I am following MIT lessons on quantum physics (Prof. Zwiebach): Part I, Lecture 6, at https://ocw.mit.edu/courses/physics/8-04-quantum-physics-i-spring-2016/lecture-notes/

Video lecture: https://youtu.be/Ex_fFlwZoM0

I understand that the normalized wave function can be written as the integral of probability density: $ N(t)=\int \rho (x,t)dx$ and if we have at the initial time t-zero $N(t_0)=1$. We can prove that probability is conserved at any time if $\frac{dN(t)}{dt}=0$ that is equal to: $$\frac{dN(t_0)}{dt}=\int_{-\infty }^{\infty } \frac{\partial \rho (x,t)}{\partial t}dx =\int_{-\infty }^{\infty }\left(\frac{\partial \psi^*(x,t)}{\partial t} \psi(x,t)+\psi ^*(x,t) \frac{\partial \psi(x,t) }{\partial t} \right)dx$$

by complex conjugating the Schrodinger equation we have: $$ \frac{dN(t_0)}{dt}=\int_{-\infty }^{\infty } \frac{\partial \rho (x,t)}{\partial t}dx =\frac{i}{\hbar}\left(\int_{-\infty }^{\infty }(\psi \hat H)^* \psi dx-\int_{-\infty }^{\infty }\psi ^* (\hat H\psi)dx\right)$$ At this point to have zero we need $$ \int_{-\infty }^{\infty }(\psi \hat H)^* \psi dx=\int_{-\infty }^{\infty }\psi ^* (\hat H\psi )dx$$

This equation is valid if the general condition of hermiticity hold: $\int\psi_1^* (T\psi_2 ) dx=\int (T^\dagger \psi_1 )^* \psi_2 dx$ in our case$ \int (\hat H \psi _1)^* \psi _2 dx=\int \psi _1^* \hat H \psi _2 dx$. Now i go to my problem, I understand that the above is valid when $\psi _1 = \psi _2$ and we can prove through the boundary condition $\lim_{x\to \pm\infty}\psi (x,t)=0 ; \lim_{x\to \pm\infty}\frac{\partial \psi (x,t)}{\partial x} < \infty $ that the $\frac{dN(t)}{dt}=0$ and if this is zero the hamiltonian is hermitian.

In the video lecture, Prof. Zwiebach asks (5:10) to prove the same also for the general case when $\psi_1$ and $\psi_2$ are different functions.

Can someone help me to understand how to make this proof?

  • $\begingroup$ "the normalized wave function can be written as the integral of probability density" is not right, to obtain the total probability ($=1$ if normalised) you integrate over the probability density, which is given by $|\psi(x)|^2$ where $\psi(x)$ is the normalised wave function. $\endgroup$
    – Charlie
    Commented Jan 4, 2021 at 15:33
  • $\begingroup$ Yes, i understand. thanks. $\endgroup$
    – franchino
    Commented Jan 4, 2021 at 15:44

1 Answer 1


So you want to prove that the Hamiltonian $\hat{H}$ is Hermitian? There's two ways of answering this.

  1. By postulate: The energy is an observable. By the postulates of QM, it is described by a linear Hermitian operator on some Hilbert space. Ta-Da! Ok, this, doesn't give you much additional insight in the mathematical workings and will hardly qualify as a proof. So the other option may help you more to understand the concepts, though keep in mind that it is less fundamental/general.
  2. By using a specific Hamiltonian: Not every operator is Hermitian, so you will need to supply a specific Hamiltonian to prove the statement. You have not given us the Hamiltonian that is used in Prof. Zwiebach's lecture. In position representation, the Hamiltonian operator that will get you through most introductory lectures is probably $$\hat{H} \psi(x) = \left[-\frac{\hbar^2}{2m} \frac{\partial^2}{\partial x^2} + V(x)\right] \psi(x).$$ Just for practice, you can check the linearity of this operator yourself. You can also check that it's Hermitian fairly easy. $V(x)$ stands for a real-valued function $V: \mathbb{R}^d \to \mathbb{R}$ (it seems to be $d=1$ in your case). Therefore, $V(x) = V^*(x)$. So quite obviously $$\int_{-\infty}^\infty \textrm{d} x\ \psi_1^*(x) V(x)\psi_2(x) = \int_{-\infty}^\infty \textrm{d} x\ \psi_1^*(x) V^*(x)\psi_2(x) = \int_{-\infty}^\infty \textrm{d} x\ (V(x)\psi_1(x))^*\psi_2(x).$$ The derivative term is almost as simple, you just need to use integration by part and recall that in the Hilbert space you have, all functions and their derivatives fall off to zero at ininifty, $\psi(x) \to 0$ and $\partial_x \psi(x) \to 0$ as $|x| \to \infty$. Therefore, you can integrate by parts twice to get the desired result.

I think you did most of the work already, judging from what you say in your question. So just have at it once more, you'll get there.


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