Note 02 Back to Ch. 11 · Cascaded Stern–Gerlach Experiments
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Proof Why time evolution preserves scalar products

Time evolution preserves scalar products, ⟨αt∣βt⟩=⟨αt0∣βt0⟩\langle\alpha t|\beta t\rangle=\langle\alpha t_0|\beta t_0\rangle. We derive it here in general: two of the principles of the treatment suffice, linear superposition and conservation of the total.

We write ∣α⟩|\alpha\rangle for the initial state and ∣α′⟩|\alpha'\rangle for the evolved state.

(I) Linear superposition. The evolved of a superposition is the superposition of the evolved states:

∣γ⟩=a∣α⟩+b∣β⟩⟹∣γ′⟩=a∣α′⟩+b∣β′⟩|\gamma\rangle=a|\alpha\rangle+b|\beta\rangle\qquad\Longrightarrow\qquad|\gamma'\rangle=a|\alpha'\rangle+b|\beta'\rangle

(II) Conservation of the total. The total of a vector, that is the sum of the squared moduli of its components ⟨α∣α⟩\langle\alpha|\alpha\rangle, does not change under evolution:

⟨α′∣α′⟩=⟨α∣α⟩\langle\alpha'|\alpha'\rangle=\langle\alpha|\alpha\rangle

for any vector.

Of the scalar product we will use three properties: ⟨β∣α⟩=⟨α∣β⟩∗\langle\beta|\alpha\rangle=\langle\alpha|\beta\rangle^*, additivity in both factors, that is ⟨α+γ∣β⟩=⟨α∣β⟩+⟨γ∣β⟩\langle\alpha+\gamma|\beta\rangle=\langle\alpha|\beta\rangle+\langle\gamma|\beta\rangle and ⟨α∣β+γ⟩=⟨α∣β⟩+⟨α∣γ⟩\langle\alpha|\beta+\gamma\rangle=\langle\alpha|\beta\rangle+\langle\alpha|\gamma\rangle, and finally ⟨α∣cβ⟩=c⟨α∣β⟩\langle\alpha|c\beta\rangle=c\langle\alpha|\beta\rangle.

There is an apparent obstacle: we want to prove something about the scalar product of two different vectors, but (II) speaks of a single vector, multiplied by itself. The idea that resolves it is to apply (II) not to ∣α⟩|\alpha\rangle or to ∣β⟩|\beta\rangle, but to their sum: in the total of the sum, as we shall see, the two vectors mix, and precisely the scalar products we are looking for appear. Since a scalar product is a complex number, two pieces of information will be needed: the sum ∣α⟩+∣β⟩|\alpha\rangle+|\beta\rangle will give the real part, the sum ∣α⟩+i∣β⟩|\alpha\rangle+i|\beta\rangle the imaginary part.

Step 1 — the real part

By (I), the evolved of ∣α⟩+∣β⟩|\alpha\rangle+|\beta\rangle is ∣α′⟩+∣β′⟩|\alpha'\rangle+|\beta'\rangle; applying (II) to this sum vector:

⟨α′+β′∣α′+β′⟩=⟨α+β∣α+β⟩\langle\alpha'+\beta'|\alpha'+\beta'\rangle=\langle\alpha+\beta|\alpha+\beta\rangle

Expanding both sides by additivity:

⟨α′∣α′⟩+⟨α′∣β′⟩+⟨β′∣α′⟩+⟨β′∣β′⟩=⟨α∣α⟩+⟨α∣β⟩+⟨β∣α⟩+⟨β∣β⟩\langle\alpha'|\alpha'\rangle+\langle\alpha'|\beta'\rangle+\langle\beta'|\alpha'\rangle+\langle\beta'|\beta'\rangle=\langle\alpha|\alpha\rangle+\langle\alpha|\beta\rangle+\langle\beta|\alpha\rangle+\langle\beta|\beta\rangle

As announced, alongside the totals of the single vectors the cross terms have appeared, containing the scalar products we seek. The totals remove themselves: ⟨α′∣α′⟩=⟨α∣α⟩\langle\alpha'|\alpha'\rangle=\langle\alpha|\alpha\rangle and ⟨β′∣β′⟩=⟨β∣β⟩\langle\beta'|\beta'\rangle=\langle\beta|\beta\rangle by (II), so they cancel between the two sides, leaving

⟨α′∣β′⟩+⟨β′∣α′⟩=⟨α∣β⟩+⟨β∣α⟩\langle\alpha'|\beta'\rangle+\langle\beta'|\alpha'\rangle=\langle\alpha|\beta\rangle+\langle\beta|\alpha\rangle

Since ⟨β∣α⟩=⟨α∣β⟩∗\langle\beta|\alpha\rangle=\langle\alpha|\beta\rangle^*, each side has the form z+z∗=2 Re⁡ zz+z^*=2\:\operatorname{Re}\:z, hence

Re⁡ ⟨α′∣β′⟩=Re⁡ ⟨α∣β⟩(A)\operatorname{Re}\:\langle\alpha'|\beta'\rangle=\operatorname{Re}\:\langle\alpha|\beta\rangle\qquad\text{(A)}

The real part of the scalar product is preserved — and since nothing was assumed about ∣α⟩|\alpha\rangle and ∣β⟩|\beta\rangle, this holds for every pair of vectors.

Step 2 — the imaginary part

The imaginary part remains: to extract it, we use i∣β⟩i|\beta\rangle in place of ∣β⟩|\beta\rangle. No new calculation is needed — (A) holds for every pair of vectors, so it holds for the pair (∣α⟩, i∣β⟩)(|\alpha\rangle,\ i|\beta\rangle), whose evolved, by (I), is (∣α′⟩, i∣β′⟩)(|\alpha'\rangle,\ i|\beta'\rangle):

Re⁡ ⟨α′∣iβ′⟩=Re⁡ ⟨α∣iβ⟩\operatorname{Re}\:\langle\alpha'|i\beta'\rangle=\operatorname{Re}\:\langle\alpha|i\beta\rangle

By the property ⟨α∣cβ⟩=c⟨α∣β⟩\langle\alpha|c\beta\rangle=c\langle\alpha|\beta\rangle we take ii out of the scalar products:

Re⁡ (i⟨α′∣β′⟩)=Re⁡ (i⟨α∣β⟩)\operatorname{Re}\:(i\langle\alpha'|\beta'\rangle)=\operatorname{Re}\:(i\langle\alpha|\beta\rangle)

For a complex number, Re⁡(iz)=− Im⁡ z\operatorname{Re}(iz)=-\:\operatorname{Im}\:z; applying this to both sides and changing sign:

Im⁡ ⟨α′∣β′⟩=Im⁡ ⟨α∣β⟩(B)\operatorname{Im}\:\langle\alpha'|\beta'\rangle=\operatorname{Im}\:\langle\alpha|\beta\rangle\qquad\text{(B)}

Conclusion

Two complex numbers with the same real part and the same imaginary part are the same number. From (A) and (B), therefore:

⟨α′∣β′⟩=⟨α∣β⟩⟺⟨αt∣βt⟩=⟨αt0∣βt0⟩\langle\alpha'|\beta'\rangle=\langle\alpha|\beta\rangle\qquad\Longleftrightarrow\qquad\langle\alpha t|\beta t\rangle=\langle\alpha t_0|\beta t_0\rangle

which is what we set out to prove. ∎

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La Quantistica · Technical note No. 02 · Rev. 2026 F. Palma