Dirac notation: state space and dual space 

You are absolutely correct, thanks for finding the typo!

Whew...I watched that 3 times scratching my head and was about to give up until I saw this comment!

did not compute xD. Thanks that got me hung up for a second too. Glad someone pointed it out here

@@ProfessorMdoesScience Can you pin this comment, please? Excellent quality of videos in general, thank you!

@@universemaster Pinned! :)

Thanks so much -- this really encourages us! We're very glad it is helpful!

Glad you like it!

Thanks, comments like yours motivate us to keep going!

Thanks for the kind words, they certainly motivate us to keep going!

Thanks!

Glad you like our approach! :)

Glad you like our videos! :)

Glad you like the videos! :)

Great to hear! :)

This is great to hear! :)

Thanks for your kind words! Where are you studying your undergraduate?

Glad you think so, and thanks for watching!

Glad you find our videos useful! What book are you using?

Glad to hear that this is useful, and thanks for watching! :)

Really glad you like our approach! Where do you study?

Glad you think so!

Glad it was helpful! May we ask where you study?

Haha. Wait there! The abstract stuff is extremely important as well. In a first course, you might think you don't need all of that stuff but as you progress further, you will realise the importance of it.

Thank you *so* much for your feedback - this was exactly our intention and we're thrilled that you like it!

Glad you like it!

Glad you like it!

Thanks for watching!

Glad you like it!

Thanks for watching! :)

Glad you like it!

Glad to be helpful!

Thanks! :)

Thanks for watching!

We hope so, although we would first like to make some more progress on the quantum mechanics series.

This is a great suggestion, thanks! We are preparing extra material (including questions+solutions) and will also look into a Q&A.

To fully answer this question one would need to delve into the mathematics of inner product vector spaces. But for our purposes, the fact that we have an inner product vector space allows us to use notions such as the length of a vector (for example to work with normalized states) or the angle between two vectors (for example to compare how similar two states are). I hope this helps!

We like a range of books, including: "Modern Quantum Mechanics" by Sakurai, "Quantum Mechanics" by Cohen-Tannoudji, and "Principles of Quantum Mechanics" by Shankar.

Yes, we are working with a complex vector space, and scalars are complex numbers in general.

The various playlists do follow a good order in which to watch the videos. And amongst those, a good order would be the following: 1. The postulates, which gives you the basic mathematical background for quantum mechanics: ru-vid.com/group/PL8W2boV7eVfmMcKF-ljTvAJQ2z-vILSxb 2. The quantum harmonic oscillator, a good example of applying the postulates for a simple system: ru-vid.com/group/PL8W2boV7eVfmdWs3CsaGfoITHURXvHOGm 3. Angular momentum, which extends the ideas above to 3D: ru-vid.com/group/PL8W2boV7eVfmm5SZRjbhOKNziRXy6yIvI 4. Central potentials and the hydrogen atom, good applications of the ideas behind 3D quantum mechanics: ru-vid.com/group/PL8W2boV7eVfkqnDmcAJTKwCQTsFQk1Air and ru-vid.com/group/PL8W2boV7eVfnJbLf-p3-_7d51tskA0-Sa There are a few other topics (like time dependence or uncertainty principles) that you can watch in parallel to the above. With these, you should have a basic grasp of quantum mechanics (about 2nd or 3rd year undergraduate student). After this, more advanced topics include things like identical quantum particles: ru-vid.com/group/PL8W2boV7eVfnJ6X1ifa_JuOZ-Nq1BjaWf or second quantization: ru-vid.com/group/PL8W2boV7eVfnSqy1fs3CCNALSvnDDd-tb or density operators: ru-vid.com/group/PL8W2boV7eVflL73N8668N0EQUnID1XaEU We are still working on adding more videos, and also working on a website to facilitate self-study, so stay tuned for that! And hope this helps!!

Glad you like our videos! Quantum mechanics textbooks we like include those by: Sakurai, Cohen-Tannoudji, Shankar, and Merzbacher. I hope this helps!

You are absolutely correct, this is a typo!

@@ProfessorMdoesScience Great videos btw.. thanks

This sounds correct. Additional comments to make are, first, that state space is a complex vector space, as opposed to real for the Euclidean space. And second, state space can be finite or infinite, it depends on the quantum system of interest. For example, if the quantum system of interest is the spin of a single electron, then the state space happens to be 2-dimensional only. I hope this helps!

Oooh i understand know. Yupiiiiii

|psi>

For a "physics" approach to these concepts, the quantum mechanics books I like include: Chapter 1 of "Modern Quantum Mechanics" by Sakurai, Chapter 2 of "Quantum Mechanics" by Cohen-Tannoudji, Chapter 7 of "Quantum Mechanics" by Messiah, or Chapter 1 of "Principles of Quantum Mechanics" by Shankar. I hope this helps!

@@ProfessorMdoesScience Thx

Thanks for the feedback -- all I wanted here was a simple analogy with a familiar vector space, and many students have not yet covered Lagrangian or Hamiltonian classical mechanics when they first start studying quantum mechanics, hence the comparison with a 3D Euclidean space.

@@ProfessorMdoesScience thanks for the reply

I've heard there is a formula like = which is used for obtaining an expectation value of operator B, which is based on the scalar product, but... what's a practical meaning of this after all

We cover the meaning of all these quantities in our videos on the postulates of quantum mechanics, which you can follow in this playlist: ru-vid.com/group/PL8W2boV7eVfmMcKF-ljTvAJQ2z-vILSxb Talking specifically about the expectation value of an operator B in a state |a>, , then you can find the relevant video here: ru-vid.com/video/%D0%B2%D0%B8%D0%B4%D0%B5%D0%BE-rEm-Ejg5xek.html This has applications, for example, in the study of the Heisenberg uncertainty principle: ru-vid.com/video/%D0%B2%D0%B8%D0%B4%D0%B5%D0%BE-pfjQtyLBBHw.html I hope this helps!

​@@ProfessorMdoesScience Ok, thank you a lot, will try to figure out

We like a range of books, including "Modern Quantum Mechanics" by Sakurai, "Principles of Quantum Mechanics" by Shankar, and "Quantum Mechanics" by Cohen-Tannoudji; but in the videos we try to use what we think is the most effective approach, which doesn't necessarily follow a specific book. I hope this helps!

Yes, like any definition this is a choice we make. The important feature is that by defining one argument to be linear then we ensure that we can define a usual norm that is always positive or zero. If both arguments were linear, we could not ensure a positive norm. I hope this helps!

I am not sure I have a good answer for this. Quantum theory does require an infinite dimensional space to describe even a single particle in one dimension. There are many ways in which one can put this, for example, the canonical commutation relation [x,p]=ihbar is not well-defined in finite dimensional Hilbert spaces. So, in a sense, the answer is: because this is the only way in which we can make it work...

@@ProfessorMdoesScience it was helpful, thank you very much:)

Unfortunately we don't have written notes, but if you think they would be useful we can add this to our list of things to do moving forward... thanks!

@@ProfessorMdoesScience Yeah, that would be very helpful actually, it takes some time to rewrite it, so I think it would be better to have more time of recalling in mind, what we have on paper :) or in pdf, whatever. But very nice videos! I'm going to check them all to my QM course ;)

Thanks for the clarification! We actually explore this idea in some of the subsequent videos in the playlist on the postulates of quantum mechanics.

Except that none of this has anything to do with "the state of the physical system". The easiest way to build a proper intuition is to consider the equivalent case for dice. Dice can be in two physical states: they can be resting on the table, in which case they have zero kinetic energy in the rest system of the table and then "their state" can be characterized by one of the outcomes "1" to "6" or they can be rolling, i.e. their kinetic energy is not zero and then their physical state has nothing whatsoever to do with the mathematics of the probabilistic outcome space that is spanned by the six outcome vectors. To talk about "the state of rolling dice" is completely meaningless within the framework of probability theory. We can, however, talk about the state of the dice throw ensemble, which for fair dice would be the equal superposition of all six outcome vectors. If you transfer this to quantum mechanical systems, then it's fair to say that the wave function (or, more generally, the density matrix) describes the state of our knowledge about the quantum mechanical ensemble, whereas the individual measured outcome tells us the outcome of the energy transfer in an individual element of the ensemble (unlike dice quantum systems can transfer different amounts of energy to the measurement system before they "come to rest"). What we can't say is that an individual quantum system has "the state of the wave function". That leads to all kinds of problems (mostly of the philosophical kind) that are not covered by any known experimental result.

Yes, a is in general complex!

Yes, the Euclidean vector space is an example of a Hilbert space.

R3 is indeed a Hilbert Space. However, in the infinite-dimensional case, the existence of the inner product is not enough, to be a Hilbert space every Cauchy sequence of vectors has to converge to an element of the given space.

You are correct that in the matrix formulation of quantum mechanics, the dimension of the state space is equal to the length of the column vectors that represent kets. The dimension of the state space depends on the system, for example it is 2-dimension for a spin-1/2 system, and infinite dimensional for a particle moving in 3D space. What system are you investigating that has a 3-dimensional state space?

@@ProfessorMdoesScience Thanks for the reply. It wasn't any particular system that I had in mind, I just remember solving QM problems with finite-dimensional state spaces so the 'Infinite D' statement confused me. To clarify, if you had a particle moving in 3D space would the classical description be 3D, i.e. could be represented by a 3D column vector with real number entries? Whereas the QM state space would be an infinite-length column vector (also with entries from the reals?)?

In the matrix formulation, a 3D particle would indeed be represented by an "infinite" column vector. However, note that the entries could in general be complex numbers. In practical calculations, we cannot use infinite dimensional bases, so they are invariably approximated with finite-dimensional basis. I hope this helps!

@@ProfessorMdoesScience Thanks a lot and nice videos!

Applying an operator on a vector is not equivalent to calculating the inner product. An operator A applied to a vector |psi> gives another vector, A|psi>=|phi>. By contrast, the inner product between two vectors gives a scalar, =c. We go into some detail about operators and their properties in this playlist: ru-vid.com/group/PL8W2boV7eVfnb10T_COKPozxEYzEKDwns

@@ProfessorMdoesScience But why it is necessary to define an inner product to describe operations on vectors?

First, a disclaimer: I am no expert in the basic mathematics of this. But let me just say that the inner product space in quantum mechanics is used so that we can have a self-adjoint operator (i.e. a Hermitian operator). For more general operators, I imagine that it is possible to define them without reference to an inner product.

Interference in quantum mechanics is indeed obtained directly from the wave function of the system. It arises from the so-called "cross terms", which are essential to reproduce the correct interference pattern for any quantum particle. We briefly touch on interference in this video: ru-vid.com/video/%D0%B2%D0%B8%D0%B4%D0%B5%D0%BE-LR5kfhrs4Cc.html I hope this helps!

@@ProfessorMdoesScience Thanks for response. Is there a mathematical basis to only retain the Re term of the cross product.( I hated it during High School that Physics Prof goes that as we live in the real world, we retain the real terms.....It is fine if we confine to only positive real numbers from the beginning.). Without understanding the maths involved, it appear to be similar to curve fitting, at least for me. One more thing, do you have a source which which explains in detail the correlation between the actual interference pattern and wave function. (just like the century classic Hyugens wave circles and interference pattern ) Thank in advance

The key is that we do not only retain the real part, we must include the full complex nature of the wave function to get the correct answer, so your high school teacher was wrong on this point. The probability of the particle being somewhere (the intensity you are referring to) is equal to the absolute value squared of the wave function. This quantity is real, but if you calculated it only from the real part of the wave function you wouldn't get the correct answer because you'd be missing the intereference terms. As to resources, I think most quantum mechanics textbooks will have a discussion of this, but a good concrete resource may be the Feynman lectures. I hope this helps!

@@ProfessorMdoesScience Just to sure we are on the same page, I rephrase my query. At time 7.28 of video ru-vid.com/video/%D0%B2%D0%B8%D0%B4%D0%B5%D0%BE-LR5kfhrs4Cc.html , you state that only the Real part of the 3 rd term is valid.term. Why do we only include the Real part and ignore the Imaginary part ? Other than textbooks declaring that we take only the Real part as we live in the Real world and such results correspond to expirements, is there a valid mathematical reason to ignore the imaginary part?. I am perfectly happy to use this convienient trick in electricity but have difficulties to accept in in Quantum mechanics Cheers

Ok, I see. I am *not* taking the real part and ignoring the imaginary part in that step, we are including all terms but the final expression just happens to depend on the real part only. To see this, let me first simplify notation. What we have to evaluate is the absolute value squared of the sum of two complex numbers: |z1+z2|^2, where z1=a1+ib1 (the c1 in the video) and z2=a2+ib2 (the c2 in the video). This is equal to: |z1+z2|^2 = (z1+z2)* x (z1+z2) by definition of absolute value squared, and where "*" means complex conjugate and "x" means multiplication. For the next step, we multiply the terms through (this is the step missing in the video): z1* x z1 + z1* x z2 + z2* x z1 + z2* x z2 (1) I am calling this "equation (1)", because we will come to it later. The first term is simply z1* x z1 = |z1|^2 and the last term is z2* x z2 = |z2|^2. To see that the two middle terms combine to something real, we calculate: z1* x z2 + z2* x z1 = (a1 + ib1)* x (a2 + ib2) + (a2 + ib2)* x (a1 + ib1) where I have expanded z1 and z2 in terms of their real parts (a1 and a2) and their imaginary parts (b1 and b2). Then we can use the definition of complex conjugate to change i to -i and get: (a1 - ib1) x (a2 + ib2) + (a2 - ib2) x (a1 + ib1) Then we can carry out the multiplication to get: a1xa2 + a1 x (ib2) - ib1 x a2 - i^2 b1xb2 + a2xa1 + a2x(ib1) -ib2 x a1 -i^2 b2xb1 Using i^2=-1, we get: a1xa2 + i a1 x b2 - i b1 x a2 + b1xb2 + a2xa1 + i a2xb1 -ib2 x a1+ b2xb1 The second term (ia1xb2) cancels with the 7th term (-ib2xa1) and the third term (-ib1xa2) cancels with the sixth term (ia2xb1), so we end up with: a1xa2+b1xb2+ a2xa1+ b2xb1 = 2 (a1xa2+b1xb2) Note that we have not made any approximation. This term is equal to 2xRe[z1xz2*], which is what I use in the video. To see this, note: z1xz2* = (a1+ib1)x(a2+ib2)* = (a1+ib1)x(a2-ib2) = a1xa2 -i a1xb2 + ib1xa2 -i^2xb1xb2 = a1xa2 +b1xb2 + i (b1xa2 - a1xb2) where in the last step I re-ordered some terms and used i^2=-1. From this expression, we get: 2xRe[z1xz2*] = 2(a1xa2 +b1xb2). Finally, we can combine this latest result with the one above, to see that the two middle terms in equation (1) above are equal to 2xRe[z1xz2*]. Therefore, without neglecting any terms, we have shown that: |z1+z2|^2 = |z1|^2 + |z2|^2 + 2xRe[z1xz2*] This is the relation I used in the video, and it is exact, I have not neglected any imaginary parts. I hope this helps clarify that maths!

We are not familiar with any, but there likely are some on RU-vid; let us know if you find them! :)

Note that the scalar product in a complex vector space is antilinear in the first argument. This means that to calculate the scalar product of (0,i) with itself, you have (using row and column vectors): row(0 -i) column(0 i) = 0 -i*i = 1 Hope this helps!

​@@ProfessorMdoesScience Ok, thank you, then it's clear, I though that it should be >=0 for any pair of vectors

Thanks for the suggestion!

In principle you could organize an infinite range of numbers in matrix form. However, as you probably suspect, in practice that would be impossible to do on paper or in a computer. In reality, when we solve quantum mechanical problems, even if the state space is infinite, we often have to truncate it to finite dimensions to be able to do the calculations. When we do this, the use of the matrix formulation becomes the best approach.

@@ProfessorMdoesScience okay I understand:)

There is a metric in the space of pure states, it is called the Fubini-Study Metric. It arises from the inner product.

@@zray2937 thank you for the information

Matrices are not identical to linear operators. The transition from one to the other is neither smooth nor trivial in general. A simple demonstration of that is e.g. the Gibbs phenomenon in Fourier analysis, which shows that we can not get pointwise convergence for arbitrary functions. This is all well worked out in functional analysis, of course. It's not something the physicist has to be concerned with much. The far more interesting question for physicists is where these linear spaces come from in the first place. It turns out that they derive (almost trivially) from Kolmogorov's axioms for ensembles, but that is rarely mentioned in either QM lectures or standard QM textbooks. At least I am not aware of common textbooks that take the time to derive the formalism from its actual roots.

I am not sure I follow your notation, can you expand on what you mean by each step?

@@ProfessorMdoesScience Hey Professor thanks for answering :D Amm .. yes so what i was wondering is .. generally in Quantum mechanics we used the square of the wave function to give physical meaning. And usually is written as the scalar product of the wave function with itself, these is the integral of the complex conjugate of the function and the original function.So .. by these notation .. the bra is the complex conjugate of its corresponding ket ? And would these give always a real number or a real function ?

@@chemistryscience4320 Thanks for the clarification. Yes, when written in the position representation in terms of wave functions, the bra corresponds to the complex conjugate wave function of the ket, and the scalar product of a ket with itself (its corresponding bra) is always a real non-negative number. We explain how to build the wave function (position) representation from kets and bras in the following playlist: ru-vid.com/group/PL8W2boV7eVfnHHCwSB7Y0jtvyWkN49UaZ I hope this helps!

@@ProfessorMdoesScience Thank you very much :D :D Since i found your videos my rigurosity in Quantum Physics has been improving :) So thank you very much !!

Bras contain the same physical information as kets, so if I understand your question correctly, they are simply a mathematical tool (the elements of the dual space).

@@ProfessorMdoesScience Thanks, it's clear now.

At Cambridge we teach this level between second and third year undergraduate; it could be slightly different at other places...

You are absolutely correc there is a typo in that expression! We do know about this, but good that you realised too :)

Thanks for the reference and suggestion!