Pros and cons of Heisenberg and Schrodinger (and Feynman) pictures?

In summary: Stephen Hawking was still alive and teaching at Harvard. But this professor seems fairly credible to me.In summary, the professor believes that the Heisenberg picture is more useful than the Schrodinger picture and that this is due to the relativistic context. He also provides some rules of thumb that have been found by many to be useful.
  • #1
AndreasC
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So there's this professor who insists that the Heisenberg picture is all the rage and much superior in most ways to the Schrodinger picture. He compares it to how you don't use the Hamilton-Jacobi formulation of classical mechanics as much as the Hamiltonian one.

Alright, I can buy it. I looked on wikipedia and there was this quote about how it is "in some ways more natural" especially in "relativistic contexts". But no examples. I checked some older posts here, and there were some people who were saying there are some problems where one is more convenient than the other. But again no examples.

So, the result is that I am still not sure where one picture is more useful than the other and why. I know the Lagrangian (Feynman) formulation is convenient in some problems for finding the propagator. But it's a bit hard for me to see why choosing between Heisenberg or Schrodinger would provide a significant advantage.
 
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  • #2
It depends on the application, which picture of time evolution is most convenient. In usual non-relativistic QM all the pictures are equivalent mathematically. So there's no a priori preferred picture.

It's a bit like gauge invariance in electrodynamics. For some purposes the Lorenz gauge is more convenient, for others the Coulomb gauge. The physics is of course the same.
 
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  • #3
vanhees71 said:
It depends on the application, which picture of time evolution is most convenient. In usual non-relativistic QM all the pictures are equivalent mathematically. So there's no a priori preferred picture.

It's a bit like gauge invariance in electrodynamics. For some purposes the Lorenz gauge is more convenient, for others the Coulomb gauge. The physics is of course the same.
Yes, but in what kind of application is one preferred over the other?
 
  • #4
It is not about the application; it is about what makes the math simpler and easier to understand. You can transform between the two pictures at any time so there is nothing stopping you from using both in the same problem. There are also other pictures that are useful; e.g. the interaction pictures.

If it a bit similar to asking whether you should be using a Cartesian or spherical coordinate system when solving problem. Saying that one is "better" than the other is a bit silly in my view.
 
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  • #5
f95toli said:
It is not about the application; it is about what makes the math simpler and easier to understand. You can transform between the two pictures at any time so there is nothing stopping you from using both in the same problem. There are also other pictures that are useful; e.g. the interaction pictures.

If it a bit similar to asking whether you should be using a Cartesian or spherical coordinate system when solving problem. Saying that one is "better" than the other is a bit silly in my view.
Yes, but I can tell you a lot of cases where spherical coordinates are simpler to use than cartesian ones, and vice versa (for instance, problems with spherical symmetries are usually easier to deal with using spherical coordinates), while I haven't been shown so far something analogous for the Heisenberg and Schrodinger pictures. This is why I am asking.
 
  • #6
AndreasC said:
But it's a bit hard for me to see why choosing between Heisenberg or Schrodinger would provide a significant advantage.
Here are some rules of thumb acquired by a lot of experience:
- If you want to know how average value of some observable changes with time, use Heisenberg picture.
- If you want to know how probability of a given value of some observable changes with time, use Schrodinger picture.
 
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  • #7
Demystifier said:
Here are some rules of thumb acquired by a lot of experience:
- If you want to know how average value of some observable changes with time, use Heisenberg picture.
- If you want to know how probability of a given value of some observable changes with time, use Schrodinger picture.
Thanks, that's the kind of answer I expected to see!

One thing that professor said though was that there were many problems really hard to solve with Schrodinger that were easier with Heisenberg, and it sounded like he meant specific problems. Like, he said something about how Schrodinger was only good for a small set of problems, like what you learn in a first QM course, and a few others, and beyond that the Heisenberg picture is more convenient. It is hard for me to see so far why that is since I have very limited experience. As I mentioned before, he also mentioned how relativistic QM is a lot more obvious with Heisenberg.

Now, I have had some professors who said a lot of nonsense. I've had an optics professor who literally didn't remember the form of the divergence operator as well as lots of other things and made tens of mistakes, and kept asking us, the audience, for confirmation. I've had a chemistry professor who apparently thought Stephen Hawking had discovered some sort of Theory of Everything (it wasn't just an off-handed comment, it was part of the power point slides and even had a nonsense date next to the claim, I guess Hawking ended physics in the 80s) and thought Gell and Mann were two different people. I've even had a professor of statics and mechanics of materials who thought the only way you could rotate something is if not one but two forces were applied on it, he literally thought if you threw say, a ball against a broomstick in a frictionless environment the broomstick would not rotate at all no matter where you hit it, which is just... Mind boggling to me how you can become professor in his field without understanding that sort of stuff. Well, not exactly mind boggling, I know how he in particular became a professor (and even a dean) but that is another story and it's not for here.

Bottom line is, I am very skeptical of the things they say sometimes. Although this particular professor doesn't seem to be that way, his lectures are very good usually. But I got to ask for what other people think about it, and I didn't find much elsewhere.
 
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  • #8
For me the Feynman picture gives semiclassical approximations particularly for (non relativistic) scattering theory that are more intuitive.
 
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  • #9
What is the Feynman picture? If you mean the path integral, it's picture independent.
 
  • #10
I am afraid I am not sufficiently conversant in the fine points of nomenclature. I meant the path integral formulations and the use of eikonal approximations via stationary phase techniques.
If you would indulge me, what actually do you mean by "picture"?
 
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  • #11
The "picture" (or more precisely "picture of time evolution") is the choice of how to distribute the "mathematical time dependence" between the states (statistical operators/state vectors) and the operators that represent observables. The physics is of course invariant under changing the choice of the picture, i.e., you can transform from one picture to another by a time-dependent unitary transformation.

"Wave functions", i.e., if you have the state ket ##|\psi(t) \rangle## and a complete orthonormal set of eigenvectors ##|u_{\lambda}(t) \rangle##, the wave function ##\psi(t,\lambda)=\langle u_{\lambda}(t)|\psi(t) \rangle## is independent of the picture.
 
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  • #12
Thank you. That did in fact awaken some dormant neurons. It turns out they were only sleeping !
 
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Related to Pros and cons of Heisenberg and Schrodinger (and Feynman) pictures?

1. What is the difference between the Heisenberg, Schrodinger, and Feynman pictures?

The Heisenberg, Schrodinger, and Feynman pictures are three different mathematical frameworks used in quantum mechanics to describe the evolution of a quantum system. In the Heisenberg picture, the operators are time-independent while the state vectors are time-dependent. In the Schrodinger picture, the operators are time-dependent while the state vectors are time-independent. In the Feynman picture, both the operators and state vectors are time-dependent.

2. What are the advantages of using the Heisenberg picture?

The Heisenberg picture is advantageous because it allows for easier calculations and predictions of the evolution of a system. This is because the operators are time-independent, making it easier to solve equations and make predictions without having to consider the time dependence of the state vectors.

3. What are the advantages of using the Schrodinger picture?

The Schrodinger picture is advantageous because it allows for a more intuitive understanding of the evolution of a system. The time-independent operators make it easier to visualize the behavior of the system over time, and the time-dependent state vectors make it easier to see how the system changes over time.

4. What is the significance of the Feynman picture?

The Feynman picture is significant because it combines the advantages of the Heisenberg and Schrodinger pictures. It allows for easier calculations and predictions like the Heisenberg picture, while also providing a more intuitive understanding like the Schrodinger picture. It is also used in quantum field theory to describe the evolution of interacting particles.

5. Can the three pictures be used interchangeably?

Yes, the three pictures can be used interchangeably as they all describe the same physical system. However, depending on the problem at hand, one picture may be more convenient or intuitive to use than the others. It is important to understand the differences between the pictures and choose the appropriate one for the specific problem being studied.

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