# B Schroedinger Equation and Hamiltonian

1. Jun 21, 2017

### mieral

The hamiltonian is not in the wave function but only exist when the amplitude is squared. But in the book "Deep Down Things". Why is the Schrodinger Equation composed of kinetic plus potential terms equal total energy. Is it not all about probability amplitude? How can probability amplitude have kinetic or potential energy?

Bruce Schumm quoted in Deep Down Things:

"Finally, notice that the Schrodinger equation consists of three terms: two
to the left of the equals sign (separated by the “_” sign) and one to the right
of the equals sign. The first term is the mathematical representation of the
procedure that, once you know y(x), tells you how to determine the kinetic
energy possessed by the particle at any location x. The second term, to the
right of the “_” sign, is the potential energy times the value of the wave
function at the location x. The third term, to the right of the “_” sign, is the
total energy times the wave function y(x).

So, if we look at the factors that multiply the wave function in the
Schrodinger equation, we find that to the left of the equals sign we have the
sum of the kinetic plus potential energies at the point x, while to the right
of the equals sign, we have the total energy. Thus, the Schrodinger equation
is just the wave-mechanical statement that the sum of the kinetic and
potential energies at any given point is just equal to the total energy—the
Schrodinger equation is simply the quantum-mechanical version of the notion
of energy conservation. From this quantum-mechanical formulation of
energy conservation arises the full set of constraints that prescribe the possible
quantum mechanical wave functions for the object. This again illustrates
the central importance of the idea of energy conservation (note 3.11)."

2. Jun 21, 2017

### vanhees71

Hi have no clue what this strange quote wants to tell you. The Schrödinger equation describes the time evolution of the wave function. For a single particle moving in a potential it reads
$$\mathrm{i} \hbar \partial_t \psi(t,\vec{x})=\hat{H} \psi(t,\vec{x})=-\frac{\hbar^2}{2m} \Delta \psi(t,\vec{x}) + V(\vec{x}) \psi(t,\vec{x}).$$
Given the wave function at $t=0$ (initial value) it tells you, how the wave function develops with time, and $|\psi(t,\vec{x})|^2$ is the probality distribution for the particle's location as a function of time.

3. Jun 21, 2017

### mieral

For a normal wave, it can pack energy by having small wavelength. But in qm.. it's a wave of probability.. it's not even a real wave.. so how can it even store energy and potential?

4. Jun 21, 2017

### Staff: Mentor

It doesn't, and the Schumm quote you provided doesn't say that it does. The energies constrain the time evolution of the wave so that it can only develop in certain ways, namely those described by the equation.

5. Jun 21, 2017

### mieral

You mean the electric field of the atoms are not stored in the wave functions? and electric fields are the so called hamiltonians which are separate from the wave functions?

But I wonder why we give so much importance to wave function than the Hamiltonians...

6. Jun 21, 2017

### Staff: Mentor

That's right - the wave function and the electrical field are completely different things. The electrical field does affect the evolution of the wave function:
The Hamiltonian contains a term for the electrical field's contribution to the potential energy, and we use the Hamiltonian to calculate the wave function. For example, the Hamiltonian that describes an electron bound in a hydrogen atom is $H=\frac{p^2}{2m}+\frac{e^2}{4\pi\epsilon_0{r}}$; the first term represents the kinetic energy and the second term represents the electrical potential energy.
The wave function is important because it tells us how the system behaves and what's going to happen to it. The general recipe for solving any problem involving quantum mechanics is: Write down the Hamiltonian; use the Hamiltonian to find the wave function by solving Schrodinger's equation; use the wave function to calculate whatever it is that you wanted to know.

7. Jun 21, 2017

### mieral

But isn't it that electrical field is also a quantum thing. Maybe that's where quantum field theory applies.. What is the general recipe for solving any problem involving quantum field theory? Is there a corresponding Hamiltonian in QFT?

8. Jun 21, 2017

### Staff: Mentor

Quantizing the electromagnetic field takes us into quantum electrodynamics, and that is a whole different and much more complex topic than the non-relativistic quantum mechanics in this thread, the theory that starts with Schrodinger's equation, wave functions, and the Hamiltonian.

9. Jun 21, 2017

### mieral

The book Deep Down Thing is about quantum field theory and the gauge fields and how extra terms produce gauge freedom to explain the 4 fundamental forces. I'd just like to know what is the Hamiltonian in QFT so I can re-read the book and understand it with better perspective.

10. Jun 21, 2017

### Staff: Mentor

You will need a real textbook for that; there's no shortcut here.

I do not know anything that would work in a B-level thread. Lancaster and Blundell's "Quantum Field Theory for the Gifted Amateur" is one of the more gentle introductions around, and it's written for someone who has made it through an undergraduate physics degree.

11. Jun 21, 2017

### mieral

In the double slit, what is the Hamiltonian? Is there an application where you can apply it without needing to write any Hamiltonian in the first step in the general recipe?

12. Jun 21, 2017

### Staff: Mentor

To work the double-slit problem for a massive particle, you start with the ordinary free-particle Hamilton $p^2/2m$. Now the solution to Schodinger's equation is an infinite plane wave, and you can apply the standard classical wave techniques to calculate the amplitude at points behind the barrier.

This won't work for photons (many reasons - the zero in the denominator of $p^2/2m$ is just scratching the surface of the problem). You can get a feel for how the quantum electrodynamical solution works from Feynman's delightfully layman-friendly book "QED: The strange theory of light and matter".

But at the risk of repeating myself... at some point you're going to need a proper college-level textbook. There's only so far that popularization can take you. The bad news is that that's a fair amount of work, and the good news is that it is completely and totally worth the effort.

13. Jun 21, 2017

### mieral

I'll do deep soul searching whether to spend 6 years going back to college.. but for now. i'd just like to know that if one use quantum field theory on say the double slit experiment.. does one still use the ordinary free-particle Hamilton $p^2/2m$ and still work on the solution to Schodinger's equation being an infinite plane wave. Or does QFT mean one only deal with particles that change or create/annihilate and not solve problems meant for Schroedinger wave equation. I have read Feynmann book "QED: The strange theory of light and matter" but can't relate how it deals with the Hamiltonian of QM. So in this thread I'd just like to know what QFT does with the Hamiltonian of QM or whether it puts it in altogether new language like Fock space still controlled by the Hamiltonian. The answer will be just a bird eye view before I learnt the answers 6 years later (that is.. if the college would still accept me or I can pass the entrance exam).

14. Jun 21, 2017

### bhobba

See the folllowing discusssion:
https://physics.stackexchange.com/q...orys-interpretation-of-double-slit-experiment

The best explanation of the double slit at the B level, IMHO, is Feynman's classic:
https://www.amazon.com/QED-Strange-Theory-Light-Matter/dp/0691024170

Now if you have done up to multivariable calculus, Linear algeba and an introductory QM course then here is a better explanation at the I level:
https://arxiv.org/abs/quant-ph/0703126

At that level you can also go into the detail of gauge forces, Higgs, Quarks etc and its relation to symmetry:
https://www.amazon.com/Physics-Symmetry-Undergraduate-Lecture-Notes/dp/3319192000

In fact that's the deepest insight of 20th century physics - that at rock bottom its really symmetry that is at work - but the detail is beyond the B level.

Its really even the basis of Classical Mechanics but is not usually presented that way. The great Lev Landau was an exception:
https://www.amazon.com/Mechanics-Third-Course-Theoretical-Physics/dp/0750628960

Thanks
Bill

Last edited: Jun 21, 2017
15. Jun 22, 2017

### vanhees71

Don't take it badly, but from what you are asking here, I guess this book is highly confusing. It's better to take a good textbook on quantum mechanics first. There's no way to understand relativistic QFT without a very solid foundation in non-relativistic quantum mechanics. For a very gentle introduction, see L. Susskind's theoretical minimum. Otherwise my favorite intro QM textbook is J. J. Sakurai, Modern Quantum Mechanics.

16. Jun 22, 2017

### Simon Phoenix

I'm just going to re-iterate what everyone else has said here. There's a certain amount of struggle involved in getting to grips with some of these things. We're kind of programmed by evolution to think and reason in a certain way based on our everyday experience of the world. But when you get right down to it our 'experience' is actually a very limited slice of what is out there - we only see a very limited portion of the EM spectrum, for example. I certainly don't move at anything even remotely close to relativistic speeds.

All of our 'everyday' intuition is based on our interpretation of this limited slice of 'reality' that we can directly experience. So it shouldn't be too hard to accept that when we try to investigate what happens outside of this limited perspective we're might need to re-wire our intuition. It's this re-wiring that takes the effort - and it's the continual struggle NOT to interpret everything in terms of our familiar everyday intuition that gives us all of the headaches.

It seems to me that you're asking some pretty deep questions before you've got the basics properly straightened out. You've got to get to the point where you're OK with the basic technical concepts behind non-relativistic 'standard' QM before trying to understand QFT. As others have said, QFT is a whole new level requiring considerably more technical facility.

One of your first questions :

indicates to me that you haven't got a proper grasp on the basics yet. Nugatory's answers seem to me beautifully succinct and clear about how you should be thinking.

So how do you get the basics? There are different thoughts about how best to learn QM and I guess it depends on your background and the kind of explanations that work for you. There are many excellent textbooks and articles - Vanhees has mentioned one of my favourites which is Susskind's book "The Theoretical Minimum". It's an excellent place to start but you'll need to be a little bit familiar with things like complex numbers, some elementary linear algebra, a little bit of probability, and so on, and some physics background wouldn't go amiss either.

Once you've got the basic re-wiring done - when you're comfortable to some level with the notion of a quantum state as distinct from a classical state, that measureable quantities are described by linear operators - and have some facility with calculations using these ideas then you can get a bit more ambitious.

But don't be too hard on yourself - it took a quarter of a century (and more) to actually pin a lot of this stuff down - and we're talking about some very brilliant men and women who spent a lot of time scratching their heads in puzzlement. Just work through a good textbook that 'speaks' to you - line by line, exercise by exercise - and you should come out of that process with a pretty good grasp.

17. Jun 22, 2017

### mieral

I know I need to be a physicist to understand all this stuff.. but for now.. just want a recipe.. According to Nugatory.
The general recipe for solving any problem involving quantum mechanics is:

1. Write down the Hamiltonian;
2. use the Hamiltonian to find the wave function by solving Schrodinger's equation;
3. use the wave function to calculate whatever it is that you wanted to know.

Just tell me what is the counterpart or general recipe for QFT as regards the Hamiltonian.. then I'd spend a lifetime to understand the math. Thanks!

18. Jun 22, 2017

### vanhees71

Your recipe is not very helpful. First of all you need to know, what you want to calculate in the first place. What you describe is what, unfortunately, many QM 1 students take out of their lecture: "One has to solve the eigenvalue problem for the Hamiltonian to get the wave function." That's because one does a lot of problems doing just this, but instead of asking, why they are doing it, they take it as a recipe. Recipies of this kind are anti-science. It's blind solving of maybe challenging math problems without having an idea what it is good for.

First of all the wave function represents the quantum state of a fixed number of particles. In QM1 usually you start with one-particle wave functions. Then it represents the quantum state of the system. It is a complex-valued function $\psi(t,\vec{x})$ which is normalized to 1 in the sense that
$$\int_{\mathbb{R}^3} \mathrm{d}^3 \vec{x} |\psi(t,\vec{x})|^2=1.$$
Then the physical meaning of this representation of the particle's state is that $|\psi(t,\vec{x})|^2 \mathrm{d}^3 \vec{x}$ is the probability that the particle at the time $t$ is in a small volume $\mathrm{d}^3 \vec{x}$ around the position $\vec{x}$.

Given the Hamiltonian, which encodes the particle's motion in a single function (to understand this you should learn analytical mechanics, particularly the Hamilton form of the least-action principle and the related mathematical tools like Poisson brackets, which is an important step to learn why QT looks the way it is), the Schrödinger equation allows to calculate the wave function, given at some initial time $t=t_0$, at later times, i.e., you solve
$$\mathrm{i} \hbar \partial_t \psi(t,\vec{x})=\hat{H} \psi(t,\vec{x}), \quad \psi(t=t_0,\vec{x})=\psi_0(\vec{x}).$$

Now, why would you look for the eigenfunctions of the Hamilton operator? Suppose the Hamilton operator is not explicitly time dependent (which is the case for many problems like the electron in a hydrogen atom) and we look for solutions of the Schrödinger equation of the form
$$\psi(t,\vec{x})=\exp(-\mathrm{i} \omega t) \phi(\vec{x}).$$
Then we get
$$\mathrm{i} \hbar \partial_t \psi(t,\vec{x}) = \hbar \omega \exp(-\mathrm{i} \omega t) \phi(\vec{x})=\exp(-\mathrm{i} \omega t) \hat{H} \phi(\vec{x}).$$
That means $\phi$ must be an eigenfunction of the Hamiltonian with the eigenvalue $E=\hbar \omega$.

Now, as we postulated above, the physics is in the absolute value of the wave function squared, giving the probability distribution for the position of the particle. For these special solutions of the Schrödinger equation, involving an eigenfunction of the Hamilton operator, we have
$$|\psi(t,\vec{x})|^2=|\phi(\vec{x})|^2,$$
i.e., the position-probability distribution is time-independent in this case. In other words, these functions represent time-independent (stationary) states, which do not change in time (the time-dependence of $\psi(t,\vec{x})$ is in a time-dependent phase factor, which drops out taking the modulus squared!). That's why these eigenfunctions are quite important in quantum theory.

E.g., it solves the problem of Bohr's model of the (hydrogen) atom: A time-independent stable atom is described by the eigenfunctions of the Hamiltonian. These are states, where the electrons are not moving around and thus can't radiate (within non-relativistic quantum mechanics; if you also quantize the electromagnetic field, this is true only for the ground state, but let's not bother with this subtlety here). The possible energy eigenvalues determine the spectral lines emitted or absorbed by the atom: The emitted (absorbed) radiation is always the difference of two energy eigenvalues, i.e., if you shine with an electromagnetic wave of such a frequency $\omega=(E_n-E_{n'})/\hbar$ on an atom it's likely to get absorbed by the atom (photoelectric effect), if all other "selection rules" for absorption are fulfilled.

19. Jun 24, 2017

### mieral

Thanks vanhees71 for your details.

I know QFT is about operators and not wave function. But for a particle under some potential giving the Hamiltonian.. we usually solve for the wave function to know the particle position or momentum at a certain "x" at time "t". But for QFT where you don't deal with wave function but operators. Can you use this to find the location of the particle at certain "x" at time "t"? Or you only use QFT for solving those related to fields.. for example.. the electromagnetic field.. meaning QFT is never used for solving for position of particle? Just want a bird eye view of this for now.

20. Jun 24, 2017

### vanhees71

Of course also QFT admits the evaluation of probabilities for the position of a particle (provided it's a particle for which a position observable is definable in the first place, but you can do this for any massive particle). It's only not possible to do this in the "1st-quantization formalism" applicable in non-relativistic QT. The reason is that when particles interact with relativistic energy exchange, it is likely to produce new particles, distroying particles present in the initial state, etc. This cannot be described by a single wave function, and that's why it's most convenient to use QFT to formulate relativistic QT.

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