Normalisation and probability current

The Born rule requires

$$\int_{-\infty}^{\infty}|\Psi(x,t)|^2\,\dd x = 1\tag{2.2.1}$$

This is a strong constraint, and it has to hold at every instant. The Schrödinger equation tells us how $\Psi$ changes with time — what guarantees that this integral does not?

If it could change, quantum mechanics would be finished: particles would appear and disappear as time went on. This section proves that it cannot, and digs out a useful new quantity along the way.

Normalisation

Given a solution $\Psi$, linearity of the equation guarantees that $A\Psi$ is also a solution for any complex constant $A$. Choose $A$ so that the integral equals 1:

$$|A|^2\int_{-\infty}^{\infty}|\Psi|^2\,\dd x = 1\tag{2.2.2}$$

Not every function can be normalised. For $\int|\Psi|^2$ to converge, $\Psi$ must go to zero fast enough as $x\to\pm\infty$. Such functions are called square-integrable and form the Hilbert space $L^2$.

Normalisation does not change with time

Proof that d/dt ∫|Ψ|²dx = 0basic~6 minexpand

We want to show that

$$\frac{\dd}{\dd t}\int_{-\infty}^{\infty}|\Psi(x,t)|^2\,\dd x = 0\tag{2.2.3}$$

The limits do not depend on time, so the derivative can move inside the integral and become a partial derivative:

$$\frac{\dd}{\dd t}\int|\Psi|^2\dd x = \int \frac{\partial}{\partial t}\left(\Psi^*\Psi\right)\dd x = \int\left(\Psi^*\frac{\partial\Psi}{\partial t}+\frac{\partial\Psi^*}{\partial t}\Psi\right)\dd x\tag{2.2.4}$$

Now use the Schrödinger equation (take its form on trust here; section 2.4 discusses it properly):

$$\frac{\partial\Psi}{\partial t} = \frac{\ii\hbar}{2m}\frac{\partial^2\Psi}{\partial x^2}-\frac{\ii}{\hbar}V\Psi\tag{2.2.5}$$

Take the complex conjugate — and note that $V$ being a real function is essential here:

$$\frac{\partial\Psi^*}{\partial t} = -\frac{\ii\hbar}{2m}\frac{\partial^2\Psi^*}{\partial x^2}+\frac{\ii}{\hbar}V\Psi^*\tag{2.2.6}$$

Substituting back, the two terms containing $V$ have opposite signs and cancel outright:

$$\Psi^*\frac{\partial\Psi}{\partial t}+\frac{\partial\Psi^*}{\partial t}\Psi =\frac{\ii\hbar}{2m}\left(\Psi^*\frac{\partial^2\Psi}{\partial x^2}-\frac{\partial^2\Psi^*}{\partial x^2}\Psi\right)\tag{2.2.7}$$

What is left inside the bracket happens to be a total derivative:

$$\frac{\partial}{\partial x}\left(\Psi^*\frac{\partial\Psi}{\partial x}-\frac{\partial\Psi^*}{\partial x}\Psi\right) =\Psi^*\frac{\partial^2\Psi}{\partial x^2}-\frac{\partial^2\Psi^*}{\partial x^2}\Psi\tag{2.2.8}$$

(the cross terms $\frac{\partial\Psi^*}{\partial x}\frac{\partial\Psi}{\partial x}$ appear twice and cancel). Hence

$$\frac{\dd}{\dd t}\int|\Psi|^2\dd x =\frac{\ii\hbar}{2m}\left[\Psi^*\frac{\partial\Psi}{\partial x}-\frac{\partial\Psi^*}{\partial x}\Psi\right]_{-\infty}^{+\infty}\tag{2.2.9}$$

and normalisability requires $\Psi\to 0$ (with bounded derivative) as $|x|\to\infty$, so the boundary term vanishes. ∎

What did the proof rely on? Only on $V$ being real. Introduce a complex potential — which condensed-matter and nuclear physics really do, to describe absorption phenomenologically — and the normalisation decays with time. That decay is the mathematics of absorption. The absorbing boundary layer in the wave-packet simulations on this site uses exactly this trick.

The probability current

The combination that appeared in that proof is important enough to deserve a name. Define the probability current density

$$j(x,t) \equiv \frac{\ii\hbar}{2m}\left(\Psi\frac{\partial\Psi^*}{\partial x}-\Psi^*\frac{\partial\Psi}{\partial x}\right) =\frac{\hbar}{m}\operatorname{Im}\left(\Psi^*\frac{\partial\Psi}{\partial x}\right)\tag{2.2.10}$$

(the two forms are equivalent; the second is easier to remember and to compute with). The proof above then amounts to

$$\frac{\partial\rho}{\partial t}+\frac{\partial j}{\partial x}=0\tag{2.2.11}$$

which is the continuity equation, identical in form to the ones in fluid mechanics and electrodynamics.

Example: the current of a plane wavebasic~3 minexpand

Take $\Psi = A\ee^{\ii(kx-\omega t)}$. Then

$$\frac{\partial\Psi}{\partial x}=\ii k\Psi,\qquad \Psi^*\frac{\partial\Psi}{\partial x}=\ii k|A|^2\tag{2.2.13}$$$$j=\frac{\hbar}{m}\operatorname{Im}(\ii k|A|^2)=\frac{\hbar k}{m}|A|^2\tag{2.2.14}$$

Since $\hbar k/m = p/m = v$,

$$j = \rho\, v\tag{2.2.15}$$

— density times velocity, exactly “flux = density × speed”. This matches classical fluid intuition perfectly.

It also shows in passing that if $\psi$ is a real function (an infinite-well stationary state, say), then $\Psi^*\partial_x\Psi$ is real, its imaginary part vanishes, and $j = 0$. In a stationary state the distribution does not move and neither does the current.

What counts as a legitimate wavefunction

Not every mathematical function will do. The requirements come from two directions: the Born rule must be usable, and the Schrödinger equation must make sense.

Requirement	Reason
Square-integrable, $\int\lvert\Psi\rvert^2\dd x<\infty$	Otherwise it cannot be normalised and the Born rule fails
$\Psi$ single-valued	$\lvert\Psi\rvert^2$ must give a unique probability
$\Psi$ continuous	Otherwise $\partial_x\Psi$ contains a delta function and the kinetic energy diverges
$\partial_x\Psi$ continuous (for finite potentials)	Follows from integrating the Schrödinger equation; it may jump where the potential is infinite or contains a delta function

∎Key formulas

• Normalisation

  $$\int_{-\infty}^{\infty}|\Psi(x,t)|^2\,\dd x=1$$

  Once true, the Schrödinger equation keeps it true

• Probability current

  $$j=\frac{\hbar}{m}\operatorname{Im}\left(\Psi^*\frac{\partial\Psi}{\partial x}\right)$$

  Reduces to j = ρv for a plane wave

• Continuity equation

  $$\frac{\partial\rho}{\partial t}+\frac{\partial j}{\partial x}=0$$

  Local conservation of probability

?Self-check3 questions

1. 1.

   Does the condition ∫|Ψ|²dx = 1 determine the constant A in front of the wavefunction uniquely?

   • No — it fixes only |A|; the phase of A is free
   • Yes — A is uniquely a positive real number
   • Yes, because probabilities must be positive
   • No, because the integral itself diverges

   Submit
2. 2.

   Which condition does the proof of conservation of normalisation depend on?

   • The potential V(x) is real
   • The potential V(x) is time-independent
   • The wavefunction is a stationary state
   • The particle mass is constant

   Submit
3. 3.

   What is the probability current j of a real stationary wavefunction ψ(x), such as an infinite-well eigenstate, and what does that mean?

   • j = 0, because the rightward and leftward currents cancel exactly
   • j = 0, because the particle is at rest
   • j = ρv ≠ 0, because the particle moves uniformly inside the well
   • j oscillates in time

   Submit

What comes next

We have been using the Schrödinger equation without ever introducing it properly. The next section deals with the stationary case first — which is what the overwhelming majority of calculations actually amount to: solving an eigenvalue problem.