# Physical interpretation of operators in QM?

1. Jul 16, 2010

### jang

hi, I am a novice to quantum mechanics and get a lot of troubles with operators. I cannot explain why:

- why QM uses operators for observables such as position, momentum, energy, ..ect, but classical physics does not?
- what are physical interpretations of operators?
- why are operators needed to quantize observables?

2. Jul 16, 2010

### alxm

Welcome to PF, Jang.

Well, there are two equivalent formalisms here. One is the operator formalism introduced by Dirac and the other is the matrix formalism that Heisenberg invented. The two are equivalent, and once you get deeper into QM you'll be expected to be know both.

Operators and matrices in themselves have no physical meaning; they're just mathematical objects. But the reason why we use them has a very important physical reason: The uncertainty principle. In quantum mechanics, unlike classical mechanics, you can't necessarily determine two different properties at the same time. For instance, position and momentum. So if you determine a particle's position and then its momentum, you will not get the same result as if you first determined the momentum and then position.

Mathematically, this is represented by non-commutivity. Now, if you consider which mathematical objects exist that don't commute, obviously ordinary variables is out of the question (x*p = p*x). But operators have the property that they don't necessarily commute. I.e. $$\hat{A}(\hat{B}f)$$ may or may not equal $$\hat{B}(\hat{A}f)$$, depending on the operators. Analogously, matrix multiplication does not necessarily commute either.

So by using these mathematical objects to describe the system, we're allowing for the fact that different pairs of observables cannot necessarily be determined simultaneously.

3. Jul 17, 2010

### unusualname

At a simple level, in classical mechanics kinetic energy is given by:

$$E = mv^2/2$$

or in terms of momentum:

$$E = p^2/2m$$

in quantum mechanics you have;

$$\hat{E}\Psi = \hat{p}^2\Psi/2m$$

where $$\hat{E}$$ and $$\hat{p}$$ are linear (differential) operators given by the Schrodinger eqn. (they "operate" on the wave function $$\Psi$$)

The possible values of observables are now eigenvalues of the operators, which may be discrete values rather than continuous values as in the classical theory

4. Jul 17, 2010

### Fredrik

Staff Emeritus
The best discussions about observables and operators can be found in books on the mathematics of QM, like Araki and Strocchi. These are pretty advanced books, but the stuff about observables is at the beginning and should be readable even if you haven't studied functional analysis.

5. Jul 17, 2010

### Hurkyl

Staff Emeritus
"Physically", you have a product: given an observable X and a state $\rho$, you combine them to produce the "expected value" of X on the state $\rho$, notated as $\rho(X)$ or maybe $\langle X \rangle_\rho$.

There is an abstract, purely mathematical construction that effectively takes the "square root" of states to simplify arithmetic. The result of this is that states can be represented as vectors -- as kets -- and operators operate on them. If $|\rho\rangle$ is the ket in some Hilbert space representing the state $\rho$, then the expected value of X is computed by $\rho(X) = \langle \rho | X | \rho \rangle$.

As far as systems of arithmetic go, linear algebra is one of the simplest, and so this form is useful for doing calculations, making definitions, and so forth. And, historically, that's how things were originally worked out.

An example of a state space described without using this construction is the Bloch sphere (and ball) for a qubit.

You can do this with statistical mechanics too. But the linear algebra is degenerate -- in the right basis, all of your operators would be represented by diagonal matrices. So we don't bother invoking any of this and just stick to straight probability theory.

Last edited: Jul 17, 2010
6. Jul 17, 2010

### unusualname

well, if you want to get swamped in mathematical formalism they might be good books, but they hardly give an intuitive explanation.

After de Broglie suggested, ~1923, that a wave property can be associated with matter it became necessary to recast the old classical models to fit this wave property.

Rather than have equations for point particles you needed an equation to describe a wave. These equations involve differential operators, hence the introduction of operators.

The fact that the operators are linear allows for the whole bundle of mathematical theory of linear operators to be applied, and out of this arose an elegant rulebook for quantum mechanics.

But the physical motivation is that matter has a wave property that can been modelled by a linear differential equation.

That's First Quantisation - where you go from real-valued observables to operators. An even more sophisticated mathematical model has been developed by quantising fields (Second Quantisation)

In the end, these models predict probabilistic outcomes for what is essentially a wave phenomenon.

7. Jul 18, 2010

### orienst

Why we use arrow for vector? I think it’s only am instrument. Use the appropriate tools to make work clear, that’s what we should do when we learn physics. We mustn’t use operators, we can use matrix instead.

8. Jul 18, 2010

### Fredrik

Staff Emeritus
I would say that they are the only ones that do.

9. Jul 18, 2010

### unusualname

Well, I think i did a better job in two posts of explaining how operators enter our physical description of the world than they do, but obviously they give a fine explanation of how to apply that in detail.

10. Jul 19, 2010

### jang

thanks a lot. I've started get the sense of operators and observables.

11. Jul 20, 2010

### Fredrik

Staff Emeritus
I strongly disagree. The issue of "how operators enter our physical description of the world" is one of the areas where these "mathematics of QM" books do a much better job than any of the "QM" books.