# Why Psi^2 not square rooted

1. Sep 11, 2010

### sujiwun

As far I can follow it, it seems that Schroedinger's Equation uses the complex plane in a sense to allow the spatial directions to exist on the real axes (reflecting real space) and put the probability amplitude of the wavefunction on an conveniently "invented" imaginary axis - sort of an additional hyper dimension to house the amplitude.

To turn this complex number then into a real probability the Psi wavefunction is squared.

I'd like to know why this result isn't then required to be square rooted, to get the correct real number probability magnitude?

For example.
For two possible states with probability amplitudes summing to 1 on the imaginary axis, they might have the following complex values;

2i/3 and i/3

squaring and taking the moduli gives the following real numbers

4/9 and 1/9 which don't sum to 1, but will if square rooted. Why is this?

I think it has something to do with the Normalization process, where the integral of the Psi^2 between - infinty and +infinity must equal 1. But I don't understand why this is Psi^2 and not just the integral of Psi between + & - infinity.

Last edited: Sep 11, 2010
2. Sep 11, 2010

### alxm

Because the probability is a real-valued number, and the total probability must equal 1. (E.g. in position-space, the particle must have a location somewhere)

The integral of Psi OTOH isn't real-valued, nor has any direct physical interpretation.

3. Sep 13, 2010

### sujiwun

Thanks - perhaps it was a dumb question, or just the way I phrased it was dumb. Perhaps I erred in thinking that the real part of the wave function represented positions and/or momenta whilst the the imaginary part represented the probability amplitude of those energy states through all space and time.

I reconized that psi when muliplied by its conjugate becomes a real number, but on its own is complex, and that it is the integral of psi x psi* between + & - infinity that gives the probability density.

It seems to generally put forward that this was always the case, that this property was built into psi deliberately from the outset.
But from what I have read of the history of its development, it was Max Born who came up with this integral of psi x psi* as the real statistical probability density after Schrodinger had originally introduce it in his famous equation. Schrodinger, apparently didn't allude to this property of psi x psi* and seems to have even taken issue with Born over it.

I am just trying to get my head around the maths (not my strong suit).
So perhaps what I was really asking was, 1) what did Schrodinger originally mean by psi (if not probability plotted on an imaginary axis) when he first employed it in his equation? and 2) How did Born get from Schrodinger's psi to the Born rule?

4. Sep 13, 2010

### Demystifier

He thought that |psi|^2 is the charge density of the electron.

At that time, a probabilistic interpretation of psi seemed to be the only interpretation compatible with experiments. Why |psi|^2 and not |psi|? Because the integral of |psi|^2 over space does not depend on time.

5. Sep 13, 2010

### sujiwun

Ok, so that means the need for the integral of |psi|^2 was in effect built-in from the beginning, and it was the interpretation of what that would eventually mean in real physical terms that changed from being a real value for the actual charge distribution to a probability density for existence.

So, my last questions then is why the original need for psi to be complex within Scrodinger's equation? What benefit did employing the imaginary axis bring, why not use just use another real variable for charge distribution within the Schrodinger equation and not have to go through the process of normalizing |psi|^2?

6. Sep 13, 2010

### Demystifier

No!
The Schrodinger equation has been derived from completely different demands. AFTER that, it was realized that |psi|^2 can be consistently interpreted probabilistically.

7. Sep 13, 2010

### Demystifier

If you want the classical equation
E=p^2/2m
to be obtained as a dispersion relation of a wave equation, then the wave equation must be complex, not real. This is related to the fact that the equation above is linear in E, not quadratic. Indeed, in the relativistic case you have a purely quadratic relation
E^2=p^2+m^2
(with c=1), which can be obtained from the real Klein-Gordon equation.

8. Sep 13, 2010

### unusualname

The Born Rule for interpreting the wave function as a probability density by squaring the modulus can also be justified by considering interference patterns.

In a double slit experiment you get an interference pattern which is not simply the sum of two waves considered independently going through each slit but the square of the modulus of the sum: (sorry, tex formatting seems to have a cache problem, so have to use ascii)

|Psi_ab|^2 = |Psi_a + Psi_b|^2 = |Psi_a|^2 + Psi_a.Psi_b* + Psi_a*.Psi_b + |Psi_b|^2

If it were just the modulus, you would expect the interference to be |Psi_a + Psi_b| but experiments show that is not the case. In fact, a recent experiment claims to also have ruled out any higher order terms (such as cubic terms due to three paths in a triple slit), although it's not clear if this was not already known indirectly:

Ruling Out Multi-Order Interference in Quantum Mechanics

9. Sep 13, 2010

### zonde

I think it's more useful to represent it using http://en.wikipedia.org/wiki/Euler_formula" [Broken].
You have radius in complex plane and phase in complex plane. Just take radius (square of modulus) as real valued and phase as "invented" dimension.
At the end you always take square of modulus. So final result is just changed from sum of squares by interference term of two phases. So take sum of squares as "real" but interference term as "not real" (looking at unusualname's equation).

Last edited by a moderator: May 4, 2017
10. Sep 13, 2010

### Tac-Tics

A good attempt at trying to rationalize the use of complex numbers, but totally wrong ;(

Here's why.

The wavefunction, Ψ, is a complex-valued function on the state space. That means the input is the state of a system (which is the combination of the position, time, energy, momentum, or any "classical" quantities you are modeling). The output is a complex number.

In the case of spin, you have Ψ(s), where s can be one of UP or DOWN (so, Ψ(UP) and Ψ(DOWN) are the only two points on the function, and the whole wavefunction can be summarized in a 2x1 column vector).

In the one-dimensional case, you have Ψ(x), which takes only the position of one particle. Ψ(x) is a function from real numbers to complex numbers.

If you have two particles in 3D space, you have Ψ(x1, y1, z1, x2, y2, z2). Clearly in this second case, you couldn't possibly use the real part to store all the information of BOTH particles in all threes dimensions in a single real number.

So the position (and all other state) are encoded in the parameters (the input) of the wavefunction. It's the output which is complex-valued. The meaning of the output is that it is the amplitude of that state.

And the amplitude is just a souped up version of probability. If you don't like the complex numbers, you can always imagine, instead, the amplitudes are instead arrows (on a plane). They interact in more interesting ways than probabilities. Adding two probabilities together gives you a BIGGER probability. Always. But with amplitudes, adding two together might make it bigger, if they are pointing the same way. It might also make it smaller, it they point opposite each other. Or it might make it a little bigger, but not too much bigger, if they are pointing at sharp angles to each other. (Amplitudes also have more interesting multiplication, which is handy too).

At the end of the day, though, we need a probability. The simplest way to do this is to take the "length" of the final amplitude. Of course, mathematically, it's super-duper easy to get the SQUARED length of a vector (just take the inner product of it with itself). And so nature, being lazy, settled on that.

11. Sep 13, 2010

### unusualname

I realised my explanation referring to interference isn't really answering the question here.

In fact DeMystifier answered it in post #4, the reason you use |Psi|^2 is because the (normalised) integral is constant in time (this is not true for the integral of |Psi|), and conservation of probability is required.

The proof can be found in most standard Quantum textbooks under a section explaining "conservation of probability" or similar, or via google eg http://en.wikipedia.org/wiki/Probability_current#Derivation_of_continuity_equation

12. Sep 14, 2010

### sujiwun

Hmmm, all very confusing, but then QM was never meant to be a walk in the park.

I agree Euler's formula seems the best starting point from the all the other much appreciated explanations. If its good enough for Feynman then its good enough for me...

Richard Feynman called Euler's formula "our jewel"[2] and "one of the most remarkable, almost astounding, formulas in all of mathematics."

I still can't fully appreciate why Schrodinger would go with complex numbers, but this much is obvious.....

psi = e^ia = cos a + isin a where a=kx-wt
then
psi x psi* = (cos a + isin a)(cos a - isin a)
therefore
|psi|^2 = cos^2a+sin^2a = 1 hence a ready made "wave" function with unity built in - so perhaps that was his rationale behind using it.

also
e^ix=cosx+isinx has a fairly straight forward and acceptable proof.

Last edited by a moderator: May 4, 2017
13. Sep 14, 2010

### unusualname

e^ia describes a phase since it is periodic with period a=[0,2pi] and has constant modulus=1.

You also need an amplitude R to describe a simple plane wave as R*e^ia.

For arbitrary waves you add combinations of such plane waves, and in fact, using fourier analysis you can prove that any continuous function is the limit of such sums.

Schrodinger didn't just "go with" complex numbers, he derived his wave equation by using de Broglie and Einstein relations plugged into an equation for a wave described in terms of a fourier integral.

The imaginary i arises by differentiating the e^i... like terms.

Once he had the equation it was not obvious what a complex valued wave represented but it enabled some accurate physical calculations to be made which agreed with experiment.

It was about 1 year later that Born suggested the probabilty amplitude interpretation so that Psi*.Psi = |Psi|^2 is interpreted to be the probability (because it satisfies conservation of probability as explained above)

Attempts to derive a real valued wave function fail for various reasons, there's some analysis in David Bohm's book on Quantum Theory.

14. Sep 14, 2010

### akhmeteli

I disagree based on Shroedinger's article (Nature, v.169, p.538(1952)). See e.g. my post https://www.physicsforums.com/showpost.php?p=1539835&postcount=20 . The same can be done in a general case for the Dirac equation as well.

15. Sep 14, 2010

### unusualname

The Klein-Gordon equation doesn't define an invariant positive-definite probability density and the Dirac equation has 4 complex components, but I've no doubt there are clever ways to construct useful equations so that the complex-valued property isn't explicit. After all the complex nature of the Schrodinger Eqn isn't anything mysterious beyond a mathematical formalism.

16. Sep 14, 2010

### planck42

The probability amplitudes do not have to add up to 1 in any plane. For a 50/50 system e.g. a coin, the amplitudes associated with its states sum to $$\sqrt{2}$$, but the probabilities sum to 1 like they should.

17. Sep 14, 2010

### sujiwun

I disagree. Would you suggest that Schrodinger used de Broglie and Einstein and from that, just happened to stumble upon Euler's equation as his wave function in the process of formulating his own wave equation?

Nah, obviously he took Euler's formula and ran with it - he needed a generic wave function, and crow barred it together with the Hamiltonian for a classical oscillator. Pretty slick, but probably just throwing what he thought might work together and seeing what stuck.

18. Sep 14, 2010

### akhmeteli

That does not mean that "Attempts to derive a real valued wave function fail"

Again, why "Attempts to derive a real valued wave function fail"?

19. Sep 14, 2010

### unusualname

Er, ok, attempts to derive a real valued wave function with the properties enjoyed by Schrodinger's complex wave function fail for various reasons, such as not providing a suitable probability density eg see sec 4.5 in D Bohm's Quantum Theory.

20. Sep 14, 2010

### unusualname

Well actually, Schrodinger's original formulas didn't have a complex valued wave function, and were derived from his masterful knowledge of classical hamilton-jacobi theory and differential equations, I think he only introduced the modern complex-valued wave equation in a later paper published the same year (1926).

The derivation was quite imaginative (and much more involved than I suggested above), you can read all about it in a historical survey such as J. Mehra's 'The Golden Age of Theoretical Physics' (Vol 2)