# How to go from Heisenberg operators to Schrödinger operators

1. Sep 15, 2013

### snoopies622

It is obvious to me how

$$\hat {x} = x; \hspace{5 mm} \hat {p}_x = -i \hbar \frac {\partial} {\partial x}$$

implies

$$[ \hat {x} , \hat {p}_x ] = i \hbar$$

and I can accept that these two formulations are mathematically equivalent, but I do not know how in general (or even in this specific case) to go in the opposite direction, that is, to start with operators defined solely in terms of their commutation relations and change them into forms which show what they do to wave functions in the position basis.

Is this always possible, and if so, how?

2. Sep 15, 2013

### atyy

Last edited by a moderator: May 6, 2017
3. Sep 15, 2013

### Ravi Mohan

Consider a state vector $|\alpha\rangle$.
Obtain its expansion
$$|\alpha\rangle = \int dx\langle x|\alpha\rangle |x\rangle .$$
Apply commutator operator
$$[ \hat {x} , \hat {p}_x ]|\alpha\rangle = [ \hat {x} , \hat {p}_x ]\int dx\langle x|\alpha\rangle |x\rangle$$
and take its projection on $|x'\rangle$
$$\langle x'|[ \hat {x} , \hat {p}_x ]|\alpha\rangle = \langle x'|[ \hat {x} , \hat {p}_x ]\int dx\langle x|\alpha\rangle |x\rangle .$$
On LHS you have
$$\langle x'|[ \hat {x} , \hat {p}_x ]|\alpha\rangle = i \hbar \alpha (x').$$
On RHS you have
$$\langle x'|[ \hat {x} , \hat {p}_x ]\int dx\langle x|\alpha\rangle |x\rangle = \int dx\langle x|\alpha\rangle \langle x'|\left( \hat {x} \hat {p}_x - \hat {p}_x \hat {x} \right)|x\rangle ,$$
$$\int dx\langle x|\alpha\rangle \left(x' \langle x'| \hat {p}_x|x \rangle - x \langle x'|\hat {p}_x |x\rangle \right).$$
This means
$$x' \langle x'| \hat {p}_x|x \rangle - x \langle x'|\hat {p}_x |x\rangle = i \hbar \delta (x - x') .$$
Now consider the operation of momentum operator on the state and its projection on position basis
$$\langle x'| \hat {p}_x|\alpha\rangle = \int dx\langle x|\alpha\rangle \langle x'|\hat {p}_x|x\rangle .$$
Thus
$$\langle x'| \hat {p}_x|\alpha\rangle = -i \hbar \int dx\langle x|\alpha\rangle \frac{\delta (x - x')}{x-x'}.$$

There is better formulation in terms of translation operator which also proves the commutation relation. You can check it in J.J Sakurai chapter 1 or (if you don't have access to it) http://www.hep.upenn.edu/~rreece/do...entum_operator_in_position_representation.pdf

[EDIT]
For $\hat{x}$ it is trivial.
$$\hat{x}|x\rangle = x|x\rangle$$
Take projection on $|x'\rangle$
$$\langle x'|\hat{x}|\alpha\rangle = x'\alpha (x')$$

Last edited: Sep 15, 2013
4. Sep 15, 2013

### snoopies622

Thanks guys, great stuff. Wonderful detail.

I was wondering if, in addition to the position / momentum case, this sort of thing is possible with any pair of operators. That is, given only their commutation relations, can they be changed into forms like those shown in the second line of the OP (the first line of equations)?

Last edited: Sep 15, 2013
5. Sep 15, 2013

### wotanub

It's interesting to note that historically, Heisenberg was not motivated by the idea of a wavefunction. The idea of matrix mechanics was to write a formulation of quantum mechanics that had an obvious connection to things we actually measure (e.g. Energy as opposed to something called a "wavefunction.") He started wth something really abstract (states) and went to the ultimate concreteness. He along with Pascal, Jordan and Dirac really put linear algebra in the spotlight as being important to physics, which in my mind at least seems more impressive that just using differential equations "again." (To be fair, there were some weird imaginary things in the SE that was kinda different.)

6. Sep 16, 2013

### strangerep

One needs a little more meat in the assumptions. Note that the exposition above works because the (generalized) eigenvectors of the (self-adjoint) $\hat x$ operator are assumed to span the (rigged) Hilbert space in a suitable way. Hence any state can be expressed in terms of them, and that (potentially) allows other operators in the commutator algebra to be represented as above.

More generally, we start with a dynamical algebra of (in general noncommuting) observable quantities, then find a maximal mutually-commuting subset therein, then find their spectrum and construct a (rigged) Hilbert space in terms of the associated common eigenvectors of those observables. Then verify that the rest of the observables in the dynamical algebra can also be represented as operators on the space so constructed.

HTH.

7. Sep 16, 2013

### Ravi Mohan

Plus

To see more examples see J.J Sakurai ch-3 (Theory of Angular Momentum). The operators used are $\hat{L}_z$ and $\hat{\phi}$.

[EDIT]
If position operator is self-adjoint then its eigen vectors do span the Hilbert space (spectral theorem). I don't think there is a need for this assumption seperately (or am I missing something?).

Last edited: Sep 16, 2013
8. Sep 16, 2013

### snoopies622

9. Sep 17, 2013

### strangerep

I was just trying to phrase things cautiously. For unbounded operators like $\hat x$, its (generalized) eigenvectors can't be normalized, hence they're outside the Hilbert space, the standard spectral theorem doesn't apply without extra care, hence I added the qualifier "rigged", and then one can rely on the Gel'fand-Maurin nuclear spectral theorem. But extending operators continuously to such a Gel'fand triple built from a nuclear space involves a lot more math if performed rigorously.

So when I said "assumed", I meant "assume that all this has already been done".

10. Sep 17, 2013

### strangerep

For the EM field, one can take the $A^\mu$ potential as a canonical configuration variable, and the electric field as its canonically conjugate momentum. (Actually, one must take more care to handle the EM gauge freedom suitably, but the basic idea is still to arrive at canonical variables satisfying Poisson bracket relationships similar to the usual position and (ordinary) momentum variables. Then one passes to a unitary representation of this algebra via operators on a Hilbert space, and off we go...

(Translation: I'm not sure whether you still have a question.)

11. Sep 17, 2013

### snoopies622

All set for now, thanks strangerep. :)