Comment on “Could ordinary quantum mechanics be just fine for all practical purposes?” by Alexey Nikulov

Abstract: Contrary to the claim by Alexey Nikulov it is argued that there is no error in Landau and Lifshitz derivation of magnetic coupling formula (113.2)

In [1], Sec. 4.3, “Quantum mechanics cannot describe both opposite cases”, Alexey Nikulov claims that Landau and Lifshitz have made an error deriving, in their quantum theory textbook [2], the formula for the coupling of the angular momentum to the magnetic field in their chapter on “An atom in a magnetic field”. More specifically Nikulov claims in [1] that the formula (113.2) in [2] is obtained by Landau and Lifshitz from (113.1) by an “illegal substitution” or, by an “elementary arithmetic error” [3].

While it is true that Landau and Lifshitz could have been more explicit in their derivation, I will derive their (113.2) from (113.1) in all detail – showing that their is no “arithmetic error”, and that it is, in fact, Alexey Nikulov who made an error misunderstanding the arguments in Ref.[2] and “overshooting” with his criticism.

As it is very important to pay attention to details, let us start with an exact image (see Fig. 1) of the starting formula (113.1) in Ref. [2]:

Fig. 1. Landau-Lifshitz p.461, Eq. (113.1)

A. Nikulov is questioning the derivation, given in Ref. [2], of the following formula copied here from Ref. [2] in Fig. 2 below.


Fig. 2, Landau-Lifshitz p.461, Eq. (113.2)

The Bohr magneto \mu_B is defined here as \mu_B=|e|\hbar/2mc. We will neglect spin \hat{\mathbf{S}} and the scalar potential U, and write down the the starting formula just for one electron, setting c=1:

Remark: These simplifying assumptions can be made without losing the essence of the argument.

Thus our starting formula, which we denote as ((113.1)) reads as follows

((113.1))   \begin{equation*} \hat{H}=\frac{1}{2m}\left(\hat{\mathbf{p}}+|e|\mathbf{A}\right)^2,\end{equation*}

where

(1)   \begin{equation*} \hat{\mathbf{p}}=-i\hbar \nabla\end{equation*}

is the canonical momentum vector operator and \mathbf{A} is the vector potential for the magnetic field \mathbf{H}:

(2)   \begin{equation*} \mathbf{H}=\nabla\times\mathbf{A}.\end{equation*}

The electron charge (negative) is e=-|e|.

In order to derive the formula (113.2) of Ref. [2] we need to calculate the Hamiltonian \hat{H} explicitly. To this end we take the square:

(3)   \begin{eqnarray*} \hat{H}&=&\frac{1}{2m}\left(\hat{\mathbf{p}}+|e|\mathbf{A}\right)^2\nonumber\\ &=&\frac{1}{2m}\hat{\mathbf{p}}^2+\frac{ |e|}{2m}\hat{\mathbf{p}}\cdot \mathbf{A}+\frac{|e|}{2m}\mathbf{A}\cdot \hat{\mathbf{p}}+\frac{|e|^2}{2m}\mathbf{A}^2.\end{eqnarray*}

Following Ref. [2] we introduce \hat{H}_0 defined as

(4)   \begin{equation*} \hat{H}_0=\frac{1}{2m}\hat{\mathbf{p}}^2,\end{equation*}

where \hat{\mathbf{p}} is the canonical momentum defined in Eq. (1).
Then, using Eq. (4) we can rewrite (3) as

(5)   \begin{equation*} \hat{H}=\hat{H}_0+\frac{ |e|}{2m}(\hat{\mathbf{p}}\cdot \mathbf{A}+\mathbf{A}\cdot \hat{\mathbf{p}})+\frac{|e|^2}{2m}\mathbf{A}^2.\end{equation*}

The vector potential \mathbf{A} is a function of coordinates, therefore in general it will not commute with the components of the canonical momentum operator. We have

(6)   \begin{equation*} \hat{\mathbf{p}}\cdot \mathbf{A}=\mathbf{A}\cdot\hat{\mathbf{p}}-i\hbar \nabla\cdot\mathbf{A}.\end{equation*}

We consider the case of a uniform magnetic field, i.e. the vector \mathbf{H} is constant. Then \mathbf{A} can be chosen as

((111.7))   \begin{equation*} \mathbf{A}=\frac{1}{2}\mathbf{H}\times\mathbf{r}.\end{equation*}

In this case, with \mathbf{A} given by ((111.7)) we have \nabla\cdot \mathbf{A}=0,
therefore \mathbf{p}\cdot\mathbf{A}=\mathbf{A}\cdot \mathbf{p}. Thus \hat{H}, given by Eq. (3), takes the form:

(7)   \begin{equation*} \hat{H}=\hat{H}_0+\frac{ |e|}{m}\mathbf{A}\cdot \hat{\mathbf{p}}+\frac{|e|^2}{2m}\mathbf{A}^2,\end{equation*}

or, making use of Eq. ((111.7)) again:

(8)   \begin{equation*} \hat{H}=\hat{H}_0+\frac{ |e|}{2m}(\mathbf{H}\times\mathbf{r})\cdot \hat{\mathbf{p}}+\frac{e^2}{8m}\,(\mathbf{H}\times\mathbf{r})^2.\end{equation*}

Using the standard vector product identity

(9)   \begin{equation*} (\mathbf{H}\times\mathbf{r})\cdot \hat{\mathbf{p}}=\mathbf{H}\cdot (\mathbf{r}\times\hat{\mathbf{p}})\end{equation*}

we can rewrite the second term to obtain

(10)   \begin{equation*} \hat{H}=\hat{H}_0+\frac{ |e|}{2m}\mathbf{H} \cdot (\mathbf{r} \times \hat{\mathbf{p}})+\frac{e^2}{8m}\,(\mathbf{H}\times\mathbf{r})^2.\end{equation*}

We now introduce canonical angular momentum operator \hat{\mathbf{L}} defined as

(11)   \begin{equation*} \hat{\mathbf{L}}=\mathbf{r}\times\hat{\mathbf{p}}.\end{equation*}

Then

(12)   \begin{equation*} \hat{H}=\hat{H}_0+\frac{ |e|}{2m}\mathbf{H} \cdot \hat{\mathbf{L}}+\frac{e^2}{8m}\,(\mathbf{H}\times\mathbf{r})^2.\end{equation*}

The above is Eq. (113.2) of Ref. \cite{LandauL} that we have derived step by step, and without any “arithmetic errors”. Therefore the part of Ref. \cite{nikulov2016} questioning the validity of this derivation is erroneous.

It is nevertheless interesting to try to understand why such an wrong conclusion could arise. The following is my guess based on the following sentence in Sec. 4.3 of Ref. [1]:
\begin{quotation}“The additional summand \mu_B\hat{L}B could appear in the relation (113.2) due to illegal substitution of P^2/ 2m by m\hat{v}^2/2 in \hat{H}_0.”\end{quotation}

While in quantum mechanics we do have the canonical momentum operator (see Eq. (1), there is no canonical velocity operator. The explicit expression for the velocity operator must be calculated for each Hamiltonian separately. Given a Hamiltonian \hat{H} (assuming not time dependent) the corresponding velocity operator is defined by

(13)   \begin{equation*} \hat{\mathbf{v}}=\frac{i}{\hbar}[\hat{H},\mathbf{r}].\end{equation*}

Its expectation value \langle \bv\rangle|_{\psi(t)} is equal to the time derivative of the expectation value of the position operator.

For a free particle, when \hat{H}=\hat{H}_0=\frac{\hat{p}^2}{2m}, we find that
the expression for the velocity operators is given by

(14)   \begin{equation*} \hat{\mathbf{v}}=\frac{i}{\hbar}[\hat{H_0},\mathbf{r}]=\frac{\mathbf{p}}{m}.\end{equation*}

For a particle in a magnetic field \mathbf{H}=\nabla \times \mathbf{A} the Hamiltonian \hat{H} is defined by Eq. ((113.1)), that is:

(15)   \begin{equation*} \hat{H}=\frac{\hat{\pi}^2}{2m}, \end{equation*}

where the kinetic momentum \hat{\boldsymbol{\pi}} is defined as

(16)   \begin{equation*} \hat{\boldsymbol{\pi}}=\hat{\mathbf{p}}-e\mathbf{A}.\end{equation*}

In this case the velocity operator \hat{\mathbf{v}} is given by a different expression than that in the free case:

(17)   \begin{equation*} \hat{\mathbf{v}}=\frac{i}{\hbar}[\hat{H},\mathbf{r}]=\frac{\boldsymbol{\pi}}{m}.\end{equation*}

In both cases the Hamiltonian can be written as the kinetic energy m\hat{v}^2/2, but the expression in terms of the canonical momenta and positions is different in each case. Nikulov seems to claim that it is impossible to derive the second term in Eq. ((113.2)), the one containing the coupling of the magnetic field to the angular momentum, from “just the kinetic energy” because he fails to notice that the kinetic energy’ has a different expression for a particle in a magnetic field than the one without. It is for this reason that the whole section of Ref. [1] needs to be completely rewritten.
References
[1] Alexey Nikulov, Could ordinary quantum mechanics be just fine for all practical purposes?, Quantum Stud.: Math. Found. (2016) 3: 41., doi: 10.1007/s40509-015-0057-3 (see also arXiv:1508.03505)
[2] Landau, L. D., Lifshitz, E. M.: Quantum Mechanics: Non-Relativistic Theory. Volume 3, Third Edition, Elsevier Science, Oxford, 1977.
[3] Alexey Nikulov, private communication

Addendum 1:
On March 9, 2018 I receive en email from A. Nikulov containing the following statements:

В разложении бинома на слагаемые (p – qA)^2/2m = p^2/2m – pqA/m + (qA)^2/2m в книге LL принимается, что первое слагаемое описывает атом без магнитном поле, а два других в магнитном поле. Это не только нельзя ничем обосновать, но и приводит к очевидной арифметической ошибки.

Translating into English:

In the decomposition of the binom into separate terms (p – qA)^2/2m = p^2/2m – pqA/m + (qA)^2/2m in the LL book it is assumed that the first term describes the atom without the magnetic field, while two other in a magnetic field. Not only there is no way to justify it, but also it leads to an evident arithmetic error.

Again A. Nikulov is making an error. The first term is nothing else but \hat{H}_0, which is the same as \hat{H} with \mathbf{A} set to zero, that is the free Hamiltonian, corresponding to the atom without the magnetic field. It is not true that Landau and Lifshitz state that two other term describe the atom in a magnetic field. They do not say so. What describes the atom in magnetic field is all three terms. All that is evident and follows from the definitions. There is nothing that need to be justified and there are no arithmetic errors.

Addendum 2:

As a reply to the Addendum 1 above, I received the following:


я не понимаю, где я опять сделал ошибку. Я согласен с тем, что The first term is nothing else but Н_0, which is the same as Н with А set to zero, that is the free Hamiltonian, corresponding to the atom without the magnetic field. Но я не могу согласится с тем, что It is not true that Landau and Lifshitz state that two other term describe the atom in a magnetic field. Даже если Landau and Lifshitz об этом ничего не написали, то что two other term describe the atom in a magnetic field очевидно из того, А в них не равно нулю. Приравняйте в выражении (8) или (10) Вашего Comment магнитное поле нулю во всех слагаемых (а не в одном!) и Вы получите выражение Н = Н_0 без арифметической ошибки (в котором кинетическая энергия равняется кинетической энергии), но и без энергии магнитного момента в магнитном поле. Мне странно объяснять, что нельзя в одном слагаемом суммы принять А = 0, а в остальных оставить ненулевое значение А. Но мне пришлось это сделать в очередном комментарии к своей статье “Could ordinary quantum mechanics be just fine for all practical purposes?” на ResearchGate. Неужели Вам непонятно, если сделать такую же глупость, какая сделана в книге Landau and Lifshitz, то мы получим арифметическую ошибку?

Below is my English translation of the relevant part, separated into two parts, intertwined with my explanations.

I do not understand when have I made again an error. I agree that “The first term is nothing else but Н_0, which is the same as Н with А set to zero, that is the free Hamiltonian, corresponding to the atom without the magnetic field.” But I cannot agree that it is not true that “Landau and Lifshitz state that two other term describe the atom in a magnetic field. Even if Landau and Lifshitz did not write it, it is clear that the two other terms describe the atom in a magnetic field, because A there is not equal zero.

My comment: Because we are discussing questions of “arithmetic errors” and “illegal substitutions” in a popular textbook, it is very important to be very precise and logically impeccable. The way it stated in the quotation above it is not logically precise and it is potentially misleading. The two terms do not constitute the full description of the atom in a magnetic field. The full description is given by all three terms together. The first term is the kinetic energy of the free particle. The second term is the linear coupling of the angular momentum to the magnetic field, the third term, quadratic in the magnetic field, is usually neglected completely for weak magnetic fields.

Perhaps it should also be stated why sometimes the talk is about the atom, while only electron figures out in the Hamiltonian. The reason is that the terms are inversely proportional to the mass, therefore the proton energy terms are small compared to the electron energy terms and can be neglected.

If you set magnetic field to zero in the expression (8) or (10) of your comment in all terms, (and not only in one!) then you will get the expression H=H_0 without arithmetic error (in which kinetic energy is equal to the kinetic energy), but without the energy of the magnetic moment in magnetic field. t is strange that I have to explain that you should not to set A=0 in one term, and to let it to be non-zero elsewhere.

The sentence above does not make sense. In fact H_0 has been obtained from (8) or (10) by setting the magnetic field to zero in all terms.

Spinning Top – Space Odity

I start with quoting from Wikipedia:

Peggy Annette Whitson (born February 9, 1960) is an American biochemistry researcher, NASA astronaut, and former NASA Chief Astronaut. Her first space mission was in 2002, with an extended stay aboard the International Space Station as a member of Expedition 5. Her second mission launched October 10, 2007, as the first woman commander of the ISS with Expedition 16. She is currently in space on her third long-duration space flight and is the current commander of the International Space Station.

Two weeks ago we could see live how “Kimbrough and Flight Engineer Peggy Whitson of NASA reconnect cables and electrical connections on PMA-3 at its new home on top Harmony”

That is surely fascinating, but what we are particularly interested in the spinning top in zero gravity experiments on the board of ISS in 2013

At 0:42 in this video Peggy tells us that:

“Conservation of angular momentum keeps the top axis pointed in the same direction”

We look at it, and we compare with the other video featuring the Dzhanibekov effect

The same effect I have modeled with Mathematica:

is even better visible here:

There is also conservation of angular momentum there, but the axis of rotation evidently is not being kept all the time in the same direction. Once in while, quasi-periodically, it flips.

What’s going on?

The answer is: it is, as Chris Hadfield sings, a Space Odity

Can you hear me, Major Tom?
Can you “Here am I floating ’round my tin can
Far above the moon
Planet Earth is blue
And there’s nothing I can do

Indeed, there are laws of physics and sometimes they are odd. Major Tom can do nothing about it except of just watching:

Major Tom: And there’s nothing I can do

But no, not exactly. We can do something about it. We can try to figure it out, why things happen the way they happen. Mathematics will help us when physical intuition does not suffice.

And that is our plan for the future. As P.A.M. Dirac wrote it on the blackboard during his lecture in Moscow in 1956:

Physical law should have mathematical beauty

Physical law should have mathematical beauty and we are watching this beauty while playing with geodesics of left-invariant metrics on Lie groups.

We will be looking into the spinning top – but relativistic one. These relativistic tops fly somewhere in space, but they are not yet mass-produced in factories in China. But soon….

It is all about “attitude”. Mathematically the attitude matrix satisfies nonlinear differential equations, and they have their odities. And, as Chris Hadfield explains it in his book “An Astronaut’s Guide to Live on Earth”:

In space flight, “attitude” refers to orientation: which direction your vehicle is pointing relative to the Sun, Earth and other spacecraft. If you lose control of your attitude, two things happen: the vehicle starts to tumble and spin, disorienting everyone on board, and it also strays from its course, which, if you’re short on time or fuel, could mean the difference between life and death. In the Soyuz, for example, we use every cue from every available source—periscope, multiple sensors, the horizon—to monitor our attitude constantly and adjust if necessary. We never want to lose attitude, since maintaining attitude is fundamental to success.
In my experience, something similar is true on Earth. Ultimately, I don’t determine whether I arrive at the desired professional destination. Too many variables are out of my control. There’s really just one thing I can control: my attitude during the journey, which is what keeps me feeling steady and stable, and what keeps me headed in the right direction. So I consciously monitor and correct, if necessary, because losing attitude would be far worse than not achieving my goal.

Geodesics of left invariant metrics on matrix Lie groups – Part 2 Conservation laws

In the last post, Geodesics of left invariant metrics on matrix Lie groups – Part 1,we have derived Arnold’s equation – that is a half of the problem of finding geodesics on a Lie group endowed with left-invariant metric.

Suppose G is a Lie group, and g(\xi,\eta) is a scalar product (i.e. a nondegenerate bilinear form) on its Lie algebra Lie(G). Then, using left translations g defines a left invariant (Riemannian or pseudo-Riemannian) metric on the whole group G. If t\mapsto a(t)\in G is a path in G, we use left translations to define the image \omega(t) of the tangent vector \dot{a}(t) in Lie(G)

(1)   \begin{equation*}\omega(t)=a(t)^{-1}\dot{a}(t).\end{equation*}

Usually it is written more carefully, using L_{a^{-1}} instead of a^{-1}, but I am using a simplified notation, well adapted to dealing with problems for matrix groups.

If a(t) is a geodesic for the metric g, then \omega(t) satisfies the Arnold’s equation

(2)   \begin{equation*}\dot{\omega}=B(\omega,\omega),\end{equation*}

where

    \[B:Lie(G)\times Lie(G)\rightarrow Lie(G)\]

is defined as

(3)   \begin{equation*}B^k_{ij}=g^{kl}B_{ij,l}=g^{kl}C_{jl,i}=g^{kl}g_{im}C_{jl}^m,\end{equation*}

and C_{jl}^m are the structure constants of G, that is

(4)   \begin{equation*}[\xi_i,\xi_j]=C_{ij}^k\, \xi_k\end{equation*}

where \xi_i form a basis for Lie(G) and

(5)   \begin{equation*}B_{ij,l}=g_{lk}B_{ij}^k,\,C_{ij,l}=g_{lk}C_{ij}^k.\end{equation*}

Equation (2) is a system of nonlinear ordinary differential equations with constant coefficients. In this form it can be found in Arnold’s 1966 paper “Sur la g\’eom\’etrie differentielle des groupes de Lie de dimension infinie et ses applications \`a l’hydrodynamique des fluides parfaits”, Ann. Inst.
Fourier (Grenoble) 16 (1966), 319-361.

In his blog post “The Euler-Arnold equation” Terrence Tao mentions that some people call this equation the “Euler-Arnold” equation, while some other prefer to skip Arnold’s name completely and call it “Euler-Poincare equation”. Go-figure!

Solving equations (2) is just one half of the whole problem of finding a geodesic. Once \omega^{i}(t) are known, we need to solve the linear differential equation with variable coefficients that results from the definition of \omega (??):

(6)   \begin{equation*} \dot{a}(t)=a(t)\omega(t),\end{equation*}

For the case of the rotation group and free rigid body we did it in Taming the T-handle continued.

B. Kolev in his paper “Lie groups and mechanics. An introduction” shows that for the rotation group O(n) the equations of motion for the free rigid body are completely integrable. We will not need this result, but is useful to know that in general case there are always two quadratic constants of motion corresponding to “kinetic energy” and “square of the angular momentum”. We were discussing these constants of motion for the group O(3) in
“Asymmetric Spinning Top – The Hardest Concept To Grasp In Physics” – they are used in so-called Poinsot construction. We will discuss a version of it for the group O(2,1) in the future.

The first observation is that the “kinetic energy” T(t)=g_{ij}\omega(t)^i\omega(t)^j is, in fact, a constant, independent of t. To see that this is the case, we differentiate:

(7)   \begin{equation*}\frac{dT(t)}{dt}=g_{kl}\dot{\omega}^{k}\omega^l+g_{kl}\omega^{k}\dot{\omega^l}=2g_{kl}\omega^k\dot{\omega}^l.\end{equation*}

In the previous post we wrote Eq. (2) as

(8)   \begin{equation*}g_{kl}\,\dot{\omega}^l=C_{jk,i}\,\omega^{i}\omega^{j}.\end{equation*}

Substituting into Eq. (7) we get

(9)   \begin{equation*}\frac{dT(t)}{dt}=2C_{jk,i}\omega^{i}\omega^j \omega^{k}=0.\end{equation*}

The result is zero because C_{jk,i} is antisymmetric in j,k while \omega^j\omega^k is symmetric. That is an often used property: if A^{ij}=-A^{ji} and B_{ij}=B_{ji}, then the contraction A^{ij}B_{ij}=0. Indeed A^{ij}B_{ij}=-A^{ji}B_{ij}=-A^{ji}B_{ji}=-A^{ij}B_{ij}, where in the last equality we have exchanged dummy indices names i\leftrightarrow j. If a number is equal to its negative, it must be zero.

To get the formula for the second quadratic invariant we need to return to the Ad-invariant scalar product that we have denoted \mathring{g} in Killing vectors, geodesics, and “Noether’s theorem”:

(10)   \begin{equation*}\mathring{g}(\eta_1,\eta_2)=\mbox{const}\, \frac{1}{2}Re(\mbox{Tr}(\eta_1\eta_2))).\end{equation*}

The fact that \mathring{g} is Ad-invariant implies an important relation between the matrix \mathring{g}_{ij} and the structure constants C_{ij}^k
that we are going to use. Ad-invariance means that:

(11)   \begin{equation*}\mathring{g}(e^{t\eta}\xi_1 e^{-t\eta},e^{t\eta}\xi_2 e^{-t\eta})=\mathring{g}(\xi_1,\xi_2)\end{equation*}

for all t\in\mathbf{R} and \xi_1,\xi_2,\eta\in Lie(G).
Differentiating at t=0 we get

(12)   \begin{equation*}\mathring{g}([\eta,\xi_1],\xi_2)+\mathring{g}(\xi_1,[\eta,\xi_2])=0.\end{equation*}

Setting \xi_1=\xi_i,\xi_2=\xi_j,\eta=\xi_k we get

(13)   \begin{equation*}C_{ki}^l\mathring{g}_{lj}+C_{kj}^l\mathring{g}_{li}=0.\end{equation*}

Multiplying both sides by g^{im}g^{jn} we obtain

(14)   \begin{equation*} C_{ki}^n\,g^{im}+C_{kj}^m\,g^{jn}=0\end{equation*}

We can now derive the second conservation law. The angular momentum m_i is defined as

(15)   \begin{equation*}m_i=g_{ij}\omega^{j}.\end{equation*}

Notice that m is a covector, a one-form on Lie(G), it is in the dual Lie(G)^* of Lie(G). It is the metric that connects the space to its dual. While vectors in Lie(G) play an active role, they generate transformations, elements in the dual, one-forms from Lie(G)^*, are “passive”, they evaluate vectors to numbers. It is the metric that is the third element here, that allows the active principle to connect to the passive principle. The metric depends on the mass distribution. In application to rigid bodies the inertia tensor is encoded in the metric on the rotation group.

The second conservation law states that the square of the angular momentum evaluated with the Ad-invariant metric \mathring{g} is constant:

(16)   \begin{equation*} m_0^2(t)=\mathring{g}^{ij}m_i(t)m_j(t)=\mbox{const}.\end{equation*}

To verify we differentiate and use Eq. (8) rewritten as

(17)   \begin{equation*}\dot{m}_k=C_{jk}^{i}\,m_{i}\omega^{j}.\end{equation*}

(18)   \begin{equation*}\frac{dm_0^2(t)}{dt}=2\mathring{g}^{kl}\dot{m}_k m_l=2\mathring{g}^{kl}C_{jk}^{i}m_im_l\omega^j.\end{equation*}

Now, according to Eq. (14) \mathring{g}^{kl}C_{jk}^{i} is antisymmetric in (i,l), while the product m_im_l is symmetric, therefore we get zero:

(19)   \begin{equation*}\frac{dm_0^2(t)}{dt}=0.\end{equation*}

In the following posts we will first return to the case of the rotation group in three dimensions and the rigid body, and then try to apply a similar reasoning to the case of the Lorentz group O(2,1) in 2+1 dimensions.