# What is wavefunction

1. Feb 20, 2009

### electrogluon

i am a amature at quantum mechcanics unfortuatly.
i need to understand wavefunction but i cannot find a site that explains it fully
does anyone have any suggestions?

2. Feb 20, 2009

### QuantumBend

Yes, wavefunction inner tenser prodact of hilberts spacers - you must look mathematic first or no good for you. Wavefunctions have states and observerables. We say many times this.

Last edited: Feb 20, 2009
3. Feb 20, 2009

### sirchasm

Look for papers or online books for non-physicists; the QIS stuff can be a easier path - I found a few that take you through the tensors in classical theory (binary logic) and how to shift into tensors in Hilbert space; a vector is a vector even if its always 2-valued. So you start with classical probability, and move to quantum probability (you can use coin-flipping too).

Try googling "quantum information wavefunction beginners non-physicist" or something.

4. Feb 20, 2009

### wofsy

I don't know sites but I can tell you the books I am learning from. Feynmann's Lectures on Physics Book3, The Berkeley Course in Physics, Quantum Physics, and a highly mathematical treatise called Quantum Mechanics by cohen-tannoudji, diu, and laloe. The first two books motivate Quantum mechanics from experimental data and various specific situations. They are great for intuition and will tell you what a wave function is. The third book is formal and dense for me something to work into rather than startout with.

5. Feb 20, 2009

### Fredrik

Staff Emeritus
A wavefunction is a member of a Hilbert space, and a Hilbert space is a vector space with an inner product that satisfies a specific condition. (All Cauchy sequences must be complete in the norm associated with the inner product). Tensor products don't have anything to do with this, unless you want to combine two Hilbert spaces into a new one, e.g. to construct the space of two-particle states from the space of one-particle states. The words "wavefunction" and "state" are usually used interchangeably, but it makes more sense to define the state corresponding to a normalized wavefunction f as the set of all exp(it)f where t is a real number.

6. Mar 3, 2009

### Karl G.

In classical mechanics, the state of a system is given by 6 variables: its velocity and position coordinates. In QM, the state is given by an infinite number of parameters: the value of its wavefunction at each point. Although his doesn't answer you question, it is worth noting.

7. Mar 3, 2009

### v2kkim

It should be very difficult to have an idea of wave function without actually solving some Schrodinger equation. My work experience is on electron transport property in semiconductors, where we get electron movement and current etc, by solving quantum mechanical equations using computer, and we treat electron as a wave function not a particle mostly which results in pretty good prediction to actual measurement.

8. Mar 3, 2009

### alxm

That's not right. There aren't an 'infinite' number of parameters. That's like saying a classical wave requires an 'infinite' number of parameters. (which it does not)

The Schrödinger equation is just a differential equation, and like any differential equation, its solutions are characterized by a set of eigenvalues.

9. Mar 4, 2009

### Karl G.

Hmm.... That is what my QM text says (Cohen- Tannoudji, pg. 39). Or is my wording imprecise? Or is the book wrong?

Last edited: Mar 4, 2009
10. Mar 4, 2009

### QuantumBend

NO, book not wrong - you wrong.

11. Mar 4, 2009

### alxm

The book isn't wrong, but I'd say you've generalized what it was saying beyond what was intended.

The issue with this is that it's a classical analogy. And an analogy to particle systems, treated as particles, with definite locations etc. And in that analogy to location there is an 'infinite' number of locations at which the particle might exist. But the thing is, while the classical solution can be characterized in terms of location etc, the quantum mechanical one is not. It's characterized in terms of the solutions to the Schrödinger equation, which in turn are characterized by quantum numbers/eigenvalues.

Just to give an example, if you consider an electron moving around a fixed nucleus (a single-particle system), then the quantum mechanical state can be specified in only three parameters; n, l, m. (and optionally, s).

12. Mar 4, 2009

### Fredrik

Staff Emeritus
I agree completely with what Karl said in #6. Alxm, what you're saying about how a state is characterized by eigenvalues is correct too, but you need to specify an infinite number of them to specify an arbitrary state. So that stuff about eigenvalues doesn't contradict #6 in any way.

13. Mar 4, 2009

### alxm

The text wasn't talking about arbitrary states. It was talking about single-particle systems. In the case of a single particle in a potential, there's a finite number of bound states. To continue that example, give me the n,l,m numbers for an electron orbiting a point charge and I can tell you the the location probability distribution, the energy, momentum, angular momentum, and any other property. The state is entirely characterized by those three values.

On the other hand, knowing the infinite number of values for |psi|^2 doesn't really give me that information. It does not characterize the state.

14. Mar 4, 2009

### Fredrik

Staff Emeritus

15. Mar 4, 2009

### alxm

He was quoting a passage from Cohen-Tannoudji.

16. Mar 5, 2009

### Karl G.

Now that I have pondered over it, Cohen- Tannoudji was probably referring to the need of the (squared modulus of the) wavefunction to determine a particle's probable location. So his statement is most likely referring to the determinism of classical mechanics vs. the probabalistic interpretation of QM.

17. Mar 6, 2009

### bunburryist

Conceptually, does it make sense to think of the wave aspect of quantum physics as a mathematical way of moving around and developing the probability density? In other words, where as the probability density has a direct relation to reality in the sense that it allows us to make statistical predictions of actual experiments (the odds of an electron being found at a given location, etc.), is the wave aspect just a way of manipulating that density and developing it? Or can we think of it as a mathematical geometric entity that provides a structure of the probability density allowing us to define that density by location and time?

As an analogy, imagine if we were creating a mathematical model of the thickness of an expanding balloon in time. The wave equation would be analogous to the mathematical model of the expanding balloon, and the probability density would be analogous to the thickness value. The expanding balloon aspect (wave equation) would provide a mathematical structure upon which the thickness value (probability density) would ride. Does this make any sense?

Last edited: Mar 6, 2009
18. Mar 6, 2009

### alxm

No, because the spatial probability density is just a single measurable aspect of the system. The wave function contains information about every measurable aspect; momentum, angular momentum, spin, etc.

The ground-state energy for some systems can be determined from the particle density/densities. (The Hohenberg-Kohn theorem) Which is why density functional theory exists. The problem in that case is that for most real systems, nobody knows exactly what that functional is. (It's something of the Holy Grail of Quantum Chemistry)