In order to get to the parts I am stuck at, I will add the examiners' solutions to each subquestion, which is needed to get to the subquestion that I am querying.

The following is a bizarre question from a highly ranked university quantum mechanics exam:

Consider the quantity

$$\hat{\mathcal{O}}_{mn}(t)=\int {u_m}^{*}(x,t)\hat{\mathcal{O}}\,u_n(x,t)dx$$ for some operator $\hat{\mathcal{O}}$ which has no explicit time-dependence, where $u_n(x,t)$ and $u_m(x,t)$ are eigenstates of $\hat{H}$ at time $t$.

Write this expression in terms of the eigenstates $u_m(x)$ and $u_n(x)$ at time $t=0$


In the second line I believe that there is a mistake and $\int {u_m}^{*}(x,t)\hat{\mathcal{O}}\,u_n(x,t)dx$ should be $\int {u_m}^{*}(x,0)\hat{\mathcal{O}}\,u_n(x,0)dx$ since all the time dependence is in the exponentials out front.

Could someone please confirm or deny whether this is indeed correct?

What is the time-derivative? $$\frac{d}{dt}\hat{\mathcal{O}}_{mn}(t)?$$


Consider the operator $\hat{\mathcal{O}} = [\hat{A},\hat{H}]$ and find $\hat{\mathcal{O}}_{mn}(t)$ for this case. Find a relationship between $\frac{d \hat{A}_{mn}}{dt}$ and $\hat{\mathcal{O}}_{mn}(t)$.


In the first integral there may be a typo as I think that $$\int {u_m}^{*}(x)[\hat{A},\hat{H}]u_n(x,t)dx$$ should be $$\int {u_m}^{*}(x)[\hat{A},\hat{H}]u_n(x)dx$$

I think that $$\hat{\mathcal{O}}_{mn}(0)=(E_n-E_m)\langle{\hat{A}\rangle}$$ when the examiner writes "Comparing to the above" in the last line, I presume the part that is being compared is $$\frac{d}{dt}\hat{\mathcal{O}}_{mn}(t)=-\frac{i}{\hbar}\left(E_n-E_m \right)\hat{\mathcal{O}}_{mn}(t)$$

I can't figure out how to derive the relationship $$\frac{d}{dt}\hat{A}_{mn}(t)=-\frac{i}{\hbar}[\hat{A},\hat{H}]_{mn}\tag{1}$$ as I don't even think it is correct since the question asked to find $\color{red}{\hat{\mathcal{O}}_{mn}(t)}\,\color{red}{\text{for this case.}}$

It also asked for $\color{red}{\text{a relationship between}}\,$ $\color{red}{\frac{d \hat{A}_{mn}}{dt}}$ $\color{red}{\text{and}}\,$ $\color{red}{\hat{\mathcal{O}}_{mn}(t)}$ not $[\hat{A},\hat{H}]$.

I would really like to understand how to derive relation $(1)$, so if anyone could give me any hints or advice on how to go about this it will be much appreciated.

  • $\begingroup$ I don't think this is a "very strange" or "bizarre" question. It looks pretty standard to me. Yes there are a few typos in the solution, but have pity on the poor solution writer who has to deal with so many subscripts and arguments. $\endgroup$ – hft Jul 31 '18 at 0:42
  • $\begingroup$ @hft Hi, I used the words 'strange' and 'bizarre' as before this exam I've never seen operators written with a variable dependence like $\hat{\mathcal{O}}_{mn}(t)$ shown explicitly. I'm used to seeing functions written with their explicit dependence on a variable such as the eigenstates $u_n(x,t)$ but never an operator. I have no pity for the examiner that wrote the solution as their solution needs to be verified before just handing it out for students to see. Also, it makes it a nightmare trying to learn from a solution loaded with typos (especially if you fail to identify them). $\endgroup$ – BLAZE Jul 31 '18 at 2:26
  • $\begingroup$ In the "Schrodinger picture" of Quantum Mechanics all the time dependence is in the states and the operators are time-independent. In the "Hamiltonian picture" it's vice versa. In the "Interaction Picture" both the states and the operators have time dependence. All these ways of looking at the time evolution are useful, so best to just start getting used to it. ;) $\endgroup$ – hft Jul 31 '18 at 4:12

In the second line I believe that there is a mistake... Could someone please confirm or deny whether this is indeed correct

Correct. There is a mistake/typo.

In the first integral there may be a typo...


I can't figure out how to derive the relationship $$\frac{d}{dt}\hat{A}_{mn}(t)=-\frac{i}{\hbar}[\hat{A},\hat{H}]_{mn}\tag{1}$$

First consider $$ \mathcal{O}=A\;. $$ In this case: $$ A_{mn}(t)=e^{i(E_m-E_n)t}A_{mn} $$ and $$ \frac{dA_{mn}(t)}{dt}=i(E_m-E_n)e^{i(E_m-E_n)t}A_{mn}=i(E_m-E_n)A_{mn}(t)\;. \qquad (1) $$

Next consider (as you already did in the test problem): $$ \mathcal{O}=[A,H]\;. $$ In this case: $$ [A,H]_{mn}(t) = e^{i(E_m-E_n)t}(E_n-E_m)A_{mn}= (E_n-E_m)A_{mn}(t)\;. \qquad (2) $$

Comparing $i$ times Eq (1) with Eq (2), I see that: $$ i\frac{dA_{mn}(t)}{dt} = -(E_m-E_n)A_{mn}(t) = (E_n-E_m)A_{mn}(t) = [A,H]_{mn}(t)\;. $$

In other words: $$ \frac{dA}{dt}=-i[A,H] $$

  • $\begingroup$ Thank you for your answer (+1), I would just like to point out that you are missing factors of $\hbar^{-1}$ not only in the final answer but also in the complex exponentials and in the equation after $(2)$. Many thanks. $\endgroup$ – BLAZE Jul 31 '18 at 2:50
  • $\begingroup$ I have chosen units such that hbar=1. See, for example, en.wikipedia.org/wiki/Natural_units. You may restore the "missing" factors of hbar by dimensional analysis. For example, since hbar has units of energy times time, there is clearly one factor of hbar (=1) in either the numerator of the LHS of the final equation or the denominator of the RHS of the final equation... $\endgroup$ – hft Jul 31 '18 at 4:05
  • $\begingroup$ I'm sorry, still a little confused by what you mean when you say "First consider $\mathcal{O}=A$. In this case $A_{mn}(t)=e^{i(E_m-E_n)t}A_{mn}$". Could you please expand on what you are doing here? As last time I checked $\hat{\mathcal{O}} = [\hat{A},\hat{H}] \ne A$. Thanks for your help so far. $\endgroup$ – BLAZE Aug 1 '18 at 1:22
  • 2
    $\begingroup$ @BLAZE The first part of this answer is just addressing the part of the original question that asks for $d\hat{A}_{mn}/dt$. The part of the original question just before that finds $d\hat{\mathcal{O}}_{mn}/dt$, without any restriction on what $\hat{\mathcal{O}}$ is, so hft is just saying to use that result but replace $\hat{\mathcal{O}}$ with $\hat{A}$. The second part of this answer says to use a different replacement. There are really two different versions of $\mathcal{O}$. It might have been better in the original to use different letters, but hft has stuck with the original notation. $\endgroup$ – Mike Aug 1 '18 at 2:14

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