Understanding Complex Conjugates in QM (Griffiths pg. 13)

[RahSuh]

Homework Statement

Page 13 of Griffiths on Quantum Mechanics ( the part on Normalization)

Relevant Equations

Am look at the proof of normalised solns of Schrodingers Eqn stay normalised

Am looking at page 13 of QM by Griffiths - have become stuck on minor point. He is proving that a normalised solution of Schrodingers eqn stays normalised. The bit I don't get is how can you just take the complex conjugate of Schrodingers eqn and assume its true. (ie how does he get from Eqn 1.23 to Eqn 1.24) I assume it true but how ? I get that for example, the complex conjugate of the product of 2 complex numbers is the product of the complex conjugates etc - but does it apply to differential eqns. Can you just take the complex conjugates of differential eqns. Thanks for help in advance! (am sure its something very simple)

 

[topsquark]

Homework Statement:: Page 13 of Griffiths on Quantum Mechanics ( the part on Normalization)
Relevant Equations:: Am look at the proof of normalised solns of Schrodingers Eqn stay normalised

Am looking at page 13 of QM by Grif tufiths - have become stuck on minor point. He is proving that a normalised solution of Schrodingers eqn stays no normalised. The bit I don't get is how can you just take the complex congugate of Schrodingers eqn and assume its true. I assume it true but how ? (I get that you the complex conjugate of the product of 2 complex numbers is the product of the complex conjugates etc - but does it apply to differential eqns. Thanks for help in advance! (am sure its something very simple)View attachment 315309

Yes. If you have any expression $A = B$ then $A^*= B^*$. This holds true, even if we are taking the derivative with respect to a complex variable:
$\dfrac{dA}{dz} = B \implies \dfrac{dA^*}{d z^*} = B^*$ (though we usually change the notation from $z^*$ to $\overline{z}$.)

As t is usually taken to be real we don't do anything with that.

-Dan

 

[kuruman]

You can show that formally. After all, the left and right side of Schrodinger's equation are complex numbers (or expressions thereof).
One can always write two complex numbers as the sum of a real and an imaginary part,
$z_1=a_1+ib_1$ and $z_2=a_2+ib_2$.
If $z_1=z_2$, it follows that $a_1=a_2$ and $b_1=b_2$. Then
$z_1^*=a_1-ib_1=a_2-ib_2=z_2^*$, Q.E.D.