# Why do we need a wave function?

Assuming our only aim is to solve double slit experiment (or other problems that can be mapped into that). Knowing that electron does some strange thing not expected of a particle, we need a function which would give the probability distribution of the electron on the screen. Why take the pain of knowing the wave equation and then interpreting out in Born manner. Isn't the aim of the science to keep things simple? This is my first quantum mechanics course so try to be as simple as possible.

• No doubt the science try to distinguish, in a problem, the main features for eliminating un-necessary complications. But in doing that, we have to be enough wise and not loose the essential features. So, as Robin Ekman says, not any function would do the job. If you want to eat an apple, you don't pick a banana. In short, the best function for solving the 2slit diffraction is the wave-function itself. – Sofia Feb 21 '15 at 19:39

The non-negative real probability distribution can't interfere like a complex wave function can. To produce interference phenomena it is necessary for quantum mechanics to deal with probability amplitudes, not just probabilities.

• can you elaborate on what you mean by interference? – Manish Kumar Singh Feb 21 '15 at 19:22
• actually I got it so no need to explain. anyways thanks for your response. – Manish Kumar Singh Feb 21 '15 at 20:02

You say that we are only interested in the probability distribution on the screen, $\rho(x,t) = \lvert \psi(x,t) \rvert^2$, which is essentially correct. So, why do we have $\psi(x,t) = \lvert\psi(x,t)\rvert\mathrm{e}^{\frac{\mathrm{i}}{\hbar}S(x,t)}$?

Well, looking at the time evolution equation for the probability density, the continuity equation of quantum mechanics is

$$\partial_t \rho = \frac{\rho}{m}\partial_xS$$

that is, the time evolution of the probability density depends on the complex phase of the wave function. So you have two degrees of freedom that are coupled to each other - the phase and the modulus - and the most convenient way to encode that information is then in a single complex function $\psi = \sqrt{\rho}\mathrm{e}^{\mathrm{i}\phi}$.

Moreover, you observe interference - conceive of the double slit as two single slits. Now, if the particle passed through a single slit, you would have a probability density on the screen $\rho_\text{single}$. Now, if you shoot it at the two slit such that it can pass through both, you would, classically, assume that the total density to detect it somewhere on the screen is something like $\frac{1}{2}(\rho^\text{left}_\text{single} + \rho^\text{right}_\text{single})$, because the particle passes through either slit (say its 50-50, but it doesn't matter), and then you just have the single slit situation.

That doesn't happen.

You get the characteristic interference fringes, which are not the sum of the individual probability densities, because these are always positive, but for destructive interference, you need something that's positive as well as negative. So, again, the most efficient way to encode this idea is to take some function $\psi$ that can also be negative and say $\rho = \lvert\psi\rvert^2$. It then turns out that allowing $\psi$ to be positive and negative is not enough, no matter how you vary it in time and space, you don't get the $\rho$ you observe, and you finally give up and let it be complex. Now, just adding the $\psi$ that belong to the single slits works and you get a $\rho$ you actually observe.

Assuming our only aim is to solve double slit experiment (or other problems that can be mapped into that).

Actually the double slit experiment for electrons is a derivative/prediction from the quantum mechanical theory as it started with the Schrodinger equation ,its wavefunction solutions and the interpretation of differential operators with energy and momentum and angular momentum.

The Schrodinger equation solutions solved elegantly the spectrum of the hydrogen atom, once the wavefunctions were interpreted as the basis on which the energy operator operated. The known experimentally spectral series came out of operating on the wavefunctions with the energy operator . That is how the theory of quantum mechanics started and was established. The price was in interpreting the square of the wavefunction as a probability for the electron to be bound around the proton of the hydrogen atom, defining an orbital, instead of the Bohr orbits.

Knowing that electron does some strange thing not expected of a particle, we need a function which would give the probability distribution of the electron on the screen. Why take the pain of knowing the wave equation and then interpreting out in Born manner. Isn't the aim of the science to keep things simple?

The aim of science is to make observations and experiments , find theoretical models that explain the data and that predict new phenomena to be studied. Yes, in the simplest format possible, but including all the known data. The double slit is a tiny part of the experimental data that quantum mechanics explains/models.

Since the two slit experiment is a bit complicated for what I'm about to discuss, allow me to consider a simplified toy model. Consider an $N$ component state vector $| \alpha \rangle$, about to be acted on by some operation, and subsequently measured. This evolution can be represented by:

$$| \beta \rangle = U |\alpha\rangle$$

The important bit here is that forr probability to be conserved tthe matrix $U$ is unitary, that is it satisfies $U U^\dagger = I$.

Then say you're measuring an observable whose eigenvectors are $(1,0,...,0)$,$(0,1,...,0)$,...,$(0,0,...,1)$. Then the $n$th component of $| \beta \rangle$ is the amplitude for measuring the eigenvalue corresponding to the $n$th eigenvector above.

Now what you want to do is to say... screw that noise. I want to start with a probability distribution $A$, apply an operation represented by a stochastic matrix $S$, and end up with a probability distribution $B$. Obviously then to reproduce the predictions of quantum mechanics we must have this:

$$\left( \begin{array}{c} |\beta_1|^2 \\ . \\ . \\ . \\ |\beta_N|^2 \end{array} \right) = \left( \begin{array}{ccc} S_{11}& \cdots &S_{1N} \\ . &...& .\\ . &...& . \\ . &...& . \\ S_{N1}& \cdots &S_{NN} \end{array} \right) % \left( \begin{array}{c} |\alpha_1|^2 \\ . \\ . \\ . \\ |\alpha_N|^2 \end{array} \right)$$

So the first thing you notice is that $S$ will in general depend on $\alpha$ as well as the physical operation you're doing on the system. To see that this is the case consider the following $N = 2$ example:

$$\left( \begin{array}{c} 1 \\ 0 \end{array} \right) = \left( \begin{array}{cc} \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}}\end{array} \right) % \left( \begin{array}{c} \frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}} \end{array} \right)$$

Which can be written in stochastic language as:

$$\left( \begin{array}{c} 1 \\ 0 \end{array} \right) = \left( \begin{array}{cc} 1 & 1 \\ 0 & 0 \end{array} \right) % \left( \begin{array}{c} \frac{1}{2} \\ \frac{1}{2} \end{array} \right)$$

Now consider:

$$\left( \begin{array}{c} 0 \\ 1 \end{array} \right) = \left( \begin{array}{cc} \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}}\end{array} \right) % \left( \begin{array}{c} \frac{1}{\sqrt{2}} \\ -\frac{1}{\sqrt{2}} \end{array} \right)$$

which corresponds to

$$\left( \begin{array}{c} 0 \\ 1 \end{array} \right) = \left( \begin{array}{cc} 0 & 0 \\ 1 & 1 \end{array} \right) % \left( \begin{array}{c} \frac{1}{2} \\ \frac{1}{2} \end{array} \right)$$

So this is a bit weird: our descriptions of physical operations require knowledge of what these operations are acting on. We can't simply say that the time evolution operator is something simple like $e^{-iHt}$, for example.

But it gets worse.

Consider applying transformations $1$ and $2$ in sequence:

$$|\gamma \rangle = U_1 |\beta\rangle = U_1 U_2 |\alpha\rangle = U_3 |\alpha\rangle$$

Where $U_3 \equiv U_1 U_2$. So you might naively expect that constructing stochastic matrices in this fashion, the following relation would be true:

$$S_1 S_2 = S_3$$

Once more let's consider an example:

$$\left( \begin{array}{c} -\frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}} \end{array} \right) = \left( \begin{array}{cc} -\frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}}\end{array} \right) \left( \begin{array}{cc} \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}}\end{array} \right) % \left( \begin{array}{c} \frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}} \end{array} \right)$$

We already computed $S_2$ above,

$$S_2 = \left( \begin{array}{cc} 1 & 1 \\ 0 & 0 \end{array} \right)$$

Now since

$$\left( \begin{array}{c} -\frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}} \end{array} \right) = \left( \begin{array}{cc} -\frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}}\end{array} \right) % \left( \begin{array}{c} 1 \\ 0 \end{array} \right)$$

We have

$$S_1 = \left( \begin{array}{cc} \frac{1}{2} & 0 \\ \frac{1}{2} & 0 \end{array} \right)$$

So

$$S_3 = \left( \begin{array}{cc} \frac{1}{2} & \frac{1}{2} \\ \frac{1}{2} & \frac{1}{2} \end{array} \right)$$

But let's say you wanted to construct $S_3$ directly.

$$\left( \begin{array}{cc} -\frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}}\end{array} \right) \left( \begin{array}{cc} \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}}\end{array} \right) = \left( \begin{array}{cc} 0 & -1 \\ 1 & 0 \end{array} \right)$$

This seems to imply that $S_3$ should be

$$S_3 = \left( \begin{array}{cc} 0 & 1 \\ 1 & 0 \end{array} \right)$$

And indeed this is a perfectly reasonable choice. It turns out that the only way to ensure your choice of $S$ will be unaffected by arbitrary splittings of the unitary operator $U$ is to make the trivial choice:

$$S = \left( \begin{array}{ccc} |\beta_1|^2& \cdots &|\beta_1|^2 \\ . &...& .\\ . &...& . \\ . &...& . \\ |\beta_1|^2& \cdots &|\beta_1|^2 \end{array} \right)$$

In other words, the only way to define $S$ uniquely is essentially to do the calculation using amplitude vectors and unitary matrices, and then use the answer to construct the trivial $S$.

There are other problems as well but this post is already getting a bit long, so I'll refer to Scott Aaronson's explanation in the context of hidden variable theories: http://www.scottaaronson.com/democritus/lec11.html

So to cut this long story short, the answer is yes. You can avoid talking about amplitudes and unitary transformations and talk only of stochastic matrices and probability vectors. However, the theory you get is rife with nonintuitive, weird features, so you don't gain anything by doing this, and in fact, you lose the ease of calculation and conceptual clarity provided by the orthodox formalism.

The reason why they put equations in is because physics is best described in terms of equations. They tried to remove randomness with equation, ultimately believing that QM is not that random, that it is behaving according to certain parameters. They tried to set the boundaries in the randomness/unpredictability, but the equations still couldn't remove the unpredictability of QM. More powerful equation is needed to make sense of this randomness.

I think there is a certain system in this randomness IMHO.