I previously asked this question, in which I'm trying to better understand this joshphysics's derivation of an interpretation of the time-energy uncertainty principle.

And the gist of what I get is (from the answers and the chatroom) within interpretations of quantum mechanics which say the measurement is non-unitary one can definitely make the case that the derivation cannot be applied to all situations (in fact I feel the application is quite limiting).

However, it seems intuitive to me (after chatroom discussion) with a quasi-classical measuring apparatus and the Copenhagen interpretation enables one to circumvent my objection? (And also enables one to apply it to cases involving a measurement - see related/objection).

Note: The below is a labyrinth of positions one can take on this argument. And I feel there are $2$ ways out (way $3$ or way $4$)of this maze. I'm asking for which way is better? (See Questions for what I mean by "better")

Math and Context

The proof (I questioned)

So let's begin with Joshphysics's answer:

Let a quantum system with Hamiltonian $H$ be given. Suppose the system occupies a pure state $|\psi(t)\rangle$ determined by the Hamiltonian evolution. For any observable $\Omega$ we use the shorthand $$ \langle \Omega \rangle = \langle \psi(t)|\Omega|\psi(t)\rangle. $$ One can show that (see eq. 3.72 in Griffiths QM) $$ \sigma_H\sigma_\Omega\geq\frac{\hbar}{2}\left|\frac{d\langle \Omega\rangle}{dt}\right| $$ where $\sigma_H$ and $\sigma_\Omega$ are standard deviations $$ \sigma_H^2 = \langle H^2\rangle-\langle H\rangle^2, \qquad \sigma_\Omega^2 = \langle \Omega^2\rangle-\langle \Omega\rangle^2 $$ and angled brackets mean expectation in $|\psi(t)\rangle$. It follows that if we define $$ \Delta E = \sigma_H, \qquad \Delta t = \frac{\sigma_\Omega}{|d\langle\Omega\rangle/dt|} $$ then we obtain the desired uncertainty relation $$ \Delta E \Delta t \geq \frac{\hbar}{2} $$ It remains to interpret the quantity $\Delta t$. It tells you the approximate amount of time it takes for the expectation value of an observable to change by a standard deviation provided the system is in a pure state. To see this, note that if $\Delta t$ is small, then in a time $\Delta t$ we have $$ |\Delta\langle\Omega\rangle| =\left|\int_t^{t+\Delta t} \frac{d\langle \Omega\rangle}{dt}\,dt\right| \approx \left|\frac{d\langle \Omega\rangle}{dt}\Delta t\right| = \left|\frac{d\langle \Omega\rangle}{dt}\right|\Delta t = \sigma_\Omega $$

Before I go on about my objection. Let me let you know it's basic premise.

Basic premise

The measurement is cannot be modelled as a unitary transformation. $$ |a \rangle + |b \rangle \to |a \rangle $$ Cannot be done by a unitary transformation and I thought I was vindicated in this post. Which is why the common referral of the measurement as "non-unitary"

The Objection

joshphysics's answer makes use of Heisenberg's equation of motion which rely on unitarity. I am not aware of a derivation of Heisenberg's equation of motion which is applicable during a non-unitary process (for example the measurement).

See the $2$'nd equation in his derivation. This clearly relies on "Heisenberg's equation." Further the whole notion of getting a $\Delta t$ is only dimensionally true and not true even as an approximate (if this objection holds).

One way out: is your calling something velocity which has the same dimensions of velocity but is not the time derivative of position. But that is fundamentally wrong! (and abused in that case).

Another way out: is to claim that there is a kind of continuity argument but I was under the impression the measurement is a discontinuous process. So this one will fail.

The third way out: is to limit the physical scope of the applicability of the this derivation. For example, one cannot use this derivation to say the time between $2$ subsequent measurements cannot be taken to $0$. Perhaps it can (pushing it's limitations) tell you some kind of energy cost involved between $2$ subsequent measurements. But that would be at the price of presupposing "the time between $2$ subsequent measurements cannot be taken to $0$" as one would have assume that between 2 measurements unitarity must reign and the time interval cannot be taken to $0$.

So far so good?

I was on board with the third way out. In fact, any interpretation of quantum mechanics which uses the language "collapse of the wave function." I am under the impression will be sympathetic to my misunderstanding to say the least. Taken from wiki, The existence of the wave function collapse is required in

  1. the Copenhagen interpretation
  2. the objective collapse interpretations
  3. the transactional interpretation
  4. the von Neumann interpretation in which consciousness causes collapse.

The fourth way out (?)

Now, another user in the chat is of the opinion my objection is non-sensical (at least thats the impression I get) and upon conversing it seems he has a fourth way to bypass my objection!

The fourth way: In this fourth way one stops the basic premise of my objection which is the measurement being a non-unitary process! In fact this is convincingly done with:

"A measurement is an interaction between a measuring apparatus which is (quasi-)classical and a quantum system You can model this, it's in Landau's QM section 7, I don't want to quote the whole thing, but this wave function collapsing business is pretty straightforward there without all the woo"

In fact, he claims to be a Copenhagen-ist but doesn't like the words "collapse of the wave-function."

The fourth way out does manage to circumvent (if not completely destroy) my objection and also enables one to use this derivation in far more many many situations I previously see as unfathomable for example: the time between $2$ measurements cannot be taken to $0$.

In fact, I'm under the opinion the number of physical experiments the $3$'rd way out enables you to do are under $5$ (?). But the $4$'th way is clearly superior in this metric (?). A better question might be where does it not apply?

More about way $4$

For those who are shocked that Landau was on board with way $4$ and (fairly) demanding evidence before investing time for an entire section $7$ of Landau's QM.

See page 6 of the file (page $9$ of .pdf) of this where he argues why one should take way-out $4$:

’...It is in principle impossible ... to formulate the basic concepts of quantum mechanics without using classical mechanics.’ (LL2)

’...The possibility of a quantitative description of the motion of an electron requires the presence also of physical objects which obey classical mechanics to a sufficient degree of accuracy.’ (LL2)

’...the ’classical object’ is usually called apparatus and its interaction with the electron is spoken of as measurement. However, it must be emphasized that we are here not discussing a process ... in which the physicist-observer takes part. By measurement, in quantum mechanics, we understand any process of interaction between classical and quantum objects, occurring apart from and independently of any observer. The importance of the concept of measurement in quantum mechanics was elucidated by N Bohr.’ (LL2)

The quotes go on ... The above is only the beginning (In fact the .pdf is a condensed version of "pure gold")

At some risk (I'm not sure if the person who exposed me to way $4$ would agree this paper does justice to their arguments): https://arxiv.org/abs/1406.5178

To be fair a different version of way $4$ has been asked on this site:


Where the answer-er response ends with:

"Why not let the final state be in a superposition also, as quantum mechanics requires? The alive cat is not aware of the dead cat, because the Schrödinger equation is linear."

For a non-absurd association of the same words above: https://www.youtube.com/watch?v=gNAw-xXCcM8

(In light of this let me just say my question is not about the plausibility of the Unitary of the measurement but it's about if assumed can one assume this and push the number of physical situations this derivation is applicable to?)

Is a unitary measurement compatible with broadening the usage of the time-energy uncertainity principle?

Also Landau states: "From section $44$ of Landau, the $\Delta E$ in the time-uncertainty relation is the difference "between two exactly measured values of the energy $E +\varepsilon$ at two different instants, and not the uncertainty in the value of the energy at a given instant" (for a system of two parts with energies $E$ and measuring apparatus energy $\varepsilon$)"

I added this part so it didn't seem like I was skipping any logic. I'm sure Landau would have only said that if he felt his book justified the case.


Is this "fourth way" plausible? Can someone tell me an explicit case where if I assume this circumvention is indeed true what are the physical cases the this "quasi-classical measuring apparatus and the Copenhagen interpretation" seem to work for that the original objection stops (way $3$)? (I do mention one but am under-confident if it is right)

P.S: I feel this post was self-contained but to do justice to way $4$ I had to link it off. Please do not hesitate to contact me if you feel this question can be improved?

Cross-posted on PSE.SE

  • $\begingroup$ Please title and ask your question in a form that doesn't require following links to understand what the question is. $\endgroup$ Commented Sep 12, 2019 at 20:38
  • $\begingroup$ @CraigGidney in the cross-post it came to the point where glS wanted the chat transcript as well ... Also if I were do that I would also have to copy paste a derivation and comment my lack of satisfaction. The size of this post would grow massively. However, if you are of the opinion I should do it then I will. $\endgroup$ Commented Sep 12, 2019 at 20:41
  • $\begingroup$ Yes, you should quote the derivation. $\endgroup$ Commented Sep 12, 2019 at 20:43
  • $\begingroup$ @CraigGidney I have made the post self contained $\endgroup$ Commented Sep 13, 2019 at 6:58
  • $\begingroup$ Reason for downvote? $\endgroup$ Commented Sep 15, 2019 at 7:22

1 Answer 1


There are a lot of comments and objections in the question, too many in fact to go through them all. I will try to address some of the points that I think hide misconceptions, to hopefully give a clearer idea of what's going on.

About "interpretations of QM in which the measurement is non-unitary"

The non-unitarity of the measuring process is not a matter of interpretations. The mapping between a state and a post-measurement state is non-unitary, period. Interpretations, by definition, do not change any predictions of a theory, and thus won't change this fact.

To be clear, this remains true even if one somehow "explains" at exactly what point, and due to what process, the non-unitarity of the measurement process "happens" (e.g. due to some collapse model, or if one adheres to the view that the non-unitarity of measurement is an effective description of a unitary process in which some of the information is neglected, à la Zurek). This is because any such model would essentially explain how/why the mapping $$|\psi\rangle\mapsto|\psi_i\rangle\tag A$$ happens with probability $|\langle\psi_i|\psi\rangle|^2$. However, the mapping (A) will remain non-unitary, regardless of such explanations, even though one might be able to explain the non-unitarity as arising from some other principle/axiom/mechanism.

All this is to say that the statement "derivation X cannot be applied to Y due to interpretations of the measurement process" is nonsense. Nonetheless, it is true that the quoted derivation cannot be applied "to all situations".

The derivation cannot be applied in "all situations"

Sure: it can be applied in all and only the situations in which the derivation applies.

The derivation under consideration is nothing but Heisenberg's uncertainty relation for a pair of operators $A,B$, in the form $\sigma_A \sigma_B\ge |\langle[A,B]\rangle|/2$, in the special case $A=\Omega$ and $B=H$, plus some additional definitions and algebraic manipulations.

What this is telling us is that if, at any given moment $t$, we compute $\sigma_\Omega$ and $\sigma_H$ (which will require multiple repetitions of the experiment, as each experiment will produce a single sample from the underlying distributions, and $\sigma_\Omega$,$\sigma_H$ are the variances of such distributions), we'll find the relation $\sigma_H\sigma_\Omega\ge|\partial_t\langle\Omega\rangle|/2$ to bew satisfied. Again, note that to actually observe this, you would need to perform the same exact experiment multiple times, and then measure the relevant quantities all at the same time $t$ (and probably also at some future time $t+\epsilon$ to compute the time-derivative of $\langle\Omega\rangle$).

In other words, you can think of this relation as telling you what you would observe if, at any given point of the (unitary evolution), you were somehow able to know the values of the relevant variances and expectation values (which you cannot, but you can simulate the situation by doing many measurements repeating the experiment). This is, therefore, a statement about the values of the expected variances at any given point of time.

Furthermore, notice how the argument holds for any state, not just pure ones (although the interpretation of time provided in joshphysics' answer might not). This means that it doesn't care about whether measurements were performed at some point in the past of the considered instant of time $t$. The only thing that matters is the state of the system at the time $t$. Note that you can also model measurement processes as (non-unitary) maps applied to the state. For example, measuring a state $\rho$ in the computational basis can be modelled as the mapping $$\rho\mapsto\mathcal E(\rho)\equiv\sum_k \operatorname{Tr}(\rho P_k)P_k,$$ where $P_k\equiv|k\rangle\!\langle k|$. This still produces a state, on which you can apply the uncertainty realtion.

But we use Heisenberg's equation to obtain the result!

True. So I guess to be more precise, the statement refers to the relation between $\sigma_\Omega,\sigma_H$ at a given time $t$, and the rate of increment of $\langle\Omega\rangle$ at the time $t$. This increment refers to the rate of increment of $\langle\Omega\rangle$ between some $t$ and $t+\epsilon$ when the state evolves unitarity in between the two times. In other words, to compute it, you need to measure $\langle \Omega\rangle$ at time $t$, and then measure $\langle\Omega\rangle$ at time $t+\epsilon$ (and by this I mean that you repeat the measurement without measuring at $t$, but this time only measuring at $t+\epsilon$).

  • $\begingroup$ I'm re-reading your answer and I'm wondering why doesnt "effective description of a unitary process" increase the scope of applicability of this when we use "Heisenberg's equation to obtain the result $\endgroup$ Commented Sep 16, 2019 at 3:59
  • $\begingroup$ I'm guessing in the last paragraph of the answer means we have to talk about different systems. I'm getting the impression from here: "time $t+ϵ$ (and by this I mean that you repeat the measurement without measuring at $t$)" In light of the previous comment why can't I talk about the same system? $\endgroup$ Commented Sep 16, 2019 at 4:53
  • $\begingroup$ @MoreAnonymous because even if you have an "effective description of a unitary process", the only thing that changes is how you describe what happens in between having $|\psi\rangle$ and having $|\psi_i\rangle$. However, the calculation will apply unchanged when you only consider the post-measurement state (here $|\psi_i\rangle$).The way you will explain obtaining the measurement results will be different, but not the measurement results themselves ,which is what this calculation uses. $\endgroup$
    – glS
    Commented Sep 16, 2019 at 8:03
  • $\begingroup$ @MoreAnonymous I'm still talking about the same system, but here you have to be careful in how you measure things. What I mean is that you need to measure $\langle \Omega\rangle(t)$, which is obtained by evolving unitarily $|\psi\rangle$ from $\tau=t_0$ (whatever $t_0$ is here) to $\tau=t$, and then you need to measure $\langle\Omega\rangle(t+\epsilon)$, which you do evolving unitarily $|\psi\rangle$ from $\tau=t_0$ to $\tau=t+\epsilon$. However, the calculation does not apply if you first collapse the system at $\tau=t$, then evolve unitarily from $t$ to $t+\epsilon$ and then measure again $\endgroup$
    – glS
    Commented Sep 16, 2019 at 8:06
  • $\begingroup$ @MoreAnonymous or if you like, yes you are considering different copies of the same system, which is what I mean when I talk of the need of repeated measurements to actually observe this. Each system can only be made to collapse once before being irreversibly changed, so to measure expectation values and variances of operators you need to prepare the same system $|\psi\rangle$ at time $\tau=t_0$ multiple times and have each collapse at $\tau=t$ or $\tau=t+\epsilon$, in order to collect enough statistics to compute the quantities involved in the relation $\endgroup$
    – glS
    Commented Sep 16, 2019 at 8:09

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