Schroedinger Equation and Hamiltonian

[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)."

 

[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.

 

[mieral]

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.

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?

 

[Nugatory]

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?

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.

 

[mieral]

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.

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...

 

[Nugatory]

You mean the electric field of the atoms are not stored in the wave functions?

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:

and electric fields are the so called hamiltonians which are separate from the wave functions

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.

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

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.

 

[mieral]

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.

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?

 

[Nugatory]

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?

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.

 

[mieral]

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.

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.

 

[Nugatory]

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.

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.

 

[mieral]

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.

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?

 

[Nugatory]

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?

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.

 

[mieral]

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.

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 Feynman 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 learned the answers 6 years later (that is.. if the college would still accept me or I can pass the entrance exam).

 

[bhobba]

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.

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/dp/0691024170/?tag=pfamazon01-20

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/dp/3319192000/?tag=pfamazon01-20

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/dp/0750628960/?tag=pfamazon01-20

Thanks
Bill

 

[vanhees71]

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.

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.

 

[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 :

How can probability amplitude have kinetic or potential energy?

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.

 

[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!

 

[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.

 

[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.

 

[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.

 

[mieral]

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.

Just to clarify this ambiguous sentence "It's only not possible to do this in the "1st-quantization formalism" applicable in non-relativistic QT". You meant if the particle didn't interact with relativistic energy exchange.. it can't be solved using QFT? Or did you meant by that sentence that that normal non-relativistic Schrodinger equation couldn't solve for relativistic energy exchange.. of course.

Or when you wrote the first sentence. We you describing particles interacting with relativistic energy exchange? I was asking about simple non-relativistic particle in box under the action of some potential. Can QFT using operators and not wave function solve it? Can the Hamiltonian uses operators instead of wave function in solving for the particle position or momentum? (were you saying yes in your first sentences?)

 

[vanhees71]

No, what I wanted to say is that relativistic QT is most efficiently formulated as a QFT, i.e., in the "2nd quantization formalism". The wave function approach is very difficult and cumbersome compared to the QFT formulation. As far as I know it's only formulated for QED in terms of Dirac's "hole theory". For bosons it's hard to formulate at all and for more complicated models like the non-Abelian gauge theories of the Standard Model, it seems to be impossible to formulate it other than with QFT.

 

[mieral]

No, what I wanted to say is that relativistic QT is most efficiently formulated as a QFT, i.e., in the "2nd quantization formalism". The wave function approach is very difficult and cumbersome compared to the QFT formulation. As far as I know it's only formulated for QED in terms of Dirac's "hole theory". For bosons it's hard to formulate at all and for more complicated models like the non-Abelian gauge theories of the Standard Model, it seems to be impossible to formulate it other than with QFT.

Ok. That is clear. Wave function thing is too cumbersome for relativistic particles and QFT is the tool to use. But please address the following question reversed to it. For simple non-relativistic particle in a box under the action of some potential. Can QFT using operators and not wave function solve it? Can the Hamiltonian uses operators instead of wave function in solving for the particle position (some range) or momentum in the box? If not.. why?

 

[vanhees71]

Of course, you can use QFT also in the non-relativistic case. If the Hamiltonian is conserving particle number, you'll just get the same as with the Schrödinger equation in the 1st-quantization formalism. I don't understand what you mean by "the Hamiltonian uses operators". In QT (no matter whether in wave mechanics, matrix mechanics, or QFT) the Hamiltonian is a self-adjoint operator in Hilbert space.

 

[mieral]

Of course, you can use QFT also in the non-relativistic case. If the Hamiltonian is conserving particle number, you'll just get the same as with the Schrödinger equation in the 1st-quantization formalism. I don't understand what you mean by "the Hamiltonian uses operators". In QT (no matter whether in wave mechanics, matrix mechanics, or QFT) the Hamiltonian is a self-adjoint operator in Hilbert space.

Remember the student recipe. 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.

Forgive it if it's not rigorous. It's just to get initial idea.

So to solve the *same problem (that QM is meant to solve)* using QFT.. I thought the student recipe was:

1. Write down the Hamiltonian;
2. use the Hamiltonian to find the Operators by solving Fock space (?)
3. use the Operators (or Fock space) to calculate whatever it is that you wanted to know. (??)

If this is wrong. Please write the 3 simple steps (never mind if I don't understand the terms). I just want to know this as bird eye view before spending 10 years going back to college (if I got accepted).

 

[Simon Phoenix]

Remember the student recipe. The general recipe for solving any problem involving quantum mechanics is

Whilst I can certainly sympathise with the desire to see a general 'recipe', if only it were that easy! I distrust thinking in terms of 'recipes' - it's much better to focus on what you want to know and then to think about the best way to get that information.

Consider straightforward projectile motion - say a cannonball fired from a cannon. It's not too difficult to write down the differential equations to solve by applying Newton's laws of motion, and then not too difficult to solve them. But suppose you just wanted to know the maximum height for a given initial launch speed and angle? Sure you could go through all the rigmarole of solving the equations and then find the stationary points of the trajectory and so on - but it's much cleaner, when you want to know the maximum height, to use energy conservation - so the initial kinetic energy in the vertical direction equals the potential energy at the maximum height, from which the maximum height can be very simply determined.

So whilst I appreciate that you're trying to get a kind of 'road map' here - I would encourage you not to let it become too fixed.

I have often wondered when skinning cats was so popular that there were several ways to perform this grisly deed - but there is, allegedly, more than one way to skin a cat.

 

[mieral]

Whilst I can certainly sympathise with the desire to see a general 'recipe', if only it were that easy! I distrust thinking in terms of 'recipes' - it's much better to focus on what you want to know and then to think about the best way to get that information.

Consider straightforward projectile motion - say a cannonball fired from a cannon. It's not too difficult to write down the differential equations to solve by applying Newton's laws of motion, and then not too difficult to solve them. But suppose you just wanted to know the maximum height for a given initial launch speed and angle? Sure you could go through all the rigmarole of solving the equations and then find the stationary points of the trajectory and so on - but it's much cleaner, when you want to know the maximum height, to use energy conservation - so the initial kinetic energy in the vertical direction equals the potential energy at the maximum height, from which the maximum height can be very simply determined.

So whilst I appreciate that you're trying to get a kind of 'road map' here - I would encourage you not to let it become too fixed.

I have often wondered when skinning cats was so popular that there were several ways to perform this grisly deed - but there is, allegedly, more than one way to skin a cat.

Look. There are many people who don't know how to cook. So they are not looking for exact recipe of a food. They just want a sample of what it means to even have a recipe. So just give me a sample of a recipe of QFT. Maybe you mean there are different situations like in QM which needs different steps like in determining particle in a box in QM or double slit.. which are different cases? If so, please give one sample of a case or step or recipe in QFT. I just want to know of what it even means to have steps or recipe of QFT.

 

[vanhees71]

Remember the student recipe. 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.

Forgive it if it's not rigorous. It's just to get initial idea.

So to solve the *same problem (that QM is meant to solve)* using QFT.. I thought the student recipe was:

1. Write down the Hamiltonian;
2. use the Hamiltonian to find the Operators by solving Fock space (?)
3. use the Operators (or Fock space) to calculate whatever it is that you wanted to know. (??)

If this is wrong. Please write the 3 simple steps (never mind if I don't understand the terms). I just want to know this as bird eye view before spending 10 years going back to college (if I got accepted).

Your recipe is totally useless. I don't even know, which cake I should bake with it! 1. is ok. To specify the system, I need the Hamiltonian that describes the dynamics of the system (it's the generator of time evolution). But what do you mean by 2. and 3.? To solve the Schrödinger equation you need the initial state or you want to find the energy eigenfunctions to get the static states and the energy eigenvalues (e.g., to get the spectral lines of an atom, molecule, or nucleus).

For QFT the standard recipe is to use Feynman rules and calculate S-matrix elements for scattering processes.

 

[Simon Phoenix]

I just want to know of what it even means to have steps or recipe of QFT.

I'm afraid I've lost the plot of what you're asking for in this thread. Here's where we started :

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?

And now we're talking about QFT where you appear to want to be given a breakdown of the steps involved in a QFT calculation (any calculation it would seem)?

I've no real grasp of what you're asking for, or why, and I think you're trying to run before you can walk. I honestly can't see the benefit of giving you any detail of a calculation in QFT (which do tend to be really quite technical and challenging) when it's quite clear you haven't sorted out what wavefunctions, Hamiltonians and probabilities in QM are. Perhaps you should focus first on just getting the very basics in place.

Sorry to be so blunt, I don't mean to be offensive - just trying to get to the bottom of what it is you really want or need.

 

[mieral]

I'm afraid I've lost the plot of what you're asking for in this thread. Here's where we started :
And now we're talking about QFT where you appear to want to be given a breakdown of the steps involved in a QFT calculation (any calculation it would seem)?

I've no real grasp of what you're asking for, or why, and I think you're trying to run before you can walk. I honestly can't see the benefit of giving you any detail of a calculation in QFT (which do tend to be really quite technical and challenging) when it's quite clear you haven't sorted out what wavefunctions, Hamiltonians and probabilities in QM are. Perhaps you should focus first on just getting the very basics in place.

Sorry to be so blunt, I don't mean to be offensive - just trying to get to the bottom of what it is you really want or need.

I just want to know the difference between QM and QFT recipe or steps in solving for the same particle in box. But based on Vanhees71 message. I think he meant QFT is only meant for many particles.. and if you only have particle in a box. Then as Vanhees put it "If the Hamiltonian is conserving particle number, you'll just get the same as with the Schrödinger equation in the 1st-quantization formalism"?. Can you rephrase his statement.

Vanheez:

I don't understand what you mean by "the Hamiltonian uses operators". In QT (no matter whether in wave mechanics, matrix mechanics, or QFT) the Hamiltonian is a self-adjoint operator in Hilbert space.

The following is what I meant by QFT using operators instead of wave function. I read Sonverdal stuff:

"Second quantization is a somewhat misleading term to me (and many others) because it seems to imply that you do two steps of quantisation, which is not correct. What happens is this: In QM, you have a wave function that assigns a prob. amplitude to each point in space (and time). The Dirac equation or Klein-Gordon equation was initially conceived as an equation of a wave function like the Schroedinger equation. Then people realized that this is not correct (because particle number is not conserved due to creation of particles/antiparticles etc.) and that you need in fact a field theory. In a field theory, you use the same equation (for example the Klein-Gordon equation), but you now interpret this as the field equation of a classical field. Solutions of this equations look like they did before (when you thought that you were dealing with a prob. amplitude), but now the field has to be interpreted as a classical quantity (for example, the displacement of a membrane as a intuitive example). In the second step, you then quantise this field, using the standard rules of quantum theory, converting observables to operators. Since the classical field itself is an observable, it becomes a field operator. What makes things confusing is that you now usually do not deal with wave functions anymore - QFT is usually not phrased that way (laudable exception in the book of Hatfield "QFT of point particles and strings"). If you think in terms of wave functions, what QFT does is to assign a probability amplitude to each possible field configuration. (Since this is not easy to do because of the infinity of possible functions, people prefer other ways of describing QFT.) The standard way of explaining this quantisation - turning wave functions into creation/annihilation operators - is also confusing for another reason, in my opinion: Before you have a set of possible solutions, with coefficients a and b that can have any possible value. Then you do the second quantisation and these coefficients turn into creation/annihilation operators, which are precisely defined objects. Where did all the possibilities for the solution go? Answer: The 2nd quantised solution that contains operators has to be applied to a state vector, and the freedom you have is now in the state vector, not in the solution anymore. Unfortunately, this is not always explained."

So Simon and Vanhees. What I was asking was if you only need to solve one particle in a box using QFT.. What happens to the creation/annihilation operators? What is there to create and annihilate?

 

[Simon Phoenix]

What I was asking was if you only need to solve one particle in a box using QFT.. What happens to the creation/annihilation operators? What is there to create and annihilate?

Have a look at this introductory set of notes on QFT

http://www.damtp.cam.ac.uk/user/tong/qft/two.pdf

On page 43 the author, David Tong, explains how one recovers 'standard' QM from QFT, but you'll really need much of the previous material to understand it. Have a look at the first pages where, working in the Schrödinger picture he explains the process of field quantization. He does it this way so that the link with 'standard' QM is evident - and we even have the Schrödinger equation, but now the state $| \psi \rangle$ would give a wavefunction that is a functional - that is a function of every possible configuration of the field.

I confess I'm far from being an expert on QFT - I know just enough to be bloody dangerous with it so I'll defer to others like Vanhees who are considerably more expert than I.

 

[vanhees71]

I never that that QFT is only for many-body systems. I said that it's the most convenient formalism to use for many-body systems, and in relativistic QT you always have to deal with many-body systems since at relativistic collision energies there's some probability for the annihilation and creation of particles. Of course, you can use QFT also for situations, where the particle number is conserved (in non-relativistic QT). Then it's completely equivalent to the 1st-quantization formalism.

 

[mieral]

vanhees71, in spectroscopy or quantum chemistry or solid state physics.. do they adjust the values of the potential term in the Schrodinger equation? I read they mostly put the electromagnetic field or other fundamental forces in the potential terms.. but can you also include the spectroscopy or even thermal vibrations into the potential terms?

What other energy can you put in the "potential" terms of the Schrodinger equation, is there a limit?

 

[jeremyfiennes]

From the Wikipedia: "The Schrödinger equation provides a way to calculate the wave function of a system and how it changes dynamically in time. However, the equation does not directly say what the wave function is." If Schrödinger didn't know what the wave function was, how could he deduce it? What was in his mind when he derived it?

 

[bhobba]

From the Wikipedia: "The Schrödinger equation provides a way to calculate the wave function of a system and how it changes dynamically in time. However, the equation does not directly say what the wave function is." If Schrödinger didn't know what the wave function was, how could he deduce it? What was in his mind when he derived it?

Read:
https://arxiv.org/abs/1204.0653

Thanks
Bill