Expectation values in quantum mechanics

[@zetaolin7253]

The series of videos help me a lot. Thank you Professor M !!

[@ProfessorMdoesScience]

Glad to hear! :)

[@armalify]

What a great explanation ! I'm speechless.

[@ProfessorMdoesScience]

Thanks! :)

[@amaljeevk3950]

This channel is a gem❤

[@ProfessorMdoesScience]

Thanks for your support! :)

[@BlochSphere]

We are so lucky to have content creators like you!!

[@ProfessorMdoesScience]

Glad you like the videos! :)

[@firstnamelastname307]

consuming prof M content feels like eigen state eigen value :-) it meets high expectation with nearly no quality deviation ...

[@ProfessorMdoesScience]

Nice one :)

[@nasszelle534]

15.29: problem to unterstand: sigma is the sum of an operator (something like a matrix) and an expectation value (scalar). These Data types seems incompatible to me.

[@ProfessorMdoesScience]

You are absolutely correct. The proper way of writing this would be: sigma_A = A - 1, where 1 is the identity operator. However, in physics we typically use shorthands to simplify notation, and rewrite 1 = . With this convention, we omit the identity operator 1, but we implicitly understand it is there. I hope this helps!

[@anubratasaha4367]

Thanks for asking it :) I had the same question!

[@TheWingEmpire]

Really helpful and clarifying!

[@ProfessorMdoesScience]

Thanks for watching!

[@LifeIzBeautiful10]

Great presentation!

[@ProfessorMdoesScience]

Glad you like it!

[@DavidRoux-uv3bl]

The best out there. I so enjoy your content.

[@LifeIzBeautiful10]

Cool! Thank you Professor M!

[@ProfessorMdoesScience]

Thanks for watching!

[@elenaclaramaria8577]

A small consideration and a small question, which I hope to be of value to other students, especially italian students. I think Professor M defines MSD (mean square deviation) and RMSD (root mean square deviation) for a generic estimator. When the estimator is the mean, MSD is what we call "variance", and RMSD is what we call "standard deviation". All my italian textbooks of QM didn't talk about general estimators and I was left with the latter type of definitions which are also labelled differently, variance is usually just sigma(a)^2 without the "" parenthesis. A small question about . Is it always zero? Or just for symmetric distributions like the red and blue ones that were drawn in this particular example? Thanks so much, as always, for your time and this beautiful deep dive into QM.

[@ProfessorMdoesScience]

Thanks for the clarification! As to your question, is always zero even if the distribution is not symmetric. To see this, note that: sigma_A = A - . Even in an asymmetric distribution, is measuring the "average" position of the distribution, even though in that case will not be in the "middle" of the distribution. To see this mathematically: = < A - > = - But is just a scalar, and the expectation value of a scalar is the same scalar, so = , and this means =0 always. I hope this helps!

[@workerpowernow]

very clear derivations. Thanks :)

[@ProfessorMdoesScience]

Thanks for the comment! :)

[@whiteorchid98]

Hey, thanks for a great series! I'm a little bit confused about 9:09 though. I get that we can collapse the double sum into a single sum because of the Kronecker delta, but I don't get why the remaining terms in the sum equal to ||^2... What am I missing?

[@ProfessorMdoesScience]

Glad you like the videos! After exploiting the Kronecker delta to get a single sum, we end up with: Sum_(n) lambda_n The definition of the absolute value squared for a complex number z is |z|^2 = z* z, where z* is the complex conjugate. For brackets, we have the general property that =*, so we can combine the two brackets together into their absolute value squared. I hope this helps!

[@whiteorchid98]

@@ProfessorMdoesScience Oh wow, that was so simple... I really overthought things. Thank you so much!

[@armalify]

* The information captured by quantum state can be represented by probability distribution. * Expectation values and measurements outcomes are completely two different things in quantum mechanics. * In the special case of the state of the system being an eigenstate of the operator, then the expectation value is simply the associated eigenvalue. So, the expectation value does coincide with the outcome of a measurement. * If we only analyze probability distributions using expectation values there would be no way of telling these two distributions apart. * The root mean square deviation measures the width of the distribution.

[@fernandojimenezmotte2024]

Thank You professor M does Science for your wonderful series course in Quantum Mechanics. I am a scientist , that comes from Electrical Enginering course work/career and my final metamorphosis dream is to fully mutate to a mix of Teoretical Physics + Computational Mathematics scientist

[@ProfessorMdoesScience]

Glad you like our series, and fully support your aspiration! :)

[@maryamnaeem2191]

Love your videos so much

[@ProfessorMdoesScience]

Glad you like them! :)

[@oraange]

Why do we not divide the expectation value by sqrt(phi* phi) instead of phi * phi

[@ProfessorMdoesScience]

Good question! Consider an un-normalized state |phi>. Then the expectation value of operator A with respect to this state is: . Now consider the corresponding normalized state, which for clarity I will call "chi", and which we can build as: |chi>=|phi>/sqrt() The corresponding normalized bra is:

[@oraange]

@@ProfessorMdoesScience Oh! I get it now. Thanks you a lot !

[@schmetterling4477]

One should probably give a warning to the novice: while it is tempting to assume that we are measuring the state of systems directly, in reality (i.e. experimentally) we are measuring the energy differences between states. This is easiest to see in atomic physics, where we aren't observing an excited state of an atom or the ground state, but we are observing an optical line that has a photon energy given by the energy difference between the excited state and the ground state (or lower state). In the Stern-Gerlach experiment we are measuring the separation between the two spin states, which is proportional to the energy difference of the two states in the magnetic field. So while the formalism isn't wrong, it does not deliver directly what actual experiments return. A good class on QM/atomic physics should explicitly mention this, but I find that (while trivial) it's often not discussed properly.

[@ProfessorMdoesScience]

Good insight! Our approach is that we should first understand the mathematical formalism (as our videos are currently doing), but we do hope to then apply it to specific problems in, for example, condensed matter physics or quantum optics, where the precise relation with experiment will become clear.

[@schmetterling4477]

@@ProfessorMdoesScience I understand the approach, but it is problematic from a teaching perspective. Students can't make the connection between theory and actual physics on their own unless they are being shown the actual experimental facts. This may matter less in a university setting where atomic physics and QM 101 lectures are typically being read in parallel, but here on the internet it's unlikely that a student will "get it". While the formalism talks about state, nature gives us energy. State is an abstract, energy is a concrete concept. Energy is actually the main concept of physics and that does not change in quantum mechanics, either. I think students should have that understanding reinforced rather than weakened.

[@KaranveerSingh-xn4tv]

When the state of the system is one of the eigenvectors of operator, the only possible outcome (for the whole system) is the value associated with that eigenvector.. Right?

[@ProfessorMdoesScience]

This is correct!

[@geraldpellegrini2782]

What happens if you 'isolate' a quantum harmonic oscillator so it is forced (or measured) to be in an energy eigenstate. According to Ehrenfest theorem the expectation values of position and momentum are constants (i.e., no motion) and classical correspondence is broken!

[@nomanahmadkhan7791]

How can we determine the probability distribution of eigenvalues spectrum? Like, would it be normal distribution or something else etc?

[@ProfessorMdoesScience]

It will depend on which state you are in. For example, if you are in an eigenstate of the corresponding observable, then a single eigenvalue will feature, the one corresponding to the eigenstate you have. I hope this helps!

[@nomanahmadkhan7791]

@@ProfessorMdoesScience If I understand correctly then what you are saying is that the expectation value depends upon the state of the system?

[@ProfessorMdoesScience]

@@nomanahmadkhan7791 Yes, you always calculate the expectation value with respect to some state, and the actual expectation value will depend on which state you are in. I hope this helps!

[@nomanahmadkhan7791]

@@ProfessorMdoesScience Keeping in view the definition of expectation value (=sum lambda(n) ||2) and the fact that eigenvalue of a Hermitian operator is always real, can expectation value be pure imaginary or mixed complex number?

[@ProfessorMdoesScience]

@@nomanahmadkhan7791 No, both terms in the expression (eigenvalue and absolute value squared) are real, so the overall expression must also be real.