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Is the time-derivative Hermitian?

  1. Jan 12, 2015 #1
    I want to know why the time-derivative acts as though it's Hermitian under conjugation. I have read elsewhere that the time-derivative isn't a true operator in the quantum mechanical sense but I don't understand why that's the case, and if that's the case I still don't understand why [itex] \partial_t^\dagger = \partial_t [/itex]. From what I do understand, in quantum mechanics a operator acts on a state vector which then gives us a new vector. This seems to be the case for the time-derivative, at least in the Schrodinger picture. For spatial derivatives we can use integration by parts to deal with the conjugate operators of the derivatives (i.e. in a scalar product) but since we're not integrating through time I can't find such a method to deal with the time derivative. So can anyone show me explicitly why the time-derivative is Hermitian?

    Thanks
     
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  3. Jan 12, 2015 #2

    vanhees71

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    The time derivative cannot be hermitian or self-adjoint, because time is not an observable in quantum theory. It's clear, why this must be so, as was explained by Pauli in the early 1930ies already: If time was an observable it should be described by a self-adjoint operator in Hilbert space and, due to the definition of observables through their association with space-time symmetries a la Noether (the mathematical satisfactory form of the "correspondence principle" as a way to heuristically "derive" a quantum model from a classical model), the canonically conjugate momentum to time would have to be the Hamiltonian (representing energy). But then you'd have a commutation relation as for position and momentum, implying that necessarily the entire real axis is the energy spectrum, which in conclusion what not be bounded from below.

    This contradicts the most evident empirical finding for the correctness of quantum theory: that matter is pretty stable. At least it doesn't "decay" into lower and lower energy states, which is because in Nature obviously there is a state of lowest energy (called "the vacuum" in quantum field theory). So the idea that time may be an observable leads to a contradiction with basic empirical evidence about the stability of the matter around us, and thus time is a parameter in quantum mechanics as it is one in classical physics.
     
  4. Jan 12, 2015 #3

    dextercioby

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    vanhees71, your argument in the first paragraph is mathematically flawed. It's not necessary for both operators to have unbounded spectra.
    [Mentor's note: some follow-on discussion moved to https://www.physicsforums.com/threads/follow-on-to-is-the-time-derivative-hermitian.792270/] [Broken]

    Let's turn to the OP. The differentiation with respect to time is not a true operator and if it was, then going from Psi(t) to Psi'(t) is equal to H applied to Psi(t), for all Psi (t) in the Hamiltonian's domain.
     
    Last edited by a moderator: May 7, 2017
  5. Jan 12, 2015 #4
    Thank you for the explanations, and while your argument is physically intuitive I still don't understand why the time-derivative should be treated as anything other than an operator. From what I understand, mathematically a linear operator is a device that associates a new vector with every vector in the Hilbert space with a linear correspondence. The derivative has the linearity properties associated with it, and it assigns a new vector to every vector in the Hilbert space as far as I can tell. So while I know there is no time operator, I still don't understand why the time-derivative doesn't satisfy the properties of a linear operator.

    Nevertheless, this is all a bit more abstract than I need. If [itex]\partial_t[/itex] is not an operator in quantum mechanics then I don't understand the meaning of the following operation: [itex]\left[ \partial_t |\psi\rangle \right]^\dagger [/itex]. The time-derivative does not appear to be a scalar, ket, or bra. If it's not an operator, then what is it exactly? These are more or less the only quantum mechanical mathematical entities I know.

    More specifically, I know that the Hermitian conjugate of the Schrodinger equation is

    [tex]\left[ \partial_t |\psi\rangle = \frac{1}{i \hbar} H(t) |\psi\rangle\right]^\dagger \rightarrow \partial_t \langle \psi | = - \frac{1}{i \hbar} H(t) \langle \psi |. [/tex]

    Why does the time derivative in the conjugated Schrodinger equation remain the simple time derivative?
     
  6. Jan 12, 2015 #5

    dextercioby

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    I'm afraid the use of the bra-ket formalism is not clearing up things, instead this use does what it usually does: hides some delicate mathematical issues.

    Let's get it from the beginning. One of the key (not explicit) assumptions of the so-called 'orthodox' formulation of QM (due largely to Dirac and von Neumann through their 1930/1932 books) is that 'time is a parameter', not an observable. This folds nicely with classical dynamics in the Hamiltonian formulation: there, too the 't' is nothing but a parameter, not a classical observable (the phase space will not include it, instead we're interested in further parametrizing a classical state (point in the phase space) using 'time', time is then the intrinsic parameter of curves in phase space: evolution parameter of classical Hamiltonian states). So the nice Schrödinger building of a Hilbert space (ignore for now rigged Hilbert spaces or the notions of pure versus mixed states) will take 't' as a parameter (think actually of the Heisenberg picture, where the Hilbert space is <still>/<frozen>, while the observables such as momentum & position depend on time) in the following sense: At each moment in time, ## \psi (x)## is a normalized vector in a Hilbert space which is then (by the uniqueness theorem of von Neumann) safely to be taken as ##\mathcal{L}^2 \left(\mathbb{R},dx\right)## (assuming infinite motion).

    So, in what sense do we interpret the LHS of the Schrödinger equation: ##\frac{\partial \psi (x,t)}{\partial t} ##? Well, in the very old way of treating partial derivatives of independent variables in an ordinary function of 2 (3,4, etc.) variables, i.e. using limits. The Schrödinger eqn. asserts that provided that

    ## \lim_{\delta t \rightarrow 0} \frac{1}{\delta t} \left[\psi (x, t+\delta t) - \psi (x, t)\right] ## exists, then it is equal to the vector in the codomain of the Hamiltonian operator at the same moment of time.

    One can surely see the difference of 2 vectors taken before the limit as another vector whose norm then tends to 0 so that the limit is still finite (by postulation) and the result can even be normalized to unity. But one must understand that this limiting process is extraneous to the standard Hilbert space per se, i.e. this limit is not a weak, nor a strong one. Just as in classical dynamics, time acts as a parameter for curves of normalized unit vectors in the (complex, inf-dim., separable) Hilbert space. These curves are assumed to be at least of class [itex] C^1 [/itex] in the parameter (differentiable of 1st order and the 1st derivative to be continuous).

    As I wrote above, if you're willing to take the process of 1-time differentiation of a Hilbert space vector-valued function as a linear operator in the abstract sense, then, by virtue of the Schrödinger's equation, you can always compute the range of this 'linear operator': it's the vector you're getting by applying the Hamiltonian onto it.

    So (as people call it nowadays here) FAPP, this linear operator is the Hamiltonian, a genuine operator, in the sense it's a function of the usual fundamental quantum observables: position, momentum, spin. You can even consider that this linear operator which takes a vector-valued parametric function to its derivative (also vector-valued) to have a hermitian adjoint in the regular sense. This adjoint is of course the adjoint of the Hamiltonian, which is of course the Hamiltonian, because as, again by postulating, the Hamiltonian is self-adjoint. The formal manipulation in the bra-ket formulation trivially follows.
     
    Last edited: Jan 12, 2015
  7. Jan 12, 2015 #6

    strangerep

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    (I think some of the answers here might be too advanced for the OP, so I offer a simpler approach...)

    It's an operator, (loosely speaking). There are subtleties involved, as others have said, but let's just think of it as an operator that's densely defined (for present purposes, you could think of this as meaning: well-defined almost everywhere, for physically reasonable wave functions).

    Next, do you know the correct definition for "adjoint of an operator" in Hilbert space? (Simply applying a dagger or complex conjugate to everything in sight is not the correct answer.) If you're unsure, try this Wikipedia page, in particular the section on "adjoint of densely defined operators". (If you replace the x,y vectors there by wavefunctions ##\phi,\psi##, the notation might become more familiar.)

    If you understand that, then: what is the momentum operator in ordinary QM? (I hope you know this -- if not then you really do need a textbook.) Then prove from the above definitions that the momentum operator is self-adjoint. (Hint: use an explicit representation of the inner product as a spatial integral, and use integration by parts).

    If you can do that, you're well on your way to understanding the rest. (Hint: think in terms of ##i\partial_t## rather than just ##\partial_t##.) If not, say what you do/don't understand from the above.
     
  8. Jan 12, 2015 #7
    I've proved that the momentum operator is self-adjoint by showing [itex] \langle \phi | P | \psi \rangle = \langle \psi | P | \phi \rangle^* [/itex] by projecting everything into position space and integrating, however there was a very nifty integration by parts trick that allowed me to prove that. Since I never integrate over time, I don't really see how to do this for the time-derivative. I can provide more details if necessary.
     
  9. Jan 12, 2015 #8

    bhobba

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    Well, as others have pointed out, there are mathematical niceties involved that is only really resolved in so called Rigged Hilbert Spaces but I will leave that issue aside.

    You said you never integrate over time - I don't know what you mean by that - its done all the time - pun intended :-p:-p:-p:-p:-p:-p:-p:-p

    My answer though is a bit different. Have a look at Ballentine - Quantum Mechanics - A Modern Development chapter 3 where Schroedinger's equation is derived - not postulated - but derived (it assumes the symmetries of Galelaean relativity). Since the RHS is Hermitian - so must the LHS.

    Also in that chapter you will encounter the very important Wigners theorem:
    http://arxiv.org/abs/0808.0779

    Hopefully you will see what happens when you take the derivative of a unitary operator - I think its Stones theorem or something like that - but non rigorously its not hard - for a small parameter t U(t) = 1 + t dU/dt and apply UU(conjugate) = 1.

    Added Later
    I just realised - of course you don't need the full machinery of deriving Schroedingers equation - you simply need Wigners theorem - but read chapter 3 anyway.

    Thanks
    Bill
     
    Last edited: Jan 12, 2015
  10. Jan 13, 2015 #9

    strangerep

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    You don't need to. In fact, bhobba has already given away (most of) the answer: you only need consider wave functions which satisfy a Schrodinger equation. So... write down the (time-dependent) Schrodinger equation, and use it to re-express ##i\partial_t## in terms of other operators and potential. Then, can you prove that this re-expression is self-adjoint? Hint: do the integration-by-parts trick twice, or (better) apply your previous result about ##P##.

    (If/when you understand that ok, we can come back to why we need only consider wave functions satisfying the S.E.)
     
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