I am trying to prove that the normalization constant is independent of time. If we have fixed it for a particular time then it will remain constant for all time.

Suppose $\psi(x,t)$ is a wavefunction.
Let $A(t)$ be the normalization constant of $\psi(x,t)$
Then $\displaystyle A^*A\int_{-\infty}^{\infty}|\psi(x,t)|^2dx=1\tag{1}$
$\displaystyle \implies \frac{d}{dt}A^*A\int_{-\infty}^{\infty}|\psi(x,t)|^2dx=0\tag{2}$
$\displaystyle \implies\int_{-\infty}^{\infty}|\psi(x,t)|^2dx\frac{d}{dt}A^*A+A^*A\frac{d}{dt}\int_{-\infty}^{\infty}|\psi(x,t)|^2dx=0\tag{3}$

Now first analyze, $\displaystyle \frac{d}{dt}\int_{-\infty}^{\infty}|\psi(x,t)|^2dx$
$\displaystyle \frac{d}{dt}\int_{-\infty}^{\infty}|\psi(x,t)|^2dx=\int_{-\infty}^{\infty}\frac{\partial}{\partial t}|\psi(x,t)|^2dx\tag{4}$
$\displaystyle \frac{\partial}{\partial t}|\psi(x,t)|^2dx=\frac{\partial}{\partial t}(\psi^*\psi)dx=\psi^*\frac{\partial\psi}{\partial t}=\psi\frac{\partial\psi^*}{\partial t}\tag{5}$
By time dependent Schrodinger equation
$\displaystyle \frac{\partial\psi}{\partial t}=\frac{i\bar{h}}{2m}\frac{\partial^2\psi}{\partial x^2}-\frac{i}{\bar{h}}V\psi\tag{6}$
Also, $\displaystyle \frac{\partial\psi^*}{\partial t}=-\frac{i\bar{h}}{2m}\frac{\partial^2\psi^*}{\partial x^2}+\frac{i}{\bar{h}}V\psi^*\tag{7}$

So, $\displaystyle \frac{\partial|\psi|^2}{\partial t}=\psi^*\Big(\frac{i\bar{h}}{2m}\frac{\partial^2\psi}{\partial x^2}-\frac{i}{\bar{h}}V\psi\Big)+\psi\Big(-\frac{i\bar{h}}{2m}\frac{\partial^2\psi^*}{\partial x^2}+\frac{i}{\bar{h}}V\psi^*\Big)\tag{8}$
$\displaystyle \implies\frac{\partial|\psi|^2}{\partial t}=\frac{i\bar{h}}{2m}\Big(\frac{\partial^2\psi}{\partial x^2}-\frac{\partial^2\psi^*}{\partial x^2}\Big)\tag{9}$
$\displaystyle \implies\frac{\partial|\psi|^2}{\partial t}=\frac{\partial}{\partial x}\Big[\frac{i\bar{h}}{2m}\Big(\psi^*\frac{\partial\psi}{\partial x}-\psi\frac{\partial\psi^*}{\partial x}\Big)\Big]\tag{10}$
$\displaystyle \frac{d}{dt}\int_{-\infty}^{\infty}|\psi(x,t)|^2dx=\int_{-\infty}^{\infty}\frac{\partial}{\partial x}\Big[\frac{i\bar{h}}{2m}\Big(\psi^*\frac{\partial\psi}{\partial x}-\psi\frac{\partial\psi^*}{\partial x}\Big)\Big]dx\tag{11}$
$\displaystyle \implies\frac{d}{dt}\int_{-\infty}^{\infty}|\psi(x,t)|^2dx=\frac{i\bar{h}}{2m}\Big(\psi^*\frac{\partial\psi}{\partial x}-\psi\frac{\partial\psi^*}{\partial x}\Big)\Big|_{-\infty}^\infty\tag{12}$
As $\displaystyle \psi(x,t)\to0$ as $x\to\pm\infty$.
So, $\displaystyle \frac{d}{dt}\int_{-\infty}^{\infty}|\psi(x,t)|^2dx=0\tag{13}$
So, $(3)$ becomes
$\displaystyle \implies\int_{-\infty}^{\infty}|\psi(x,t)|^2dx\frac{d}{dt}A^*A=0\tag{14}$
As $\psi$ is square integrable, so $\int_{-\infty}^{\infty}|\psi(x,t)|^2dx=c$ where $c\in\mathbb R$ and $c\neq 0$
So, $\frac{d}{dt}A^*A=0\tag{15}$
$\implies |A|^2=constant\tag{16}$

I have the following doubts from the proof
(i) From $(11)$ to $(12)$, in the RHS, we have used fundamental theorem of calculus.
Ingtegrating of the derivative is the antiderivative. $\int_a^bf'(x)dx=f(x)$. But here the condition is that f has to continuous and differentiable on $[a,b]$ with $f'$ integrable on $[a,b]$.
So, in $\int_{-\infty}^{\infty}\frac{\partial}{\partial x}\Big[\frac{i\bar{h}}{2m}\Big(\psi^*\frac{\partial\psi}{\partial x}-\psi\frac{\partial\psi^*}{\partial x}\Big)\Big]dx$, we take $\Big(\psi^*\frac{\partial\psi}{\partial x}-\psi\frac{\partial\psi^*}{\partial x}\Big)$ to be continuous. As $\psi$ and $\psi^*$ is continuous, this means that $\frac{\partial\psi}{\partial x}$ is also continuous. But we know in general that first derivative of $\psi$ can be discontinuous also. So how we have used fundamental theorem of calculus here?

(ii) From $(12)$ to $(13)$, we have taken $\frac{i\bar{h}}{2m}\Big(\psi^*\frac{\partial\psi}{\partial x}-\psi\frac{\partial\psi^*}{\partial x}\Big)\Big|_{-\infty}^\infty=0$ using the fact that $\psi\to 0$ as $x\to\pm\infty$. But this also means that $\frac{\partial\psi}{\partial x}\to 0$ as $x\to\pm\infty$. But how we can be sure that $\frac{\partial\psi}{\partial x}$ is bounded?

(iii) In $(16)$, we get $|A|^2=constant$. But from this how we get $A(t)=constant$.
$|A|^2$ constant means that magnitude of the vector in complex plane is constant but it might happen that the angle of $A$ changes. This angle changes as a function of $t$. So how we conclude that $A$ is independent of time?


3 Answers 3


If the Hamiltonian is time-independent, then it seems much simpler to use $$ \vert\Psi(t)\rangle = e^{-i \hat H t/\hbar}\vert\Psi(0)\rangle\, . $$ Then it is automatic that \begin{align} \langle \Psi(t)\vert\Psi(t)\rangle &= \langle \Psi(0)\vert e^{i\hat H t/\hbar} e^{-i \hat H t/\hbar}\vert \Psi(0)\rangle\, ,\\ &=\langle \Psi(0)\vert \Psi(0)\rangle \end{align} since $e^{i\hat H t/\hbar}e^{-i \hat H t/\hbar}=\mathbb{1}$.

If the Hamiltonian is time dependent, then the evolution operator $U(t)$ is constructed to be unitary, so that \begin{align} \vert\Psi(t)\rangle &= U(t)\vert\Psi(0)\rangle \, ,\\ \langle \Psi(0)\vert U^\dagger(t)U(t)\vert \Psi(0)\rangle&= \langle \Psi(0)\vert \Psi(0)\rangle \end{align} as $U^\dagger(t)U=\mathbb{1}$ by construction.

The wavefunction version follows in the same way: \begin{align} \Psi(x,t)&=e^{-i \hat H t/\hbar}\Psi(x,0)\, ,\\ \Psi(x,t)&=U(t)\Psi(x,0) \end{align} for the time-independent and the time-dependent cases, respectively.

  • 1
    $\begingroup$ IMO, by using such abstract Hilbert-space notation, this answer skips over OP's core difficulty, which is understanding in what sense the naive position-space Schrödinger equation satisfies these Hilbert-space properties (e.g., the ability to use distributions and not just functions for $\psi$, and the precise Hermitian property of the differential operator $\hat H$). It's not quite enough to say "$\psi$ is a function" or even "$\psi$ is a square-integrable function". That's why I pointed OP to functional analysis. $\endgroup$
    – nanoman
    Jun 14, 2021 at 16:09
  • $\begingroup$ @nanoman but then this is another question altogether as it would have nothing to do with preserving the normalization over time, but rather showing that $\psi$ is a “legal” state and can be normalized at any time. $\endgroup$ Jun 14, 2021 at 16:28
  • $\begingroup$ But it turns out that preserving the normalization in position space essentially amounts to deriving the Hermitian property of the differential operator $\hat H$, so it brings to the fore the question of precisely what function space $\psi$ lives in. $\endgroup$
    – nanoman
    Jun 14, 2021 at 16:34

To make this completely rigorous, you would have to delve into functional analysis. But I will try to explain it intuitively.

(i) Arguably all derivatives of $\psi$ are continuous in reality. Non-smooth functions are an idealization, but a useful one. They can be formalized using distributions. Even when $f$ is discontinuous, $f'$ is defined to be something that integrates to $f$. The results are equivalent to considering smooth functions and taking a limit in which they develop a discontinuity or kink.

(ii) The wave function must have a real expectation value of kinetic energy on physical grounds. The kinetic energy operator is proportional to $-\partial^2/\partial x^2$ and has an expectation value proportional to $$\int_{-\infty}^\infty dx\, \psi^* \left(-\frac{\partial^2\psi}{\partial x^2}\right) = \int_{-\infty}^\infty dx\, \left|\frac{\partial\psi}{\partial x}\right|^2 - \left.\psi^* \frac{\partial\psi}{\partial x}\right|_{-\infty}^\infty.$$ The imaginary part of this is precisely proportional to your key expression, $(\psi^*\, \partial\psi/\partial x - \psi\, \partial\psi^*/\partial x)|_{-\infty}^\infty$, which must therefore be zero. (Technically, the wave function must be defined in a space where the energy operator is Hermitian.)

(iii) The Schrödinger equation fully determines the evolution of the wave function $\psi$, and what you have shown is that if $\psi$ is initially normalized ($|A| = 1$) then it remains normalized ($|A|$ remains $1$). By saying that $A$ is the normalization constant, you have only defined its magnitude. So indeed the phase of $A$ is arbitrary, but it is not part of the wave function. Given that $|A| = 1$, you can simply write the normalization condition as $\int_{-\infty}^\infty dx\, |\psi|^2 = 1$. Ultimately, $A$ is a redundant, unphysical parameter.


Regarding (i), I am still not sure what you are asking. In general, $d\psi/dx$ is continuous over an interval if the potential $V(x)$ is finite within that interval. At the boundaries, if the potential $V(x) \to \pm \infty$, then $d\psi/dx$ is no longer continuous. Therefore, the continuity of the first derivative of the wavefunction depends on the behavior of the potential.

(See, for example, Griffiths' Introduction to Quantum Mechanics Chapter 2, Section 2.5.2 The Delta-Function Well.)

Regarding (ii), for a physical system, the wavefunctions "decays" to zero as one approaches $\pm \infty$ on the real position axis -- otherwise the system is unphysical. This means the wavefunction eventually becomes flat, implying its first derivative is zero.

Regarding (iii), you are correct in that $A$ might have some "angular" dependence. This is called phase dependence and is mostly inconsequential in Quantum Mechanics. You can always isolate the phase as follows: $A(t) = |A| \, e^{i\, \phi(t)}$. After that, once you evaluate $|A|^2 = A(t)^* A(t)$, the phase factor reduces to unity.

  • $\begingroup$ I have understood your point. But I have a question that does $d\psi/dx$ has to be bounded in general also? Can you also clarify my doubt i)? $\endgroup$
    – Iti
    Jun 14, 2021 at 6:13
  • $\begingroup$ @Iti I have edited my answer. $\endgroup$
    – Yejus
    Jun 14, 2021 at 6:29
  • $\begingroup$ If $F$ is a continuous function function over $(a,b)$ and differentiable over $[a,b]$ and derivative of $F$ is integrable over that interval, then by fundamental theorem of calculus $\int\frac{dF}{dx}dx=F+c$. So by analogy with RHS of the equation $(11)$, $\frac{d\psi}{dx}$ has to be continuous. But if the potential is dirac delta function then $\frac{d\psi}{dx}$ has a jump at that point meaning not continuous. So in general how can we use fundamental theorem of calculus in $(11)$? $\endgroup$
    – Iti
    Jun 14, 2021 at 6:41
  • 2
    $\begingroup$ "the wavefunction eventually becomes flat, implying its first derivative is zero" -- this isn't strictly sufficient, because $\psi$ could conceivably approach zero with oscillations that decay in amplitude but become more and more rapid as $x \to \pm\infty$, so $\partial\psi/\partial x$ would not go to zero. $\endgroup$
    – nanoman
    Jun 14, 2021 at 7:18

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