Hamiltonian defined as 1st derivative

  • #1
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Main Question or Discussion Point

Why is Hamiltonian defined as 1st derivative with respect to time ? From the units of energy (kgm2s-2) I would expect it to be defined as 2nd derivative with respect to time.

(I'm reading http://feynmanlectures.caltech.edu/III_11.html#Ch11-S2)
 

Answers and Replies

  • #2
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In the unlabelled equation between 11.19 and 11.20, notice the presence of ##\hbar##. The reduced Planck constant has units ##Js##, which combine with the ##s^{-1}## from the derivative to leave just ##J##, the units of energy.

Explicitly, we have ##[\hbar]=kgm^2s^{-2} \cdot s=kgm^2s^{-1}##. Multiplying this by the ##s^{-1}## from the derivative gets us back to ##kgm^2s^{-2}##, the units of energy.

If you're ever in doubt about something like this, a bit of dimensional analysis like I've done here will usually solve your problem.
 
  • #3
141
3
In the unlabelled equation between 11.19 and 11.20, notice the presence of ##\hbar##. The reduced Planck constant has units ##Js##, which combine with the ##s^{-1}## from the derivative to leave just ##J##, the units of energy.

Explicitly, we have ##[\hbar]=kgm^2s^{-2} \cdot s=kgm^2s^{-1}##. Multiplying this by the ##s^{-1}## from the derivative gets us back to ##kgm^2s^{-2}##, the units of energy.

If you're ever in doubt about something like this, a bit of dimensional analysis like I've done here will usually solve your problem.
My question was deeper than that. If we had a s-2 from a second derivative then the units of the Planck constant would be adjusted accordingly.
 
  • #4
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Why is Hamiltonian defined as 1st derivative with respect to time
See Chapter 3 Ballentine - QM - A Modern Development. It will take you a day or so to carefully go through the math.

Its got to do with deep symmetry considerations you can only see in the math. Basically when you chug through it you end up with the Hamiltonian of classical physics in exactly the same form hence can be the only reasonable energy operator in QM.

That also explains where Planks constant comes from - it's basically just a conversion between units. For example, from symmetry alone, you end up with p=mv where m is some constant. Planks constant simply converts from that constant which naturally occurs in QM to our usual units of mass. But to fully understand it you need to spend the time going through the whole chapter.

Thanks
Bill
 
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  • #5
samalkhaiat
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Why is Hamiltonian defined as 1st derivative with respect to time ?
Because that is consistent with the definition of the Hamiltonian in classical mechanics. Have you seen the Hamilton-Jacobi equation: [tex]H = - \frac{\partial S}{\partial t} .[/tex] If you multiply both sides with the non-zero function [itex](i / \hbar) \exp ( i S / \hbar )[/itex], you find [tex]\frac{i}{\hbar} H e^{i S / \hbar} = - \frac{\partial}{\partial t} e^{ i S / \hbar } ,[/tex] or [tex]( H - i \hbar \frac{d}{dt} ) e^{ i S / \hbar} = 0 .[/tex]
From the units of energy (kgm2s-2) I would expect it to be defined as 2nd derivative with respect to time.
But there is also [itex]m^{2}[/itex], so why not [itex]x^{2} \frac{d^2}{d t^2}[/itex]? The choice of units is arbitrary. I can choose to measure the energy in [itex]kg[/itex], so can you tell me how you “define” the Hamiltonian in this case?
 
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  • #6
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Because that is consistent with the definition of the Hamiltonian in classical mechanics. Have you seen the Hamilton-Jacobi equation: [tex]H = - \frac{\partial S}{\partial t} .[/tex] If you multiply both sides with the non-zero function [itex](i / \hbar) \exp ( i S / \hbar )[/itex], you find [tex]\frac{i}{\hbar} H e^{i S / \hbar} = - \frac{\partial}{\partial t} e^{ i S / \hbar } ,[/tex] or [tex]( H - i \hbar \frac{d}{dt} ) e^{ i S / \hbar} = 0 .[/tex]
I'm not very familiar with classical mechanics and haven't seen that equation before. I did some searching and found the following link interesting: https://en.wikipedia.org/wiki/Hamil...relationship_to_the_Schr.C3.B6dinger_equation. However, The Schrödinger equation is chapter 16-5 in the Feynman lectures and I'm not there yet.

But there is also [itex]m^{2}[/itex], so why not [itex]x^{2} \frac{d^2}{d t^2}[/itex]? The choice of units is arbitrary. I can choose to measure the energy in [itex]kg[/itex], so can you tell me how you “define” the Hamiltonian in this case?
Not really. I just wanted to get some insight to why it is a first derivative and the second derivative was just a wild guess with very limited understanding.
 
  • #7
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For example, from symmetry alone, you end up with p=mv where m is some constant.
So... am I completely lost if I take that the Hamiltonian actually describes a momentum and you need to take the vector modulus and square it to get energy.

I should probably get the Ballentine book.
 
  • #8
Just half an hour before seeing this post, I happened to be reading Ulf Leonhardt's Quantum Optics Sec. 1.2.3 where he tries to motivate Schroedinger's equation by saying this:
Ulf Leonhardt said:
As the state vector should describe the complete physical information about the state, including its fate in coherent evolution processes, the equation must be of first order in the parameter t. Otherwise derivatives of ##|\psi>## with respect to t would be needed to completely define the initial conditions. So we are led to the differential equation: ##i\hbar\frac{d∣Ψ⟩}{dt}=\hat{H}|\psi⟩##
I felt that this was not strictly convincing, but perhaps Ballentine's explanation will be (must read).

Meanwhile, I was speculating that in a universe with a second-order time evolution, an atom would decay with an oscillating probability of emission, with an exponential envelope -- a sort of spontaneous Rabi-like oscillation? Then we would need extra Einstein coefficients to model this process (perhaps complex coefficients). Does this make sense (after allowing that it refers to an imaginary law of physics)?
 
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  • #9
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So... am I completely lost if I take that the Hamiltonian actually describes a momentum and you need to take the vector modulus and square it to get energy. I should probably get the Ballentine book.
You need to read the book.

But as a warm up also read Landau:
https://www.amazon.com/dp/0750628960/?tag=pfamazon01-20

'If physicists could weep, they would weep over this book. The book is devastingly brief whilst deriving, in its few pages, all the great results of classical mechanics. Results that in other books take take up many more pages. I first came across Landau's mechanics many years ago as a brash undergrad. My prof at the time had given me this book but warned me that it's the kind of book that ages like wine. I've read this book several times since and I have found that indeed, each time is more rewarding than the last.

The reason for the brevity is that, as pointed out by previous reviewers, Landau derives mechanics from symmetry. Historically, it was long after the main bulk of mechanics was developed that Emmy Noether proved that symmetries underly every important quantity in physics. So instead of starting from concrete mechanical case-studies and generalising to the formal machinery of the Hamilton equations, Landau starts out from the most generic symmetry and dervies the mechanics. The 2nd laws of mechanics, for example, is derived as a consequence of the uniqueness of trajectories in the Lagragian. For some, this may seem too "mathematical" but in reality, it is a sign of sophisitication in physics if one can identify the underlying symmetries in a mechanical system. Thus this book represents the height of theoretical sophistication in that symmetries are used to derive so many physical results.'

The book had a deep effect on me and the person that wrote the above. You will need to let it sink in - but once you do you will be just as amazed.

This is THE deep insight of modern physics:
http://www.pnas.org/content/93/25/14256.full

Even Einstein was awe struck when Emmy Noether wrote him a letter about her discoveries after Hilbert asked her to investigate certain issues in General Relativity:
http://xxx.lanl.gov/pdf/physics/9807044v2.pdf

It really is that deep and profound - yet - and this is the strange part - relatively unknown to the general public.

Thanks
Bill
 
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