Chapter 1: Q16P (page 22)

Show that $\frac{d}{d t} \int_{- \infty}^{\infty} {Ψ_{1}}^{*} Ψ_{2} d x = 0$
For any two solution to the Schrodinger equation $Ψ_{1}$ and $Ψ_{2}$ .

Short Answer

Expert verified

The solutions for $Ψ_{1} .$ and $Ψ_{2}$ is $\frac{d}{dt} \int_{- \infty}^{\infty} {Ψ_{1}}^{*} Ψ_{2} dx = 0$

Step by step solution

Step 1: Define the Schrodinger equation

A differential equation that describes matter in quantum mechanics in terms of the wave-like properties of particles in a field.
Its answer is related to a particle's probability density in space and time.

Determine the solutions for Ψ1 and Ψ2.

$\frac{d}{dt} \int_{- \infty}^{\infty} {Ψ_{1}}^{*} Ψ_{2} dx = \int_{- \infty}^{\infty} \frac{\partial}{\partial t} ({Ψ_{1}}^{*} Ψ_{2}) d x = \int_{- \infty}^{\infty} (Ψ_{2} \frac{\partial Ψ_{1}^{*}}{\partial t} + {Ψ^{*}}_{1} \frac{\partial Ψ_{2}}{\partial t}) d x$

But, from Schrodinger equation

$\frac{\partial Ψ_{2}}{\partial t} = \frac{i h}{2 m} \frac{\partial^{2} Ψ_{2}}{\partial x^{2}} - \frac{i}{h} V Ψ_{2}$

And

$\frac{\partial Ψ_{1}^{*}}{\partial t} = - \frac{i h}{2 m} \frac{\partial^{2} Ψ_{2}}{\partial x^{2}} - \frac{i}{h} V Ψ_{1}^{*}$

Thus,

localid="1658552483464" $\frac{d}{d t} \int_{- \infty}^{\infty} {Ψ_{1}}^{*} Ψ_{2} d x = \int_{- \infty}^{\infty} (Ψ_{2} [- \frac{i h}{2 m} \frac{\partial^{2} Ψ_{1}^{*}}{\partial x^{2}} + \frac{i}{h}] + Ψ_{1}^{*} [- \frac{i h}{2 m} \frac{\partial^{2} Ψ_{2}}{\partial x^{2}} + \frac{i}{h}] V Ψ_{2}) d x = - \frac{i h}{2 m} \int_{- \infty}^{\infty} (Ψ_{2} \frac{\partial^{2} Ψ_{1}^{*}}{\partial x^{2}} - Ψ_{1}^{*} \frac{\partial^{2} Ψ_{2}}{\partial x^{2}}) dx = - \frac{i h}{2 m} \int_{- \infty}^{\infty} (Ψ_{2} \frac{\partial^{2} Ψ_{1}^{*}}{\partial x^{2}} - Ψ_{1}^{*} \frac{\partial^{2} Ψ_{2}}{\partial x^{2}}) dx = - \frac{i h}{2 m} \int_{- \infty}^{\infty} \frac{\partial}{\partial x} (Ψ_{2} \frac{\partial Ψ_{1}^{*}}{\partial x} - Ψ_{1}^{*} \frac{\partial Ψ_{2}}{\partial x}) dx \frac{d}{dt} \int_{- \infty}^{\infty} {Ψ_{1}}^{*} Ψ_{2} dx = - \frac{i h}{2 m} {(Ψ_{2} \frac{\partial Ψ_{1}^{*}}{\partial x} - Ψ_{1}^{*} \frac{\partial Ψ_{2}}{\partial x})}_{\infty}^{\infty}$

Where this must equal zero because $Ψ_{1} .$ and $Ψ_{2}$ are normalized.

Hence, the solutions for $Ψ_{1} .$ and $Ψ_{2}$ is $\frac{d}{dt} \int_{- \infty}^{\infty} Ψ_{1}^{*} Ψ_{2} d x = 0$

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Show that $\frac{d}{d t} \int_{- \infty}^{\infty} {Ψ_{1}}^{*} Ψ_{2} d x = 0$
For any two solution to the Schrodinger equation $Ψ_{1}$ and $Ψ_{2}$ .

Short Answer

Step by step solution

Step 1: Define the Schrodinger equation

Determine the solutions for Ψ1 and Ψ2.

One App. One Place for Learning.

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Show thatddt∫-∞∞Ψ1*Ψ2dx=0For any two solution to the Schrodinger equationΨ1 andΨ2 .

Short Answer

Step by step solution

Step 1: Define the Schrodinger equation

Determine the solutions for Ψ1 and Ψ2.

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Physics Textbooks

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Show that $\frac{d}{d t} \int_{- \infty}^{\infty} {Ψ_{1}}^{*} Ψ_{2} d x = 0$
For any two solution to the Schrodinger equation $Ψ_{1}$ and $Ψ_{2}$ .