In section 6.3.1 of the following MIT Open Course Ware (PDF) (22-02, Introduction to Applied Nuclear Engineering), the author, Prof Paola Cappellaro, derives Heisenberg's equation using the definition of expectation values,

$$ \frac{d}{dt}\langle\hat{A}\rangle=\frac{d}{dt}\int d^3x\psi^*(x,t)\hat{A}\psi(x,t)\\ =\int d^3x\left(\frac{\partial\psi^*}{\partial t}\hat{A}\psi+\psi^*\frac{\partial\hat{A}}{\partial t}\psi+\psi^*\hat{A}\frac{\partial\psi}{\partial t}\right) $$ and Schrodinger's equation, $$ \frac{\partial\psi}{\partial t}=-\frac{i}{\hbar}\hat{H}\psi,\qquad \frac{\partial\psi^*}{\partial t}=\frac{i}{\hbar}\left(\hat{H}\psi\right)^* $$ Then the author uses $\left(\hat{H}\psi\right)^*=\psi^*\hat{H}^*=\psi^*\hat{H}$ to get $$ \frac{d}{dt}\langle\hat{A}\rangle=\int d^3x\left(\frac{i}{\hbar}\psi^*\left[\hat{H}\hat{A}-\hat{A}\hat{H}\right]\psi+\psi^*\frac{\partial\hat{A}}{\partial t}\psi\right)\\ =\frac{i}{\hbar}\langle\left[\hat{H},\,\hat{A}\right]\rangle+\langle\frac{\partial\hat{A}}{\partial t}\rangle $$

There are two things I don't understand in this derivation. The first is that the total time derivative $\frac{d}{dt}$ turns into a partial time derivative $\frac{\partial}{\partial t}$ when inside the integral.

The second is that the author states $$(\hat{H}\psi)^*=\psi^*\hat{H}$$

This confuses me as $\hat H$ is the Hamiltonian operator, and contains an $x$ derivative. Thus the left hand side of the above equation is another state $\phi^*$ such that $\phi=\hat H\psi$, but the right hand side is an operator: the derivative is yet to be applied to anything. This equation by itself just makes no sense to me.

Any help with either of these two issues would be greatly appreciated!


2 Answers 2


Neither left-hand side nor right-hand side are properly operators that act on states and give states. They are a form of operators that act on states and give complex numbers. We call such a thing a bra (and a state is called a ket, so when you pair them you get a bra(c)ket).

The left-hand side acts like $$((H\psi)^*)(\varphi) = \int \psi^* H \varphi \, dx.$$ If this confuses you, think about the finite-dimensional case. Then $\psi$ is a column vector and $H$ is a square matrix. While there is a complex conjugation operation for vectors, what you really want is the Hermitian conjugate, which is transpose and complex conjugation, denoted with a ${}^\dagger$. Then $(H\psi)^\dagger$ is the transpose of a column vector, so it's a row vector, and $(H\psi)^\dagger = \psi^\dagger H$ and this can act on a column vector with matrix multiplication.

For wavefunctions, $(\psi^\dagger)(x) = \psi(x)^*$ where the star is complex conjugation, and matrix multiplication is of a row vector and a column vector is integration.

  • $\begingroup$ Well if I have the operator $-i\bar{h}\frac{\delta}{\delta x}$, and apply it to $\phi (x,t)$, then I have $(-i\bar{h}\frac{\delta}{\delta x}\phi (x,t))^*=\phi ^*(x,t) i\bar{h}\frac{\delta}{\delta x}$. So the operator is no longer applied to $\phi$. But this operator (the momentum operator) \emph{does} act on a state and give a state as far as I know? $\endgroup$
    – Lammey
    Apr 3, 2014 at 17:27
  • $\begingroup$ Yes, the momentum operator acts on a state and gives a state. But to the left of it you have a bra, which acts on a state and gives a complex number. You have something of the structure (row vector) (square matrix), which is like a row vector. $\endgroup$ Apr 3, 2014 at 20:31
  • $\begingroup$ This may sound idiotic but I don't see a bra anywhere. Do you mean that $\phi^*(x,t)$ is a bra?. I don't understand how that could be the case, because $\phi (x,t)$ doesn't act on a state as far as I know? I mean if I take the product of two functions $\psi \phi$ then it's another function. $\endgroup$
    – Lammey
    Apr 4, 2014 at 11:52
  • $\begingroup$ This is why you should think of it as linear algebra! The object you want to take the conjugate of is not the point value $\phi(x)$, you want to take the conjugate of the whole vector, the ket $|\phi \rangle$, and this gives you the bra $\langle \phi |$, which does act on a state. The action is given by $\langle \phi |\psi \rangle = \int \phi(x)^* \psi(x) dx $ and this is why it is possible to confuse $\langle \phi |$ and $\phi(x)^*$. Yes, the product of two functions is a function, but the product of two vectors is something else. $\endgroup$ Apr 4, 2014 at 13:36

There are two things I don't understand in this derivation. The first is that the total time derivative $\frac{d}{d t}$ turns into a partial time derivative $\frac{\partial}{\partial t}$ when inside the integral.

Generally speaking, this is implied by the Leibniz intregral rule, which is a particular case of differentiation under the integral sign. For a function $f(x,t)$, the rule implies: $$ \frac{d}{dt} \left(\int_{x_0}^{x_1} f(x,t)dx \right) = \int_{x_0}^{x_1} \frac{\partial f(x,t)}{\partial t} dx$$ See the links for derivations, and at least note that if the limits of integration are themselves time dependent functions, their derivatives become important and you will have two additional terms outside the integral sign. The Leibniz rule typically refers to the case where the limits are constants and so their derivatives (and hence the two extra terms) are zero.

The second is that the author states: $(\hat{H}\psi)^* = \psi^*\hat{H}$. This confuses me as $\hat{H}$ is the Hamiltonian operator, and contains an $x$ derivative.

Going along with Robin Ekman's answer, you'll find that it's often more useful and illuminating to think about these expressions with your linear algebra goggles on -- that is, thinking about operators as self-adjoint matrices, rather than differential operators acting on functions in the Cartesian basis. The latter gets real ugly, real fast when you get into 3-dimensional QM. Dirac (bra-ket) notation will be your best friend, I promise.

But now that we've stressed that enough, I suppose an example might be helpful in thinking about linear differential operators. Consider the 1-D momentum operator, $\hat{p} = \frac{\hbar}{i} \frac{d}{dx}$. Let's now show that this operator is Hermitian: $$<f|\hat{p}g> = \int_{-\infty}^{\infty} f^* \frac{\hbar}{i}\frac{dg}{dx}dx$$ Now, apply integration by parts. This gives: $$\int_{-\infty}^{\infty} f^* \frac{\hbar}{i}\frac{dg}{dx}dx = \frac{\hbar}{i}f^* g |_{-\infty}^{\infty} + \int_{-\infty}^{\infty} \left(\frac{\hbar}{i}\frac{df}{dx}\right)^* g dx = <\hat{p}f|g>$$ Note that the first term must be zero, given that wavefunctions in QM are square integrable, or "normalizable," which requires that they go to zero at $\pm \infty$.

[Example source: Griffiths Introduction to QM, section 3.2.1]

  • 1
    $\begingroup$ It's strange that QM texts start with this mess of integrals, wavefunctions and PDEs. It's all linear algebra -- which is simple -- and they're hiding it under all this calculus -- which is hard. Townsend and Sakurai do it "the right way", but I don't think they're used as first texts. $\endgroup$ Apr 3, 2014 at 20:34
  • $\begingroup$ I'd agree that it's probably overemphasized to the point of confusion. My favorite reference is Shankar, which starts out with a dense 75 pages of mathematical background -- of course, almost entirely linear algebra. Nonetheless, I think calculus has it's place; I imagine it would be even stranger if students learned how to use a Clebsch-Gordan table before they knew how to solve for solutions of the finite square potential. $\endgroup$
    – RGMyr
    Apr 3, 2014 at 20:53
  • $\begingroup$ Thank you for answering the first part, I completely understand that now. I understood the bit where you showed that the momentum operator is hermitian. But the statement $(\hat{p}f(x,t))^*=f^*\hat{p}^*$ still makes no sense to me by itself. To give an explicit example, if $f(x,t)=e^{-ax^2}$ then $(\hat{p}f(x,t))^*=(-i\bar{h}\frac{\delta}{\delta x} e^{-ax^2})^*=-2ai\bar{h}xe^{-ax^2}$. This is not the same as $f(x,t)^*\hat{p}^*=i\bar{h}e^{-ax^2}\frac{\delta}{\delta x}$. It seems to me that the equality just doesn't make mathematical sense. Is it a shorthand for the integral expression? $\endgroup$
    – Lammey
    Apr 4, 2014 at 12:10
  • $\begingroup$ I don't have time to write out a very thorough explanation this morning, but take a look at this wikipedia section (particularly the first 2 subsections): en.wikipedia.org/wiki/Bra%E2%80%93ket_notation#Linear_operators $\endgroup$
    – RGMyr
    Apr 4, 2014 at 12:47
  • $\begingroup$ If you still have questions, I'll be back later this afternoon. $\endgroup$
    – RGMyr
    Apr 4, 2014 at 12:48

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