And free

Can you give an example?

Does it look like electromagnetism? I think he meant classical mechanics.

@@chuuuu1131 Not him, but in fluid mechanics, to quantify the deformation of a fluid particle in a continuous medium you need something called a stress tensor, which is a 3 by 3 matrix describing the direction of the stress imposed and on which "face" of the fluid particle it is acting on. Check out the Navier Stokes Equation expanded out.

Agreed.

True. Many applied mathematicians research in the field of fluid mechanics.

Brandon Smith maybe it was just a review which they needed, I myself got this only as a recommendation even though I’m not a subscriber to OCW, and wouldn’t have expected anyone who didn’t realize there is a second part because they weren’t told in the video to go looking for it or to recognize it if it had bit them in the nose.

They were looking for pseudoscience

Maybe they moved on to the ocw site.

How is called this teacher pelase?

@@Anb-ng2ou what are you doing here if you have problems with reading description?

@@Anb-ng2ou Dr. Barton Zwiebach

Bruh😂

Linearity depends on the dynamical variable you’re solving for. In the example the lecturer presents for Newton’s equations, you are solving a second-order differential equation for the “variable” x (which is a function of time). The equation is nonlinear *with respect to x* whenever V’(x) is nonlinear with respect to x. In the case of Schrödinger’s equation, by contrast, you are solving a partial differential equation for psi (the wavefunction), not for x. You’re right that the Hamiltonian has a potential term V that depends on x (often nonlinearly), but V doesn’t depend on psi, and it’s psi that you’re solving for.

Because potential energy is not a function of the 'wave function'- the independent variable in schrodinger's wave equation (or its derivatives); unlike in Newton's equation of motion in which P.E. WAS a function the independent variable(s) e.g. x in general.

Good eyes! Yes, this is the same room where they did the 2013 version of the course. :)

I respect that. Have fun

How evil. Torturing animals for your tongue's evil delight.

@@ProgressiveTeen ... kind Sir, sadly you are mistaken. Mac and Cheese is not an animal.

Kaushal Jain the potential in the Hamiltonian can be thought of as a set of values that span relevant space (a normal line or surface over space) that the wave function will be multiplied by at every single one of those points individually. So the wave equation (the input) will be transformed in essentially a multiplication style operation. Multiplication is linear, and so is the “potential operator”. If the potential was somehow squaring or logging the wave equation, that would be nonlinear, but that is impossible. The potential isn’t that weird. It just multiplies the wave equation by its own predetermined values, which you could do before or after multiplying by a constant and get the same result.

@@zacharythatcher7328 wow

The lecture notes are available on MIT OpenCourseWare at: ttp://ocw.mit.edu/8-04S16. Best wishes on your studies!

Hey U a physics student too?

While the potential energy term in the Hamiltonian operator of quantum mechanics can be nonlinear, the dynamics of quantum mechanics are fundamentally described by a linear equation, the Schrödinger equation. Therefore, quantum mechanics is considered a linear theory.

@@sylvenara ok understood. Thanks for the help😊

When de Broglie proposed the idea of matter waves, Schrödinger tried to find an equation which could describe these matter waves and hence came up with the famous Schrödinger equation

it is quite easy. you just have to prove that what the classical operators become in qm. like p=-ihbar d/dx or E=ihbar del/del t

@Lambda that doesn't sound easy at all, could you elaborate?

@@infinity-and-regards look finding the operators is tough but the derivation of the equation is very easy if you know the operators. okay look i will derive it for you but i will assume the operators if you want the proof for why the operators are equal to what i am assuming you will have to look it up as the proof is very long. E=KINETIC ENERGY + POTENTIAL ENERGY KE= P^2/2M. PE=V(X,) LET US QUANTIZE THIS EΨ=P^2/2M Ψ +VΨ NOW EΨ=Ih/2π dΨ/dt. and p=-ih/2πd/dx Ih/2π dΨ/dt=(-ih/2πd/dx)^2/2m Ψ+VΨ Ih(ΒΑR)dΨ/dt=-h(BAR)^2/2m d^Ψ/dx^2+VΨ AND YOU HAVE DERIVED THE SE. YOU CAN DERIVE ITIN OTHER FORMS BUT THE PROCESS IS SAME. THE REAL CHALLENGE COMES IN PROVING THE ASSUMPTIONS AND YOU TO USE BRA'S AND KET'S FOR THAT. ALTHOUGH THE PROOFS ARE GIVEN IN SOME TEXTBOOKS BUT ARE VERY COMPLEX. EVEN GRIFFITHS DOES NOT GIVE THE PROOF

I don't know enough about the Hamiltonian to directly answer this, but in general an operator is linear if it satisfies the following two conditions (O is an operator): 1.) O(f + g) = O(f) + O(g) 2.) O(c*f) = c*O(f)

@@LusidDreaming yes. I think that's the definition.

The difference is previously your solution was in terms of x(t) and the potential depends explicitly on x. Now your solution is in terms of the wave function, of which the potential is not a function. The Hamiltonian is a linear operator on the space where the wave function lives. The potential is not a function of your wave function.

@@LusidDreaming So there cannot be time compression to satisfy linearity... experiments seem to suggest that in addition to spatial nonlocality there is temporal nonlocality involved in entanglement. I doubt that changing the inertial frame of reference will get rid of such nonlinearity.

Potential V(x) is a classical quantity, whose negative derivative is force. For example: en.wikipedia.org/wiki/Gravitational_potential or en.wikipedia.org/wiki/Electric_potential#Electric_potential_due_to_a_point_charge

@@bencegabor9228 thank you sir

Not really. I am beginning to wonder how this guy made professor at MIT. ;-)

It can.. But Classical Mechanics is a good approximation and easy to use..

Initial conditions aren’t given

Graph of potential energy over time. (Potential energy meaning the work that force has to do. Force × distance) The derivative is the force acting on it at a specific time. Like if a ball is rolling down a hill, hes basically just saying that because there's mass, gravity would be pulling it down, and it loses potential energy as it gets closer to its destination and force is used. So the force is the negative of the derivative of potential energy.(someone correct me if I'm wrong)

f(x) = x² f(a + b) = a² + 2ab + b² f (a) + f (b) = a² + b² So f(a+b) is not equal to f(a) + f(b), therefore it's not additive. So it's not linear.

@@mateusmarinho72 yes, okey, thank you very much!

​@@mateusmarinho72muchas gracias, entendí la explicación.

@@damariscalleros4631 but time changes, sometimes

Velocity along the x axis

X is the Particle moving in 1d under potential v

It is linear because the quantum mechanical ensemble consists of statistically independent experiments. That's an assumption. Since we do not have an actual ensemble in real life but only the repetition of the same experiment, the statistical independence of the data set that it produces has to be tested. It will usually work out fine for e.g. the photon flux from a thermal light source, but it will show very strong temporal correlations for a laser source. So you are not completely false in saying that "we made it linear", but the naive assumption that every quantum mechanical system obeys that linearity is incorrect.

Pehle basics clear karo Bhai baadme quantum ki lectures samjhoge

@@mysteriouspandey3450 Thanks sir for your advice, please answer to bta dete doubt ka

@pippo Thanks!

Yeah, I'm also a little confused by this argument

Notice in the classical equation, m x'''(t)=-V(x'(t)), the potential is a function of the derivative of the position. While the derivative itself is linear, we don't know what the unknown potential function "V" is so we can not say with certainty that V(x'(k*t))=k*V(x'(t)). By contrast, in the Schrodinger equation the potential V is multiplied by the wavefunction psi, so V*k*psi=k*V*psi

The potential function is arbitrary, in most scenarios we approximate it to be harmonic (proportional to x^2) but it can be generally non linear. Alternatively H-hat is the Hamiltonian operator which is basically a constant time the 2nd derivative with respect to position i.e its linear. Its not the Hamiltonian itself, its an operator named after the Hamiltonian.

The quantum Hamiltonian operator acts on the state variable differently than the potential/kinetic energy (classical Hamiltonian) does in Newtonian mechanics. In particular, "x" in the potential energy function is just a parameter, not the present particle state being plugged into the function like it is in the Newtonian case with Newton's second law or Hamilton's equations, because in quantum mechanics position, momentum, etc. are not actually numbers, but "fuzzy" quantities described by probability distributions (which corresponds to reduced information, as per Shannon), and they are all wrapped up in the linear state vector, |psi>. That state vector is not a real number, but (effectively) an infinite number of them, and hence could not be inserted into the potential energy function anyways, which is expecting only one real number as input. Instead, the potential energy function _becomes_ an operator on the state vector by first considering it in the form of the positional wave function psi_x(x), which is a "basis expansion" (effectively the same thing as vector components of ordinary vectors, but with infinitely many components) and then you multiply this wave function by the potential energy function to get another wave function (i.e. form psi'_x(x) = U(x) psi_x(x)), which then represents, by going backwards through that expansion, the resulting new state vector. Since multiplication is distributive, hence linear, you get a linear action of this operator.

Because, say for one dimension, x is an independent variable of which potential energy is generally a function so that gives you a non-linear differential equation. That's it!

At your age I was learning basic astronomy, you know where I ended up, studying geological engineering, I'm in my first cycle, I like physics although it's a bit difficult to understand, but what I don't understand much is chemistry, I'm from Peru.

why pay when you can not pay? right?

Wtf

He studied at my University in Peru (well known in Peru as UNI) at the same Faculty than me and he finished (Electrical Engineer career) I believe in 1977, then he came to America to follow Master and PhD. degrees.

The moment a non linear factor is introduced, everything affected by that nonlinear factor becomes nonlinear.

dougie v yes, Aa is easy to see, but Vv will always be made using straight lines.

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