0
$\begingroup$

I have just been introduced to the wave function in my lectures. The way my book and lecturer motivate the wave function is by analogy with the light double slit experiment. The first thing that is explained to me is the fact that for light, $f_X(x)\propto|A(x)|^2$ where $f_X(x)$ is the probability density function for the random variable "X = position of photon" and A(x) is the amplitude of the light at the point on the screen. Then it is stated in class that for an electron $f_X(x)=|\psi(x)|^2$

My Questions

1.) the derivation for $f_X(x)\propto|A(x)|^2$ for light was done in a very hand wavey and confusing way (I presume it was done so to avoid as much of probability theory as possible since most people in that class dint take any probability) my question is how is this relation derived properly?

2.) how can we make such a jump and simply assume $f_X(x)=|\psi(x)|^2$ (even neglecting the proportional symbol). I understand that particles can have wave like properties and that we assign $\psi(x)$ to be the function that describes the wave however, why is this related to probability density function by squaring it ?

I have not yet been presented to the shrodinger equation, but if my question nr 2.) can be answered by using the shrodinger equation let me know and ill come back to this question after I have reached the shrodinger equation in class.

Thank You for any help

$\endgroup$
4

2 Answers 2

1
$\begingroup$

Well, for question $(1)$ I can't really give you the whole answer because then I need to introduce what is a photon formally, using 2nd quantization techniques. You are now taking your first steps in quantum mechanics so I will hand wave the answer as well. In classical EM, one can see that the intensity distribution is proportional to $|E|^2$ (the amplitude of the wave, squared). Photons are related to the energy in the wave, so their distribution should follow the same distribution as the intensity, meaning the probability distribution of a single photon will also be proportional to $|E|^2$. Again, to really prove this is possible, but only way later in your QM studies.

And well, for question $(2)$ there is no real answer without implying "This is what works, so now it is one of the axioms of quantum mechanics". One of the axioms in quantum mechanics that to describe fully the state of a system you need to know one vector, which is called "the wave function", and is often written as $|\psi \rangle$ (the notation $|...\rangle$ is just Dirac's way of writing vectors, instead of the usual $\vec{\psi}$). This is different from classical mechanics, where to describe the system you need to specify only the position and velocity of every particle (which you take as initial conditions for Newton's laws of motion, and then know everything about the system).

The Schrodinger equation is only the description of how it evolves in time, and has nothing to do with measurements and probability distributions.

To understand how this vector is related to actual measurements, there is another axiom which says that to each observable with a couple of possible outcomes (say the position of the particle can be $x_1$ or $x_2$ or $x_3$ etc...) there is a basis of vectors in the same space ($|\phi_1 \rangle$ , $|\phi_2 \rangle$ , $|\phi_3 \rangle$, etc...). The probability of measuring one of the outcomes, $x_i$, one needs to calculate (take a deep breath): the square of the magnitude of the projection of the wavefunction $|\psi \rangle$ at the instant of measurement on the relevant basis vector $|\phi _i\rangle$. That was a long sentence, but don't worry, the math is much shorter:

$$P(x=x_i) = |\langle \phi_i | \psi \rangle |^2$$

This is an axiom of quantum mechanics, which was derived after years of research in the beginning of the 20th century.

Specifically, for measuring position, the projection is just the value of the wavefuncion at a certain point in space, meaning:

$$P(x=x_i) = |\langle \phi_i | \psi \rangle |^2 = |\psi(x_i)|^2$$

One can give intuitions and show you some experiments so you could get a better grasp about how this axiom and its consequences were achieved, like your textbook tried to give you. For me personally, I understood the concept of collapse of the wave function only after learning about the Stern Gerlach experiment. But in the end of the day, this can't be derived without assuming some axiom, and the axiom nowadays is what I tried to introduce in the paragraphs above.

I hope I didn't confuse you too much, and even helped a bit :)

$\endgroup$
1
$\begingroup$

In the paper “Is quantum mechanics an island in theory space” we do get a proper argument for the first question.

The idea is this: you want a wavy probabilistic way of looking at the world, so you want vectors $\mathbf p=(p_1, p_2, \dots p_n)$ corresponding to an event having $n$ possible outcomes. You would also like to say that these probabilities sum to 1, for which you want some $L_k$-norm, $$1=|p_1|^k + |p_2|^k +\dots+|p_n|^k.$$ But you also want to have groups of linear processes (because waves should be superimposable, at least if they have low amplitudes!) that transform these probability vectors to other vectors that sum to 1.

It turns out that you have just specified the system very precisely and now you can only have classical probability with $k=1$ or quantum probability with $k=2.$ Furthermore I think the $k=1$ case only works if the numbers are nonnegative, in which case the linear transform group is the group of stochastic matrices.

So the only way to achieve a linear probabilistic theory of any interference effects is to have an algebra if probability amplitudes which square to become actual probabilities.

Furthermore the $L_2$-norm is now well-known to subsume the $L_1$-norm case. Consider the set of positive semi-definite $n\times n$ matrices with trace 1 and embed the 2-norm vectors here by forming $\mathbf p~\mathbf p^\dagger$, then the algebra of such “state matrices” also contains as a subset the $L_1$-norm as state matrices which are diagonal.

So this squaring of amplitudes rule is in some sense the only way it could possibly work—at least, perhaps, without nonlinearities which would make our heads explode.

As for question (2), for probabilities we do try to make these two things equal. There are two caveats.

  1. Many numbers in the works are not unitless like probabilities are. If you are measuring one of them then if course you are stuck with a proportionality constant to carry the units, though you can always choose units to make that constant look like 1.

  2. Often we wish to deal with continuous ranges of outcomes, not discrete individual outcomes. For this we invent probability density functions. If $f(x)$ is the probability density for a random variable $X$ then that means that the probability that $X$ occurs between some value $x$ and the very slightly larger $x+\epsilon$ is the value $f(x)~\epsilon$, so it is proportional to the interval size for tiny intervals. This often forces the values of the probability density function to have weird irrational maximums and so forth, because the actual thing which sums to 1 is the area under the curve.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.