Psi & Probability: Linking Complex Concepts

[MHD93]

I wonder and feel like knowing how the product of psi with its complex conjugate represents the Probability Density for the object to be there? how two (apparent) different concepts are linked that way?

 

[lightarrow]

I wonder and feel like knowing how the product of psi with its complex conjugate represents the Probability Density for the object to be there? how two (apparent) different concepts are linked that way?

Essentially because the product of a complex number by its coniugate is the square modulus of the complex number.

 

[MHD93]

but what makes it concern probability?

 

[Bill_K]

ψ(x) is by definition the probability amplitude for finding the particle at point x. That is all it is. What other interpretation did you have in mind?

 

[bobbytkc]

but what makes it concern probability?

It is called Born's rule, which says that the square of the probability amplitude gives the probability of an outcome. This is an implicit assumption in quantum mechanics, in fact, it is part of the axioms that make up quantum mechanics.

If you are still confused, the fact is that the wavefunction is related to the probability distribution will not be found in the mathematics of quantum mechanics. It is a physical fact that is put into the mathematical framework of quantum mechanics by physicists, and we know it is correct because it predicts the right experimental outcomes when physical experiments are performed in the laboratory..

 

[A. Neumaier]

I wonder and feel like knowing how the product of psi with its complex conjugate represents the Probability Density for the object to be there? how two (apparent) different concepts are linked that way?

Suppose you have a system with N energy eigenstates |k> (k=1:N) of energies E_k, E_1<...<E_N. The Hilbert space spanned by these states can be identified with C^n, by taking |k> to be the k-th unit vector. The Hamiltonian is then the diagonal matrix H=Diag(E_1,...,E_N) with diagonal entries E_1,...,E_N. A general state of the system is described by a density matrix rho, a semidefinite Hermitian matrix of trace 1. In particular, the diagonal elements p_k:= rho_{kk} are nonnegative and satisfy sum p_k = Tr rho = 1. Thus they look like probabilities. Observables are represented by arbitrary Hermitian matrices X, and their expectation in the state rho is defined to be <X>= Tr rho X.

A classical system corresponds (in some sense) to the case where the only allowed states and observables are diagonal. Thus rho=Diag(p_1,...,p_N) and X=Diag(x_1,...,x_N), giving <f(X)>=Tr rho f(X) = sum p_k f(x_k). This is precisely the formula for the expectation of a function f(x) of a random variable x that takes the values x_k when the random event k happens with probability p_k. (In measure-based probability theory, one would write omega in place of k, call it an elementary event, organize the possible events in a sigma algebra, and write x(omega) in place of x_k, thereby turning the random variable into a functions of elementary events.)

The quantum case is therefore just the generalization of classical probability calculus to the case where densities and observables may be matrices rather than functions.

A very special case of states are the so-called pure states. These are characterized by the fact that all their columns are proportional to the same unit vector psi (called the state vector of the pure state). Because the density matrix must be Hermitian and have trace 1, it is not difficult to conclude that in this case rho=psi psi^*, where psi^* is the conjugate transpose of psi. Therefore, the diagonal elements are
p_k = rho_{kk}=psi_k psi_k^*=|si_k|^2,
which is the Born rule.

Thus nothing fancy is going on. But typical introductions to quantum mechanics make it unnecessarily mysterious by starting with the special case of pure states rather than with the (in reality much more frequent) case of a general (mixed) state. A notable exception is my online book
Arnold Neumaier and Dennis Westra,
Classical and Quantum Mechanics via Lie algebras, 2008.
http://lanl.arxiv.org/abs/0810.1019

 

[bobbytkc]

In particular, the diagonal elements p_k:= rho_{kk} are nonnegative and satisfy sum p_k = Tr rho = 1. Thus they look like probabilities. Observables are represented by arbitrary Hermitian matrices X, and their expectation in the state rho is defined to be <X>= Tr rho X.

The requirement that the trace is equal to 1 is the assumption of conservation of probability, so there is already an implicit assumption of the probabilistic interpretation in that statement.

 

[A. Neumaier]

The requirement that the trace is equal to 1 is the assumption of conservation of probability, so there is already an implicit assumption of the probabilistic interpretation in that statement.

Of course there are assumptions, as in any theory. But one can choose which assumptions to make the basis for deriving everything else. In particular, the requirement Tr rho = 1 can be taken as fundamental (as I did), and then the probabilistic interpretation follows.

Actually, in quantum optics one does _not demand_ Tr rho = 1 in order to be able to discuss matters of efficiency of detection. Then r:= Tr rho denotes a rate of detection, and <X>=Tr rho X / Tr rho is the conditional expectation <X> given a detection. The probability interpretation still follows. And the derivation of Born's rule is still valid except that now the wave function is related to the density matrix by rho = r psi psi^*.

 

[alxm]

Agreed, I don't see how that amounts to assuming 'conservation of probability'. What you're calling 'conservation of probability' would be better described as 'conservation of normalization'.
If you assume the Born rule, then the normalization condition naturally follows as the total probability must be 1. That's how the normalization condition is usually introduced,
but I don't believe the probabilistic interpretation is required here.

If you totally disregard the Born rule and assume nothing about probabilities and measurements, you still have the Schrödinger equation giving a set of energy eigenstates for the system and their time evolution.
Normalization and its conservation is then required for conservation of energy.

 

[A. Neumaier]

If you totally disregard the Born rule and assume nothing about probabilities and measurements, you still have the Schrödinger equation giving a set of energy eigenstates for the system and their time evolution.
Normalization and its conservation is then required for conservation of energy.

Actually, whatever the value of psi^*psi is, it is conserved by the Schroedinger evolution.
Thus normalization is optional, and a change of normalization is just like a change of global phase.

But if one uses unnormalized wave functions, the Born rule must be modified to say p_k= |psi_k|^2/(sum |psi_k|^2).

 

[alxm]

Thus normalization is optional, and a change of normalization is just like a change of global phase.

Sure, but two different wave functions describing physically-identical systems should reasonably be expected to have the same observables.
So while the normalization is arbitrary, it only really makes sense to have it set to 1.

 

[A. Neumaier]

Sure, but two different wave functions describing physically-identical systems should reasonably be expected to have the same observables.
So while the normalization is arbitrary, it only really makes sense to have it set to 1.

In Hamiltonian quantum mechanics, yes. My point was that conservation f normalization is a consequence of the Schroedinger dynamics, and must not be _required_, as you had formulated it.

Things are different when dissipation must be taken into account. In wave-function-based models, this is usually done with a superimposed stochastic framework. Then psi^*psi decreases with time; it represents something with an observable meaning; see, e.g., equation (89) in http://arxiv.org/pdf/quant-ph/9702007