@Abcd Why you think like that? see some example of implementation of linked lists.. the name itself is very suggestive that it will look like a linked chain

We use user defined data type here.. as we need at least two space in a unit(what we call a node ), one to hold data and one to hold address of next node..

@Blue You are talking abt Java implementation ig

@taritgoswami we should probably discuss your question here rather than in the problem solving room.

@Blue Ooh

@JohnRennie Ok

Are you free now?

@taritgoswami particles are neither particles nor waves, they just sometimes behave like particles and behave like waves. The best description we have of particles is from quantum field theory, and that describes particles as states of quantum fields.

States of quantum fields means? can you explain it with some more details?

Are you an undergraduate, still at school, or at some other point in the education system?

I am an Undergraduate

OK, do you know what a Fourier transform is?

Yeah

Don't know very well actually

OK. Suppose you have some function that is a function of position in space, $f(x,y,z)$. We can Fourier transform $f$ and what this does is express $f$ as an infinite sum of plane waves.

That is, we write $f$ as something like:

$$ f(\mathbf x) = \int a(\mathbf k) \sin(\mathbf k\cdot \mathbf x) $$

where $\mathbf k$ is the wave vector.

Ok

Now, if you remember back to when you first learned about quantum mechanics you probably started with the Schrodinger equation for a free particle, and you discovered its wavefunction was an infinite plane wave, just like the infinite plane waves in the Fourier transform above. Yes?

Yeah

OK. What we do in quantum field theory is to say that the function $f$ represents a quantum field, and the plane waves in the integral are kind of the wavefunctions for free particles. So our quantum field is built up from the states of free particles.

By function of position in space do you mean the wave function itslef?

These individual states in the integral are called Fock states and they behave like quantised simple harmonic oscillators. The ground state of a Fock state is when no particles are present.

The first excited state is when one particle is present, the second excited state is when 2 particles are present and so.

So quantum field theory describes particles as the excitations of the Fock states from which the quantum field is built. Each Fock state describes particles with a particular momentum $\mathbf p$.

So you can create a particle of momentum $\mathbf p$ by adding energy to the correspnding Fock state to excite it to a higher energy level.

If you're now thinking WTF I sympathise :-)

Yeah I will need some time to digest it :p

When you get the hang of this it makes a lot of things suddenly a lot clearer.

For example suppose you collide two electrons in a collider, how do extra particles get created in the collision?

@JohnRennie No idea :p

It happens because the Fock states describing the incoming electrons can transfer their energy to the Fock states of other quantum fields, and this creates new particles corresponding to those Fock states.

So the creation and annihilation of particles is just shuffling energy around between different Fock states.

The individual states in the integral are Fock state, and that's the quanum state of free particle .. means Fock states are quantum states of Free particle.. am I correct? please correct me

I find examples illuminating: if you solve the equations for a free real scalar field (Klein Gordon) you get something like $$ \int \frac {d^3p}{(2\pi)^3} a_{\mathbf{p}}e^{-ipx}+a_{\mathbf{p}}^\dagger e^{ipx}$$

whoops I pressed send way before my message was ready lol

didn't mean to intrude myself so rudely

@taritgoswami yes, though this is all rather arm waving and approximate. But you can think of the Fock states as describing free particles.

anyway if you then calculate the hamiltonian you get $$\int \frac{d^3 p}{(2\pi)^3}a_\mathbf{p}^\dagger a_\mathbf{p}+\frac{1}{2}[a_\mathbf{p}, a_\mathbf{p}^\dagger]$$if you interpret $a$ and $a^\dagger$ as the usual ladder operators you can see that the field is a continuum of quantum harmonic oscillators, and that the operators can create particles at any momentum

@user2723984 I suspect this might be a bit much for @taritgoswami :-)

@JohnRennie Yeah :(

@taritgoswami I'm giving you a oversimplified and potentially misleading "pop science" description in an attempt to give you that basic ideas. @user2723984 is doing it properly :-)

oh sorry

no it's just that having already took quantum mechanics

I thought seeing that the hamiltonian of a certain field is an integral over hamiltonians of harmonic oscillators might make clearer what you meant by what you said before :)

@JohnRennie I am not getting this line - chat.stackexchange.com/transcript/message/48476681#48476681

but if not ignore what I wrote

@taritgoswami if you're just calculating the wavefuction of a free particle you get a plane wave, and you get the amplitude of the plane wave by normalising the wavefunction. Yes?

Yeah

You normalise to one because you know there is one particle and you know the particle must be somewhere, so the integral over the whole wavefunction has to be one.

Yes

The Fock state is a bit different. It can represent one particle or two particles or $n$ particles for any integer $n$. So when you normalise it the result is one, or two, or $n$ depending on how many particles are present in that state.

Ok

@taritgoswami anyway, the point is that particles are complicated things in QFT, so you have to be careful about describing them as waves or particles.

They are objects that can behave like waves or behave like particles depending on the system. It is sometimes convenient to use an approximate description of the article as a wave, and sometimes use an approximate description of the particle as a particle.

Polarisation is easy to understand when describing the particle as a wave, so that's the wave we'd normally describe it.

@taritgoswami could you clear my doubt regarding that?

@Blue How do you know that : "it simply remains as a variable of class type which can hold addresses"

@Blue isnt constructor automatically created for ever obj??

Oh I see^

What is happening here^?

@Blue What about the data inside start? Will t take that also?

OK

public void insertAtFirst(int val)
    {
        Node n;
        n = new Node();
        n.setData(val);
        n.setNext(start);
    }

  //  And in Node class we have:
    public void setNext(Node n)
        {
            next = n;
        }

@Blue Another variant^

n.setNext(start)??

next = n; ??

:48477224 class Node
{
    private int data;
    private Node next;
    public Node()
    {
        data = 0;
        next = null;
    }

    public Node(int d, Node n)
    {
        data = d;
        next = n;
    }

    public void setData(int d)

    {
        data = d;
    }

    public void setNext(Node n)
    {
        next = n;
    }

    public Node getNext()
    {
        return next;
    }

    public int getData()
    {
        return data;
    }
}

@Blue Do you want linked list code also

@Blue what does this line do^^

@Blue 1) n is Node's object 2) yes

@Blue ya

@Blue its just declared as private Node start

@Blue what does n.next actually mean??

@Blue no without analogy please

otherwise i get lost in the analogy and forget the original thing

@Blue I am facing trouble in distinguishing address and object

@Blue start = n; Does this give start full n including its data or only the address of n?

@Blue yes its helpful. Let me complete it and ill ask you if i get more doubts.

@Blue See if we have a pointer t pointing at some object

Then that's a pointer. How can it access the object's stuff using it?

Like how can we say t.getNext() or t.setNext(n);

The fact that Java only has pointers to objects has always vaguely annoyed me. C++ is much better in this respect.

@JohnRennie Ehhhh...I'd argue that you really shouldn't need to manipulate any object directly

@Blue whats the difference between variables of class type and object

@ACuriousMind it should be up to me if I want to pass an object by reference or by value.

@JohnRennie Is there any good introductory book you prefer for QFT which explain concepts clearly and suitable for an Undergrad

@taritgoswami the problem with all QFT books is that to do QFT properly is really hard. The maths involved is pretty deep. So there is no such thing as a simple QFT book.

Not need to be very simple, but a step by step approach maybe

If you're an undergrad I would recommend you don't try to learn QFT unless you plan to study theoretical physics.

Does your college offer a QFT course? It would probably be a final year option.

No, there is no QFT course

Yeah I am planning to study ThPh

Maybe

Separate mean?

I am doing dual degree in Math and Phys

Yeah

@Blue Yeah, needs lot of efforts

I have told you before also ig @Blue

@taritgoswami have a look at:

But bear in mind that you're letting yourself in for quite a lot of work.

@JohnRennie Ok, if it seems too hard then I will leave it for now and will try to understand later

@JohnRennie Thanks

I think the best way of studying quantum field theory is to join a research group in the themes of quantum field theory you are intereted in.

because every professor researching on quantum field theory teaches different contents of quantum field theory, probably because they all teach based on what are used in their research.

@CaptainBohemian well that's if you want to really go in depth, to begin with if you are comfortable with quantum mechanics and special relativity you can simply follow one of the standard textbooks

Hi

I have a doubt which I dont think would be suitable if I ask in physics SE

Can a system have an non-zero angular acceleration even if the magnitude of angular velocity be constant? It seems quite trivial question.

Like if a body A is rotating with some angular velocity X and another body B is rotating with some angular velocity Y, axis of rotation of A is perpendicular to the axis of rotation of B. Then it is asked to me find the relative angular acceleration? Intuitively it seems that the direction of angular velocity of changing, but how can I express it in equation, the definition β = dω/dt seems insensitive of that change

@Blue Are you free for 30 seconds?

int pop() // Function to delete element in Stack
    {
        if(top == -1) // Condition for Underflow
        {
            System.out.println("UNDERFLOW");
            return -999;
        }
        else
        {
            int val = ST[top]; // Storing the element which will be removed
            top = top - 1;
            return val;
        }
    }

@Blue I feel that to pop element from stack we should do ST[top] =0; (that is we should set the top element to zero and then decrement top by 1) but why do people do that^^^

@taritgoswami Student friendly quantum field theory is a good book

@AjayMishra See physics.stackexchange.com/q/105096/50583

it was written by Robert D, Klauber

@Blue why is it not necessary??

@Blue empty states of stack are denoted by 0 right?

@Blue when you implement stack using array, initially all elements are 0

and then we add elemnts to stack, we add non 0 numbers

So 0 denotes empty state

@Blue K thanks.

@ACuriousMind I didnt get the time derivative which is expressed as a cross product in the answer by brian moths

That expression seems dimensionally incorrect by the way.

@AjayMishra See en.wikipedia.org/wiki/…

@AjayMishra The equation is correct - the cross product of two angular velocities has units of inverse time squared.

anyone with the book a guide to feyman diagrams in the many body problem? im going through it and i have a few questions

p 43 the last 2 mathematical expressions (hamiltonians) are supposedly equal, but i find that they are not equal

is there a typo somewhere or am i missing something

basically it boils down to check whether p^2/(2m_0) is equal to p^2/(2m)- m_ep^2/(m(m_e+m)) where m_0 = m+m_e. i verified that it's not equal unless m_e = 0 or something like that, but we assume that the masses are different from 0 and infinity, so the equality does not hold

i wonder whether im missing somethng

i found no errata of the book . i found a typo and possibly many more. im just extremely unsure

it's the 2nd edition so i assume someone went through the book

basic electrodynamics question

say we have a circular wire with an emf producing a current

is it possible to help the emf with a magnet? by applying a B field at some part of the wire, with the lorentz force we have a term that goes like vxB which is perpendicular to both v and B. so at first, it seems that the electrons are not going to be pushed along the wire

but as soon as they are deflected, their direction v' is not along the wire anymore. so, $v' \times B$ could be in part along the wire?

i.e. we would be able to help the emf with a magnet?

is that better?

i don't see in latex

thanks but id rather use my brain as decoder rather than installing a firefox extension which would increase my fingerprint uniqueness

but ok ill use mathjax from now on

@tttt Blue gave me this, it worked for me and it's not an extension, just a thing you put in your bookmarks

that looks like magic lemme try

does not seem to work $\text{test}$

does not work.... what the heck

i restarted firefox

i have the 2 things in my bookmarks . to no avail

i dont have the bookmarks bar. but i right clicked and added them in the bookmarks, as mentioned

"^^drag this^^ to your bookmark bar or right click on it to add it as a bookmark. " i've done the latter

damn it. no freaking way im gonna add that useless bar eating precious vertical space of my small 1080p monitor

you have to click on the bookmark to render mathjax

i had clicked on one of them (not sure whether to start or not)

ah lemme retry

that works oOOOOOooo

wow thanks a lot

do i need to manually click on the bookmarks everytime i restart firefox?

@tttt exactly my reaction lol

eh, more often than that

ok i just tested. yes i need to reclick on the bookmarks

it keeps it rendered for some time, I don't know for how much exactly

ok then ill use my brian as decoder, it will be faster in the end

I just click it whenever there are equations I want to read

Mmm ...

What does normal incidence mean in optics?

@JakeRose It is perpendicular to the surface

I.e. parallel to the normal line

@SirCumference I got a question about normal interference in thin film interference. How would that even work?

*incidence

Is it good to use desk lamp in dark room

I can't seem to wrap my head around this: on a smooth manifold with an arbitrary connection, what does the parallel transport of a vector along a curve represent? I can take some Christoffel symbols and a curve and solve the equation and I find myself with a vector that varies as a function of the parameter of the curve, say like $v(\lambda)$. Should I interpret this as $v(\lambda_1)$ is "parallel" to $v(\lambda_2)$? What does parallel even mean in this context?

is this simply the definition of what it means to be pointing in the same direction on the manifold?

@user2723984 Yes.

oh ok, then I already had it, thanks

@ayc fyi science backs it up, heres a brand new study. some philosophies are healthier than others. they can have causative effects on health. The Power of Purpose and Meaning in Life— A new study reveals the pervasive effects of feeling that life is worthwhile. psychologytoday.com/au/blog/brain-waves/201901/…

and what about the Lie derivative? Why bother with a connection and not use the Lie derivative to define change of direction? I found this question but the answer just talks about the metric, I'm wondering about an arbitrary tensor field

@user2723984 How would you define it using the Lie derivative?

Well if $\mathcal{L}_X Y=0$ then $Y$ is parallel transported along the flow of $X$

hey guys, what is a good (introductory) book on quantum information?

we got a reader, but they don't explain anything, it's just exercises

@user2723984 But we want to say what it means to parallel transport a vector along an arbitrary curve.

@ACuriousMind I'm not sure I understand, usually we say that if $\nabla_X Y=0$ then $Y$ is parallel transported along the flow of $X$, how is that different?

the vector $X$ is arbitrary, we could take whatever curve and find the vector field of which it is the flow

Not every curve is the flow of a vector field

ok that surprises me, but then the covariant derivative has the same problem

@Blue ah nice, thanks

mmh if I have any smooth curve $\gamma$ on the manifold, I can define a vector field by $X_{\gamma(t)}f = (f\circ \gamma)'(t)$

and then $\gamma$ is the flow of $X$, or am I missing something?

mmh maybe it's not obvious that $\gamma$ should define a one parameter group of diffeomorphisms

@user2723984 Sorry, I misunderstood what notion of connection you are using. A formalistic reason is that the Lie derivative doesn't fulfill the axioms for a connection. For more, see physics.stackexchange.com/q/245675/50583 and the links in the answer.

@user2723984 I would use "flow" only for a vector field defined on the entire manifold

@ACuriousMind the links in the question have some interesting answers, thanks. What I gather from this is that we use the covariant derivative because it's a closer fit to what we want physically

still if someone (like, completely unrelated example, a professor in an exam) asked me to explain what does it mean that $\mathcal{L}_X Y=0$ I'd answer "the direction of $Y$ doesn't change along the flow of $X$"

Might be the funniest paper I've read

money well spent

"I'll do it for 100$ in the Blue's Journal of Advanced Computer Technology"

One of the few times I find something worthwhile on my quora feed :P

@Blue Didn't even notice the axes labels on the "graph"

@Blue About that cheating issue, I told his advisor. He immediately called him, expelled him from the lab, and told him he has to email Max Planck Institute and tell them he can't come for personal reasons.

does anyone know why they write $M_m^+M_m$, and not just $\langle\psi\vert M_m\vert\psi\rangle$?

@Mo_ woah

Assuming this isn't all a random internet story (which is not clear at all haha), just think how bad it is for someone to do something like that, there could literally be up to 100 PhD students applying for and willing to take that one place, takes months to apply and wait, then in the time wasted waiting for the person to accept all those people could have their plans totally changed etc, it would be simply terrible if the story were true

How could anybody even waste their time doing something so stupid, the story is so ridiculous, how could anybody even think it could work out

@Blue He defended his these 3 months ago, but has not yet graduated. I think he will have problems with that too for some other (cheating-related) reasons.

@ShaVuklia because if $|\psi_i\rangle=\sum_n a_n |n\rangle$ then $M_m |\psi_i\rangle=a_m|m\rangle$ and $\left(a_m|\psi\rangle\right)^\dagger \left(a_m|\psi\rangle\right)=|a_m|^2$ which is the probability you want

@bolbteppa I don't know what would have happened If I didn't see the transcript he sent to Max Planck. He could get expelled from there, or maybe he could somehow forge fake (hard copy) transcripts for them

But this is a true story! I was struggling with it and what to do for 3 days now

^ Oh I meant his thesis (typo!)

I will never join the military :)

still applying

No (I can do that but haven't yet). I mean I will start my PhD; here in the worst case.

Yes. Until Fall 2019 if I get admission from a university abroad

@user2723984 oh whoops, I was thinking we were dealing with the expectation value. thanks!

Our university is pretty good (ranked 1 nationwide and 45 in ARWU engineering), but I prefer another country for reasons Indian students probably know :)

@ShaVuklia there's a typo in there but you probably noticed, it should be $a_m|m\rangle$ not $a_m|\psi\rangle$

@user2723984 oh, yea I read past that, but thx

@user2723984 oh wait, why don't we have $M_m\vert\psi_i\rangle=ma_m\vert m\rangle?$

because $\vert m\rangle$ is the eigenvector of $M_m$ with eigenvalue $m$ right?

I don't think I understand the subscript for the measurement operator

@Mo_ Did you come clean and tell the advisor that you were actually helping the guy cheat?

@ShaVuklia I just called the eigenvalues $a_m$ instead of $m$

no wait

I have to look at the equation in the book

@ShaVuklia Sorry, I misunderstood the notation and I told you something wrong: here $M_m$ is a measurement operator, how these are defined is given at page 84

If $M_m$ is a normal projective measurement then you have $M_m^\dagger M_m=M_m^2=M_m$ and the probability is the one you expected

sorry, I should have thought more

@alarge We didn't bring that up, but I could have said I didn't know that question was for his application. I made a big mistake, but as I said, if I knew there would be even the slightest chances of him being accepted I would have never helped him.

If I have a amplitude that has a complex phase shift, is the intensity the square or the magnitude?

Do you address a graduate lecturer as Mr. or use their first name?

(I just started their class)

One of these professors that I contacted is getting very friendly in the emails; I'm thinking about using "Later alligator" as closing in one of my future emails to him

@SirCumference I think that is very culture specific. At my grad school I got a little of both plus things like "Teacher" and even "Teach".

QFT for dummies is a good book

Or QFT for your dog. Even better.

@Mo_ Sir sounds even more formal than Mr. :P