violation need occur. Furthermore, since it is possible to know the number of spots on both the top and bottom faces with complete precision, there is no violation of the uncertainty principle to contend with. Nonetheless, this example provides an excellent framework in which to discuss the questions posed by the EPR paradox.

60

The die example is also an effective mechanism for discussing the superposition principle because students are likely to have an intuitive understanding of the system used. They will intuitively know how to calculate the average value rolled on a single die over a large number of rolls. The limited number of terms in the summations helps in their evaluation a great deal. Students are also likely to understand the simple manner in which opposite sides of the cube are correlated. So while the Old One may not play dice, we can still make good use of them. Literature Cited 1. Schrödinger, E. Naturwissenschaften 1935, 23, 82. 2. Einstein, A. Letter to Max Born, December 4, 1926. Excerpt reprinted in Baggot, J. The Meaning of Quantum Theory; Oxford University Press: New York, 1992; p 29. 3. Neto, B. de B. J. Chem. Educ. 1984, 61, 1044. 4. Einstein A.; Podolsky, B.; Rosen, N. Phys. Rev. 1935, 47, 777. 5. Aspect, A.; Grangier, P.; Roger, G. Phys. Rev. Lett. 1982, 49, 91 is one example.

Journal of Chemical Education • Vol. 78 No. 1 January 2001 • JChemEd.chem.wisc.edu