Quantum Mechanics Concepts: 2 Photon Polarisation (continued)

[@nicksu332]

sir, your paper never runs out! shows that you are actually a divine physicist

[@oscargallego8952]

Time has passed, still the best of the best. Sir, you are a marvellous, generous professor. Cannot express well enough my gratitude.

[@juliogodel]

The series is awesome.The derivations and explanations are crystal clear. Thank you very much for taking time to do this. The videos are great and you are an excellent teacher. Please, keep going!

[@brendawilliams8062]

Even an amateur mathematical involved person can follow your lectures with effort. Thankyou

[@TekCroach]

I must say these set of videos on quantum mechanics is the bestEST videos (and tutorials) I have ever ever seen. Great.

[@edlynnnau536]

TekCroach: Absolutely Agree. Things that have been so murky he makes clear. He doesn’t skip a lot of steps so if we haven’t reviewed our Maths in a number of years then, not skipping steps, helps us remember. Love it!

[@rbettsx]

The best ad for Berol felt-tips I've ever seen

[@najeebullah7415]

The first ever incredible video on quantum mechanics I found,...even better than MIT lectures. THANK YOU for such a great and simplified presentation.

[@FalconX88]

As a chemist who never had proper (mathematical) training in quantum chemistry and wants to understand things like hartree fock calculations these videos are gold! Thanks for producing them.

[@princeistalri7944]

Thank you so much for uploading these videos. Your teaching style is fantastic. That and the accent.

[@jimdogma1537]

I did persevere, and it turns out that those first ten minutes were the only bumpy part of the trip. Good video, looking forward to the rest. Thanks!

[@skasmosAE]

damn!!!wish my professor could teach us these stuff so elegantly as you do!!!and i pay $4,300/semester for my classes!!!

[@matthewmoreman4000]

Thank You so much for this great series of videos. They are immensely helpful. This is the best way I know to learn more advanced physics, short of classes at a university.

[@satyamkumar567]

Best teacher in the world

[@Umrahhhh]

ChatGPT recommended me this channel and it's so good 😍❤️

[@sayandatta7685]

simply a great series of lectures! Thank you so much!

[@DrPhysicsA]

When we take the complex conjugate of the vector we take the Hermitian conjugate (ie the transposed complex conjugate) of the matrix/ operator

[@ranjitsarkar3126]

Ooooo

[@lukealucard]

what a magnificent teacher... congratulations DrphysicsA

[@bernatjv19]

I'd love if you did some videos like these about quantum electrodynamics. By the way these helped me a lot thank you very much

[@aeroscience9834]

A simple solution to stop confusing squaring with multiplying complex conjugates is to right bars (modulus) around the number before squaring.

[@VanWarren]

is to write bars (modulus) around the number before squaring is to apply the modulus operator to the number before squaring is to square the modulus of the number

[@akhilreddy3771]

this is the best series i ever see

[@naimulhaq9626]

I cannot wait to view Dr. PhysicsA's video on Maldecena's conjecture and how the universe is a quantum computer. Dr. Physics is a magician.

[@AlchemistOfNirnroot]

? He hasn't uploaded in years tho ?

[@DrPhysicsA]

Hope you manage to persevere. Part 2 seeks to show what the matrix is for a polariser at a general angle. Part 3 which I uploaded today may be a little easier.

[@DrPhysicsA]

They are both. You can consider / and \ to be superpositions of vertical and horizontal basis vectors, or they can be basis vectors in their own right (since they are of course simply a rotation of the horiz and vert vectors by 45 degrees). But they have a different Hermitian matrix to that which applies for horiz and vert.

[@momentumj4425]

Thanks for your lecture, I don't have to get a doctor's degree for my paper. Just watching yours is enough.

[@Quantiad]

In the beginning, I think it's worth clarifying that the Hermitian matrix is an Operator, NOT an observable. The resulting eigenvalue is the observable associated to the operator. It's the 'lambda' that represents the observable, such as momentum or position. The Operator is the tool for finding it.

[@DrPhysicsA]

Depends what you mean. if you mean a polariser with cross wires then none of the photons would get through because it would be possible for electrons to move in the horizontal and vertical planes.

[@elinhagglund33]

Thank you Dr Bob Eagle (Ive heard that is your name). I am studying a bachelor in physics (and actually philosophy to, im doing two bachelors at the same time) and I absolutely loooove your videos. They are so easy to follow and fun to watch. Thank you very much for making them for all of us. Very inspiring =)

[@fragarretu]

Just awesome lecture!. Congratulations and many thanks

[@kadenkwan1157]

DR.Physics A, it is a really good series. I have got a question at 6:24. The complex conjugate of a hermitian matrix does not necessarily equal to itself but the complex transpose does. can you explain why you can do that? many thanks

[@clivegoodman16]

Kaden Kwan Because he reverses a and b, he is effectively transposing the operator.

[@newtonianartist7088]

there's something, in the last video you said that a hermitian matrix is a matrix that equals to its "dagger" form which is transposed complex conjugate. but in the working at 6:25 you only used complex conjugate and that hermitian matrix remained unchanged. why is that ? (its certainly very convenient but i cant see how)

[@peterclark5244]

Because a Hermitian matrix is one where the matrix M is equal to M dagger. When you multiply your M Hermitian to a ket vector you get the vector multiplied by the eigenvalue.

[@remavas5470]

+Peter Clark i think the question is still open, because M = M(dagger), but why can he replace that M = ~M(conjugate)?

[@alanzelicoff3421]

I agree this is still an open question. H(dagger) is not the same as H(complex conjugate), What am I missing?

[@SergioNasi]

It may have to do with the way you handle and conjugate the 3-factors inner product: If you apply the complex-conjugate operator you may have to apply it twice to the middle term H so that in the end it remains unchanged, while the extremes (a and b) get swapped. I tried to derived the formula of * resulting in to no avail so far, thus I'm just speculating

[@joefagan9335]

I agree that it needs to be explained why in lect 1 H being hermitian says its equal to its own dagger, then in lec2 its equal to its own complex conjugate. Also, does mean we work right to left so equals and not

[@DrPhysicsA]

What I try to show is that the probability that a photon which is polarised, say vertically, will then pass thro a polariser at angle alpha to the vertical is cos^2 alpha. So on average for every 100 photons cos^2 alpha will get thro and that equates to a reduction in energy of cos^2 alpha.

[@skypickle29]

at 6:25 you are evaluating * I assume the next step is Then you say 'the complex conjugate of a Hermitian matrix is itself' but a Hermitian matrix was defined earlier as a matrix H, which is equal to Hdagger, where Dagger is the 'transpose of the complex conjugate' and you gave the example of matrix,m | 1 1+i | | 1-i 3 | so m* is | 1 1-i | | 1+i 3 | m.dagger is | 1 1+i | | 1-i 3 | so m=m.dagger, so m is a Hermitian but m is not equal to m*, which is what you said at 6:25 the reason this is a problem is with evaluating * you proceed to say * goes to because H=H* which I just showed above it is not. so,,, help?

[@byronhale9874]

Problems with character representation: Make that result |a> is orthogonal to |b>.

[@Jipzorowns]

I can't thank you enough for these video's!

[@isrcast01]

Thank you for making the videos, clear and nice way of teaching.

[@AndreKowalczyk]

At 6:30 H is by definition equal H dagger, which is not equal to just its complex conjugate. So, I think, you need to first apply H to |a>, and only after that perform complex conjugate on the result.

[@LaureanoLuna]

Incredible videos. 0:11:40 I'm a little confused about whether forward slash and backward slash are superpositions or eigenstates. THey don't seem to behave as eigenvectors of the H introduced earlier. Also, they are not orthogonal to the horizontal and the vertical states. Is something like a change of basis involved? 0:11:50 How do we know that the vector with components 1/sqrt 2 and -1/sqrt 2 has length -1? The Pythagorean theorem is compatible with both 1 and -1. Thanks.

[@IndiraChowdhuryMusic]

Excellent video !! Thank you sir.

[@jihadbasuni23]

You have my respect, thank you

[@jurrie1998]

Are the eigenvalues always 1 or -1 and nothing else?

[@Nobody5599]

I think it is mostly derived from empirical evidence and classical physics. The thing about quantum mechanics is that people can't wrap their heads around it. Physics is about understanding the nature and coming up with theories to describe the nature. That's why people describe momentum p=mv and the probability of your question with working equations like p=cos^2(alpha) (p as in probability). The equation itself is part of the classic mechanical physics.

[@sandorfogassy3007]

Great introduction. Thank you.

[@Quantiad]

In the final part, when you need to take the norm squared of (1/sqrt 2)cos + (i/sqrt 2)sin, I think it's easier to factor out the (1/sqrt 2) and recognise that cos(theta)+isin(theta) = e^(i*theta). It's very quick and intuitive to calculate zz* from here: [(1/sqrt 2)e^(i*theta)]*[(1/sqrt 2)e^(-i*theta)] = (1/2)*e^0.

[@rosanella8]

Very clear explanations! BIG THANK YOU! By any chance, DrPhysics, could you explain to us Ladder Operators with various examples using creation and annihilation operators, please?

[@damiangowor1404]

6:12 The Hermitian Matrix is equal to its own conjugate transpose, not its own conjugate. Is there any other assumption here in relation to H in this example?

[@arlenestanton9955]

What is one complete muke?22:25?

[@edwardarruda7215]

I wish this video was available when I was taking Physical Chemistry years ago.

[@gk89300]

Another great video, thank you. I am really enjoying this series. I was wondering whether a cross-slit (2 slits at a 90 deg angle in 1 plate) will let through every single photon. I would assume that 100% gets through this way, is this a correct assumption? And another question: is it true that approx 97% of the photons will go through a series of 90 slits (1 - 90 deg theta), so that there is a high efficiency in polarizing the light from horizontal to vertical? (cos(1)^2)^90 = 0.97.

[@dr.binodgogoi3642]

It is nice. Thank you for uploading.

[@Gruemoth]

at 44:31 why can't alpha and beta be 1/sqrt(2)?

[@Unknownxyz100]

ok can somone explain the hermitian operator for me? so why do we need the operator when we already have the does the eigen vector represent i know it's the state but what state? how is it different from the state of polarized light like normal ket vector, i know the operator means it's observable but don't we already calculate it through probability amplitude?

[@BlackEyedGhost0]

33:10 There's a trick to remembering (or quickly deriving) these identities using complex exponents: e^it = cos(t) + i*sin(t) e^i(x+y) = cos(x+y) + i*sin(x+y) e^i(x+y) = (e^ix)(e^iy) e^i(x+y) = (cos(x) + i*sin(x))(cos(y) + i*sin(y)) e^i(x+y) = cos(x)cos(y)-sin(x)sin(y) + i*(cos(x)sin(y)+sin(x)cos(y)) cos(x+y) = cos(x)cos(y)-sin(x)sin(y) sin(x+y) = cos(x)sin(y)+sin(x)cos(y) It's just these three simple identities combined: e^iθ = cos(θ) + i*sin(θ) a^(b+c) = (a^b)(a^c) (a+ib)(x+iy) = (ax-by) + i(ay+bx)

[@haakonvt]

Great series!

[@d_9696]

45:20 Since you are starting to use complex numbers, the actual condition changes to |α|² + |β|² = 1. Note that α² is not (necessarily) equal to |α|².

[@lastresort2446]

Thank you so so so much for such an excellent video. you are excellent teacher. :-)

[@minafathy83]

excellent sir, i read messiah qm beside your explanation. many thanks

[@alexbenfield3374]

Why did you add the hat onto the Hermitian Matrix? Does that serve any importance?

[@MrFeanaro9]

The hat is just to signify that the Hermitian is an operator.

[@hagglefaen]

I'm really confused about the inner product of the left and right vectors at 48:41, why didn't you square it? I mean, the number makes sense, it should be zero, I guess, but If you square -0.5 you will get +0.25. Squaring 0.5 gives you +0.25. 0.25 + 0.25 gives you 0.5.. so now the answer is 0.5 which doesn't make sense to me. I realized this when I went through the calculations in your electron spin video, where you squared it before adding them together at 26:15 in the part 3 video (0.5 + i/2 was squared, that is, multiplied with its complex conjugated). Why aren't you doing the same here? Sorry, if I'm splitting hairs, it's just that doing it in two different ways makes me confused.

[@CandidDate]

I'm wondering if you had |a> + |b> + |c> where you do three tests instead of two. Would the results use the same Pauli matrices?

[@00058000]

Nice to know that even experts like yourself have trouble memorizing trigonometric identities, I could never keep those straight either.

[@nishadingale5408]

i am wondering how it was concluded at 44:38 ,that the probabilities(alpha^2 + beta^2) won't add up to 1, due to which "i" was chosen as a possible solution?

[@jomen112]

It is somewhat complex to motivate, and maybe somebody else have a more correct motivation than I do. I am not sure I got it right even, but in short, and according to my understanding, i reflect the fact that circular polarization has one more degree of freedom. Circular polarization is composed of two phase shifted waves. That means its representation is time dependent. This time dependency is represented in the complex plain. The long story: Circular polarization is a helix (cork screw) that goes along the z-axis. This helix can be decomposed in two orthogonal waves which are phase shifted 90 degrees. If you look down the z-axis, i.e the time axis, the helix can be represented as a vector V which rotates in the x-y plane, i.e. it rotates in space only. The decomposed helix can be represented by two vectors v1 and v2 (right now I am not sure if they can be identified with Ex and Ey, but don't think so). These vector is orthogonal and represent the two waves making up the component waves of the helix. The magnitude of the vector V is set to 1 (I guess it might reflecting the fact that the wave has a persistent physical existence, i.e if you check if it exists, it should answer yes 100% of the times). The vectors v1 and v2 have the same magnitude (I haven't checked but I suspect the conservation of energy might be violated or something else funny happens if they did not have the same magnitude). This tells us the length of v1 and v2 must be 1/SQRT(2) in order for V to become 1. I guess this is where the term 1/SQRT(2) might come from, or it might just reflect the fact that v1 and v2 is orthogonal. Not really sure here, since I have not been thinking very much or deep about this. I any case, the main point is: since the phase shift between the two waves are represented as two orthogonal vectors in the complex plane, the x-y plane, it means i would represent the phase shift, i.e. time differences along the z-axis, between the two composite waves of the helix. That is, the imaginary part would be time projected into a space representation.

[@beehappy5677]

On min 12.10 i got a bit confused when all four states were presented together asociated to eigenvalues 1 and -1. But if i remember well the second pair were obtained with a different operator and then the eigenvalues -1 and -2 obtained with that second matrix belong to a different set???

[@JoshuaAalampour]

12:40

[@joannalada5815]

Thank you so much for making all of these videos; you're so great at explaining complex concepts in simple terms. I just have a quick question: For |\> i.e. the backslash ket vector, could you also call it (-1/rt2 1/rt2) rather than (1/rt2 -1/rt2)?

[@ardiris2715]

At 6:30, does that mean = directly? Working this out would have been helpful. Thank you.

[@AlchemistOfNirnroot]

He did in Quantum Concepts 1. Although has been a while since I've watched it so yh.

[@aryamanprajin7225]

No he complex conjugated it. Watch it carefully and you will notice that.

[@Downlead]

Why are the Pauli Matrices only applied onto the 3 pairs of coupled orthogonal eigenvectors, which their eigenvalues are always +1 and -1: The 1st Pauli Matrix, the eigenvectors are the Vertical and Horizontal Linearly Polarized Lights. The 2nd Pauli Matrix, the eigenvectors are the Left and Right 45 degree Linearly Polarized Lights. The 3rd Pauli Matrix, the eigenvectors are the Clockwise and Anticlockwise Circularly Polarized Lights. What about the other tetha degree pairs of coupled orthogonal eigenvectors? I know the other pairs are applied to the general form of Hermitian Matrix, but What makes the Pauli Matrices different?

[@lamjingbakhangembam130]

Thank you for valuable information... Now I will do master in physics

[@absolutezero3512]

Ummm, professor. Isn't |z|^2 = zz* ?? You have asserted at 45:20 that it is z^2 = zz* . How can be i^2 = 1?? Please correct me if I am wrong.

[@absolutezero3512]

Could it be that we have to square and add the modulus of the eigenvalues?? like |alpha|^2 + |beta|^2 ??

[@clivegoodman16]

[Absolute Zero] Because multiplication is commutative z*z=zz*. i^2 = - 1.

[@hooinkyoma8045]

First of all, it is a very good video. However, there is a minor mathematical mistake that might confuse some people: |z|^2 means something very different than z^2. You write the second expression but say you mean the first. In this case what you mean is quite obvious from context, however I recommend always using correct notation if you are teaching. (Otherwise students will adopt it and make mistakes...)

[@danielblumowski34]

[Absolute Zero] |z|^2 is the square of a absolute value (a modulus) of a complex number and it's equal to zz*, but zz*=/=z^2 unless z is a real number

[@Unidentifying]

brilliant! thank you so much

[@fadibitz]

Fun Fact: To create circular-polarized light: 1. First constrain the light to a linear 45° polarization. 2. Send the linear-polarized light through a "quarter-wave retarder. NOTE: A quarter-wave retarder retards one component of the light so that one component of the light that emerges 90° behind the other, thus 90° out of phase. To reverse the polarization, rotate the retarding medium. Circular-polarized photons reverse their polarization when reflected.

[@byronhale9874]

Still succinct and excellent. The result that lambda(a) =|= lambda(b) => |a> _|_ |b> seemed to get lost. Still, a lot of work to get this so organized.

[@nan9849]

How is he choosing the matrix for states? As a beginner, I really have no clues at it.

[@sparkliang]

A very nice video!

[@abhishekshah11]

Can anybody point out where does the Hermitian operator from @33:46 come from?

[@RangaNanayakkara]

+Abhishek shah you can check it for your self its for any value of teeta and any value other makes teeta and its orthogonal. he just put it there but you can see for your self that its correct. which means that will help with any direction.

[@LaureanoLuna]

Thanks for the clarification. Now, I see my second question came out of a mistake on my part.

[@arlenestanton9955]

20:48 isn’t that -cos theta,(-x/r), sin theta,(y/r)

[@DrPhysicsA]

The polarisation here is simply restricting the oscillation of the E field to one direction. Nothing to do with charge.

[@archishmore6276]

how do know that the if the eqution is not satisfying the probability the to divide both therms by 2^0.5

[@deltalima6703]

Omg. Could do circular polarization clockwise at various x positions, along with anticlockwise ones, and get a zoo, all perfectly under control using QM.

[@SamsherSinghJoon]

very interesting video sir.

[@thoseini]

@6:15 of the video you mention H="complex" conjugate of H since H is a Hermitian operator, while for a Hermitian operator we have: H= "Transpose Conjugate" of H.

[@ronniedahlgren2733]

I have the same question...

[@camerondavis6607]

You mentioned that you cannot prove why the inner product is the probability amplitude, but is it possible to prove that it is unprovable. (Sort of a fermats last theorem idea)

[@Gwunderi25]

The Eigenvalues +1 and -1 remind me of the slope of two orthogonal straight lines, where if the slope of the first is m, the slope of its orthogonal is -1/m. Has this +1 and -1 someting to do with it, or has it an analogy? (For vertical lines there's no slope, for you cant take diff(y) /diff(x) when diff(x) is 0). But apart from this special case, is there an analogy?

[@jomen112]

Being orthogonal is being at right angle and is represented by 0. The meaning of a minus sign in physics is most often understood as 'the same but opposite', e.g. the velocity v and -v has the same magnitude but in opposite directions. What "opposite" and "same" stands for is depends on the context, therefore it is vital to know what the objects represent before you can tell the meaning of a minus sign before it. Example: the meaning of a sloop m vs. a sloop -n might be understood as someone travel up a hill at the angle given by m. The meaning on -n then becomes someone travel down the hill at the angel given by n. Thus the minus is to be understood as "the opposite direction" when -n is in relation to m, and "the same angle but opposite direction" when -n is in relation to n, i.e. modeling the fact that the direction cos(x) = -cos(x+Pi) - well almost the same.

[@오현우3605]

Um.... Professor, I think the signs on the vectors for the RCP and LCP should have the opposite signs for the complex components.

[@davidhand9721]

Isn't there something like a rotating phase? I believe it would happen if the magnetic and electric waves are out of phase, or something similar? Would that take some of the spooky magic out of the serial filters |, /, -?

[@anjanjana79]

The proof of hermitian matrix Can you make another video on its proof. I can't understand it

[@DRsidd5]

Best Best Best Best The Best

[@axl768]

Yes, I absolutely agree! You are an educational genius. And yes, it is a pleasure to hear your clear spoken English!

[@novotnyingersol9200]

5:50 is from an ancient Egyptian hieroglyphic cartouche!

[@raminnikzad6317]

Hi, something doesn't look right to me. lets call the thing you drew in time 42:56 (clockwise thing), lets call it CW. So then you argue it must be 2 because of ZZ* . Based on your last lesson i can say"II squared" equals Probability. Now because ICW> is [(1-i)transposed ] THEN

[@jomen112]

_"Is my thinking right?"_ No. It goes wrong in this step _"And Then its squared would be 4 !"_ and this _"must be 2 because of ZZ*""_ It "must" not be 2. ZZ* IS 2 if Z = 1+i. Given the equation a^2 + b^2 = c^2, tells us that a^2 + b^2 = 2, when a=1 and b = i. This does not mean c is equal 2 but SQRT(2). The number 2, i.e. c^2, is the sum of the squared probabilities. Z^2 = ZZ* is a rule. It tells us how we should evaluate the expression a^2 + b^2 , namely as .aa* + bb*. Now it just so happens that a^2 + b^2 = 2 and ZZ* = 2. The mistake I think you did was to confuse the rule Z with a and b think in terms of calculating a^2 + b^2 as Z^2 + Z^2 = ZZ* + ZZ*, which indeed is 4. That's how I think you got to the number 4 in the first place. It might help if one think about this problem in geometrical terms instead. The task is to calculate the _squared_ distance between a and b. Geometrically we have a triangle with corners in origo (0,0) and the coordinates (1,0) and (0, i), where the corner in origo is at right angel. The hypotenuse spans the distance from (1,0) and (0, i) along the diagonal line f(x) = 1-x. The squared distance of the hypotenuse must then be 2, since both catheti are of unit length.

[@alanlao5112]

The z z^* is the square of the modulus of z instead of square of z for any complex number z.

[@selsebilx]

Could you explain this; What does it mean ' -1 ' of eigenvalue?

[@RangaNanayakkara]

+Sapiens Sapiens its just a representation of observable. if you have two slits if a photon goes from first you put a red bulb to indicate it, as here it can represent by +1 and the other -1. or up down spin of a electron.

[@jietzemiedema8002]

Is eigen an unit or word in a language

[@jimdogma1537]

Oh man, what happened? I understood the entire 1st episode of this series and now I'm completely lost, and I'm not even 10 minutes in. What happened to the matrices and their relation to the pretty red and green lights with the slits? This the same problem I had with the Einstein field equations video. I got it up until the Christoffel symbols when no mortal could keep track of all those dancing indices without a connection to a real-world model, like the bumpy lawn. Just some fan feedback.

[@alwaysdisputin9930]

1:10 "the Hermitian operator is the measurable" 2:10 "the eigenvalue is the measurable" Please can someone help resolve this apparent contradiction?

[@lunagh6149]

no it's the eigenvalue is the measurable, and hermitian operator represents a physical quantity ( observable)

[@alwaysdisputin9930]

@@lunagh6149 Thank you for your reply. I am going to study these videos hard so I can make sense of all this.

[@zombieblitz8305]

The hat on the Hermitian matrix is there to denote the fact that it is an operator.

[@HarutoSato01]

Great video thank you!

[@martinsupinda]

Is this second year or first year undergraduate physics? Thanks I loved these videos. You are a real hero in my mind.

[@DrPhysicsA]

I suppose it depends how the courses are structured. I suspect its more likely to be second year.

[@martinsupinda]

Hi Bob About 35 years ago I went to university to study Physics and ended up with an ordinary degree with millions of holes in my knowledge and I have always wanted to know what I had missed out on!! Thanks to you I now understand so much more than I did and I have achieved certain goals which I had set myself. I have a better understanding of the formalizations of quantum mechanics speed of light Maxwell em wave theory. Lagrange was vey interesting too I did that question without understanding or ever hearing about Paths of Least Energy!!!. Mostly at the moment I am plodding away at general relativity. I find your videos relaxing and an important part of the time I reserve for my holistic development program.!! Thanks again. Best wishes Martin Back to the garden....holidays!!!

[@pepaxxxsvinka3379]

@@martinsupinda Hi! How are you doing?

[@AlchemistOfNirnroot]

52mins~ in and it doesn't make sense? How can it fall of by the same amount regardless of angle?

[@odedbasis4423]

what I didnt get was how to build the Hermitian operator for the circular polarization

[@lawrencemwangi8846]

how do you know what hermitian matrix to choose for the operation? Somebody help please

[@powerblo5019]

You set the Hermitian matrix first, then you set the eigenvalues and eigenvectors.