I often think of basic of QM (although it wasnt discovered that way) is that we have a physical parameter/observable and the favourite one is the displacement $x$ of a particle $P$. Its conjugate variable is its velocity $v$ (although they are vectors, i am dropping the hat for convinience and also stick to 1 dimension.

Classical Mechanics says that given $x$ and $v$ at say time $t=0$, and the potential $V$, we can determine $x$ and $v$ at all other times $t>0$. It means we can determine the time evolution of the displacement and velocity of the particle.

In quantum mechanics, we don't have any definite values for $x$ and $v$ of the particle. At any given time say $t=0$, all we know is that there is a state function $\Psi(x,0)$ whose magnitude gives probability density function for $x$ and its Fourier transform gives probability density function for $v$ at any time $t=0$, to begin with. We are not allowed to ask for exact values of $x$ and $v$, as they are not part of the theory. Given the potential $V$, the Schrodinger equation determines the time evolution of the state function, that is $\Psi(x,t)$ at all future times. Thats about the basic difference...

This QM gives a very accurate description of our world at microscopic level.

My question :

In QM and CM we begin with the observables $x$ and $v$ at any given time, and then make a law for their time evolution.

Why cant we begin with $x(t)$, the displacement as a function over $t$ as the observable, meaning we don't observe $x$ at only a given time, but over a period of time. That is our observable and thats what we intend to measure. We don't measure displacement at only a given time, but over a preiod of time. instead of single value for displacement, we measure a function over time. In this formulation, our observable is not a value but a function/signal.

Here again we have classical and probabilistic formulations. In classical world given $x(t)$, we can uniquely determine $v(t)$ and potential $V$.

How can we formulate this in a quantum mechanical way? Would this lead to a new useful theory?

  • $\begingroup$ You can measure the position of particles in quantum mechanics all day long, as often as you like. Why would you think that one can not? Each measurement will simply change the momentum of the particle and screw up your next measurement. If you do two consecutive precision measurements of x on the same particle, dx won't tell you much about the particle's environment, it will simply tell you by how much your measurement has changed the quantum state of the system, which, unfortunately, is not what you are after. There are better measurements, than that, in QM, but not for x and p. $\endgroup$
    – CuriousOne
    Sep 8, 2014 at 8:09

2 Answers 2


We are very well allowed to ask for exact values of $x$ and $p$. We can just measure them. What we can't expect is that if we reprepare the same state, we'll get the same answer, if we measure again. We can also measure the position and the momentum and will obtain a specific answer - once again, if we reprepare the same state, we'll not get the same answers, no matter how much effort we put into preparation (and measurement). Your basic "Heisenberg Uncertainty" only tells you that the product of the variances of $x$ and $p$ measurements is bounded from below.

By the way: This is not the same as saying that a position measurement disturbs the momentum (also known as Heisenberg microscope). Maybe it does, but that's nothing you can conclude from the usual Heisenberg uncertainty, since that statement doesn't involve measurements (it's about preparations).

Bearing this in mind, you don't need new schemes for measurement. Also, how do you measure a function, if not by measuring at certain points and interpolating? Maybe the formalism you want is somewhat incorporated by what is known as "weak measurements", but I don't know enough about this highly controversial topic to elaborate.

  • $\begingroup$ "We are very well allowed to ask for exact values of x and p. We can just measure them." I really dislike statements like this. What does it mean that you can "ask for" something but not "measure" it? $\endgroup$
    – DanielSank
    Sep 8, 2014 at 20:43
  • $\begingroup$ I don't quite get your comment, sorry - could you elaborate? I'm just saying that you can indeed measure exact values for x and p for a given instance of a state. $\endgroup$
    – Martin
    Sep 8, 2014 at 22:09
  • $\begingroup$ Oh, jeez. I misread your post. Nevermind. $\endgroup$
    – DanielSank
    Sep 8, 2014 at 22:22
  • $\begingroup$ @DanielSank: No problem! I was just wondering. $\endgroup$
    – Martin
    Sep 9, 2014 at 6:51

What you are proposing is known as the Heisenberg picture of quantum mechanics which had a primitive formulation even before the Schrödinger formulation in the form of Matrix mechanics. The linked wikipedia article is very well written, so I think me giving a detailed description would be redundant.

In this picture, all the evolution is transformed into the observables and you input only the initial state to find out distributions of observables after a certain time-evolution. The observables such as $x(t)$ are however not simple functions, but "matrices" or operators evolving in time. However, with some clever manipulation similar to the ones shown in the last paragraph of the wikipedia article, you can get all the observational information you need without ever referring to a state vector.

  • 1
    $\begingroup$ He's not so much describing the Heisenberg picture as the path integral formulation. Note the emphasis in the question on regarding $x(t)$ as the object which the theory describes. $\endgroup$
    – DanielSank
    Sep 8, 2014 at 18:56
  • $\begingroup$ Well, I think Rajesh should comment on that. The path integral crossed my mind but the sentence "we measure a function over time" felt like a similar stream of thought as the original matrix-mechanics motivations. In the case of a path integral we have $x(t)$ as an important object but our resulting "observable" is only the propagator. I.e. $x(t)$ isn't a signal we measure. I think the closest analogy of a measurable signal would be $\langle X(t)\rangle$ as provided by the Heisenberg picture. $\endgroup$
    – Void
    Sep 8, 2014 at 19:20
  • $\begingroup$ A downvote because you do not agree with the interpretation of the question? Really? $\endgroup$
    – Void
    Sep 8, 2014 at 19:21
  • $\begingroup$ Don't make assumptions about who votes in various ways. I didn't down vote anything. $\endgroup$
    – DanielSank
    Sep 8, 2014 at 20:07
  • $\begingroup$ Sorry, that wasn't meant in your direction. I was just writing the comment and at the same moment a downvote to my answer and the upvote of your comment appeared, so I assumed the voter shared your objection. $\endgroup$
    – Void
    Sep 8, 2014 at 20:26

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