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This blog is for comments and reports of errors on my site. I will also post notifications of new articles published on the main site using this blog, so if you want  to be kept up to date, check here regularly, or else  subscribe to it for email notifications. I realize this may look like a roundabout way of organizing things, but I don’t trust WordPress with anything substantial any more.

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35 Responses to Comments for

  1. Chris Kranenberg says:

    Griffiths Intro to QM Solution 3.37 b): the solution can be taken a little further by replacing the vector notations with their matrices, perform the matrix addition to obtain a matrix solution with cosine and isine expressions.


    • growescience says:

      If you mean eqn (15), yes, you could factor out the {e^{-iat/\hbar}} and then substitute from (8) and (9) to get the answer in terms of sin and cos of {bt/\hbar}.


  2. Nick says:

    Griffiths Quantum, problem 4.18.

    I can’t follow from eq. 12 to 13. What justifies adding or subtracting 1 from m on the f state?


  3. Aaron Stevens says:


    I am referring to the article at:

    I am specifically referring to problem 11.4.2 in Shankar’s Quantum Mechanics book (around equation 33 in the above link).

    The final equation 38 you arrive at is valid, but then you say,

    “Since the cosine is periodic, we can’t actually calculate a unique value for
    its average, although if we do the average over an exact number of periods,
    the average is still zero. I have a feeling that I’m missing something obvious
    here, so any suggestions are welcome.”

    Let me see what you think:
    The final expression that results in the expectation value of the cosine function is not the same thing as the average of the cosine function. The only time the expectation value is equal to the average is when the probability distribution in question is the uniform distribution. In general, this expectation value is state dependent. It is a measure of what we would “expect” this cosine function to be given the probability distribution of finding the particle between x and x+dx in space (psi* times psi). We could contrive a state that gives 0 for this expectation value, but I doubt that the state will then evolve according to the Schrodinger’s equation in such a way as to keep the expectation value set at 0.

    To have momentum conservation, we need the expectation value of the commutator [P,H] to be 0 always, since momentum conservation is a property of the physical system in question, not on the states within that system. So since we get a “final answer” that is not identically 0, we must say that momentum is not conserved.

    Another way I thought to solve the problem that is simpler but I am unsure is valid is that since the potential (and therefore Hamiltonian) is not invariant under infinitesimal translations that we cannot say momentum is conserved. Or thinking a little bit differently, just because we have found points in space where the potential energy is the same does not mean momentum is conserved. It is like if you were to push a block down and back up a hill with friction so that it begins and ends at rest. If you are considering just the block as your system and then saying since it starts and stops with the same energy that its energy must have been conserved the entire time, you would not be correct.

    I grade for a quantum class that uses this book, and I think a lot of the students refer to these solutions, so this is how I came across this. Thanks!


  4. Danyel Cavazos says:


    I’d like to ask something in the page

    How do we go from eq. 22 to eq. 23? I.e., how do we know that when we evaluate Tr(A_i s_i B_j s_j) we can take A_i and B_j out of the trace operation?

    Thank you so much!


  5. Kevin says:

    On Shankar’s exercise 7.3.2, you were asked to verify that Hermite’s polynomials follow the recursion relation 5.3.15 (on the book), you proved they follow the recursion in equation 5.3.25


  6. hladacpravdy says:

    Example 2.25

    Equation 21:
    = (m * alpha / hb)^2

    From this follows:
    sigma_p = sqrt( – ^2) = m * alpha / hb

    Unit analysis:
    m … kg
    alpha … J
    hb … Js
    m * alpha / hb ……. kg/s
    SI unit for momentum is kg m/s. Where I have done mistake ?


  7. tomas says:

    I have a simple question about the Electrodynamics Spherical potential example 1.
    If the path integral goes from infinity to r, why doesnt the dl point inwards? It would appear an extra minus which will make it all wrong. But aren’t the Electric Field and the dl pointing in opossite directions?



    • growescience says:

      I think it’s just convention that if you choose your reference point to be at infinity, then {d\mathbf{l}} always points away from the origin. You can always fix the sign of {V} if you’re calculating it from {\int\mathbf{E}\cdot d\mathbf{l}}, since you know that you must have {\mathbf{E}=-\nabla V}. Thus if {\mathbf{E}} points outwards (away from the origin), then {V} has to decrease as you move away from the origin.


  8. Victoria Jenne says:

    Your logic is wrong in problem CHEMICAL POTENTIAL OF AN IDEAL GAS problem # 3.37 Thermodynamics. The book defines U as internal energy so you can’t add potential energy to kenetic energy without previously using a quantum mechanics proof to alter the multiplicity which would entail altering the volume veriable. Because we are obviously not changing the internal energy of the system your proof in not valid, even if you did get the correct answer.

    Best Wishes,



    • growescience says:

      If I understand what you’re saying, I don’t think there’s a problem. The derivation of the Sackur-Tetrode equation given by Schroeder in section 2.5 counts the number of states available to an ideal gas by using the quantum uncertainty principle, and assumes that the gas is contained in a volume in free space (that is, no gravitational field). This leads to Schroeder’s equation 2.40 which is in turn used to derive equation 2.49. The energy {U} in this derivation is entirely kinetic, since it’s obtained from the ‘volume’ in momentum space. If we apply a gravitational field by placing the gas a distance {z} above the Earth’s surface, we don’t change the number of states available to the gas; all we do is shift their total energy by an amount {mgz} per molecule. Thus the derivation of the Sackur-Tetrode equation would still be valid, except that the kinetic energy in the formula would now be {U_{K}=U_{total}-Nmgz}.


  9. Anonymous says:

    I believe your equation 0.7 for Shankar’s problem 1.8.9 should be the ith component of L, not the L vector as a whole.


  10. Arpon says:

    I have some confusion about your solution to problem 14.05.03, Principles of Quantum Mechanics, Shankar. (Link: )
    “Now suppose we return to the case where the first apparatus transmits only spin z of +h/2 but the second (aligned along the x axis) transmits everything (no blocked beam) into a third apparatus, which is aligned again along the z axis, but now transmits only particles with spin z of -h/2. In this
    case, the middle (x axis) apparatus has no effect since it doesn’t filter the particles at all, with the result that we’re feeding a stream of +h/2 particles into an apparatus that detects only spin -h/2. In this case, nothing will get through.”
    My opinion is: When the spin +z electrons pass through the second SG (along x axis) apparatus, the outcome is electrons in the +x and -x spin state. The state is no more +z spin. So when the +x and -x spin electrons pass through the 3rd SG apparatus (along z axis) 50% of them should get through.


    • growescience says:

      I originally thought the same thing, but I think the point is that we aren’t allowed to look at the output of the middle detector, so we don’t ever measure a particle’s x-spin. In that case, all the particles remain in the +z state, so they all get blocked in the third detector. I suspect you’re right that if we did measure the x-spin in the middle detector that would place the particle in either the +x or -x state, with z-spin undetermined, so that 50% of them would indeed get through the third detector.
      Shankar gives zero as the answer at the back of the book, so my guess is that’s what’s happening.


  11. incomprehensiblething says:

    Just to say thank you for all your work on PhysicsPages, past and present. Your site is much appreciated, but don’t overstress yourself getting it back online.


    • growescience says:

      Glad you find it useful. I’m trying to get it restored ASAP, since once it’s done, I’ll be able to get back to actually writing something about physics…


  12. Electro says:

    sir when will your answers for griffiths electrodynamics be ready. It was very very helpful and i also need it now. Your explanations were brilliant. Please we all need that fast sir.


    • growescience says:

      It will take some time for me to restore everything. In the meantime, there is an archive of the old site available which contains most of the content up to March 2017. See the home page for the link, as it doesn’t seem possible to insert links in comments here.


  13. Will says:

    Thank you!!!


  14. Hello. There is a web page called the Wayback Machine ( They take snapshots of the internet at various times and store them for access. There is a nearly 100% functional version of the old Word Press site from June of this year available. I have been using it to view content. You can find it here:

    It might be a good option to share with people until you get the new website up and running? Just an idea.

    I’m an undergrad Physics student. Three semesters to go (including this one). Your website is extremely valuable, and I thank you very much for the work you have done.



  15. Danyel Cavazos says:

    Although I don’t know extremely much about web design, I have done a little before. This page is and has always been extremely useful, please let me know if I can help you in any way!


    • growescience says:

      Thanks for the offer. I think the basic design is OK – I want to keep it simple so it’s easy to maintain. The only thing that would be useful is some way of reliably displaying PDF files in the various browsers, although I suspect this might not be possible. Chrome has removed support for the Adobe plugin. On Firefox, the equals signs in the equations don’t show up. And so on and on…
      I think the best solution is for users to just download the PDFs and view them offline.


      • Danyel Cavazos says:

        What were you using to display equations before? I’m sure there are ways of presenting LaTex eqns in HTML code via extensions


      • growescience says:

        I was using the Latex “Beautiful Math” plugin from Jetpack in WordPress. I think there are Latex extensions (Mathjax?) for displaying math in HTML, but as I use the Lyx editor for writing the posts in Latex and it has a Latex–>PDF converter, it’s easiest to just create PDFs and upload them.
        Thanks anyway.


  16. Antonio Victor Nascimento says:

    i knew the website the last wek, it is very great and important. I am physics student of UFES in brazil.
    i hope that the website normalize soon.
    sorry my english..
    hugs of brazil.


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