To show that the operator A = d/dx is not Hermitian, we need to show that it does not satisfy the property of a Hermitian operator. A Hermitian operator A satisfies the property that the inner product of (ψ, Aφ) is equal to the complex conjugate of the inner product of (Aψ, φ) for all ψ and φ in the space of square integrable functions.

The inner product of two functions f and g in the space of square integrable functions is defined as the integral over all space of the complex conjugate of f times g.

So, we need to show that ∫ψ*(dφ/dx) dx is not equal to the complex conjugate of ∫(dψ/dx)* φ dx.

Let's compute these two quantities:

∫ψ*(dφ/dx) dx = ψ*φ | from -∞ to ∞ - ∫(dψ/dx)* φ dx

The first term on the right hand side, ψ*φ | from -∞ to ∞, is zero because ψ and φ are square integrable, which means they go to zero faster than 1/√x as x goes to ±∞.

So, we have ∫ψ*(dφ/dx) dx = - ∫(dψ/dx)* φ dx.

This shows that the operator d/dx is not Hermitian because its action on the inner product does not satisfy the property of a Hermitian operator.

Question

To show that the operator A = d/dx is not Hermitian, we need to show that it does not satisfy the property of a Hermitian operator. A Hermitian operator A satisfies the property that the inner product of (ψ, Aφ) is equal to the complex conjugate of the inner product of (Aψ, φ) for all ψ and φ in the space of square integrable functions.

The inner product of two functions f and g in the space of square integrable functions is defined as the integral over all space of the complex conjugate of f times g.

So, we need to show that ∫ψ*(dφ/dx) dx is not equal to the complex conjugate of ∫(dψ/dx)* φ dx.

Let's compute these two quantities:

∫ψ*(dφ/dx) dx = ψ*φ | from -∞ to ∞ - ∫(dψ/dx)* φ dx

The first term on the right hand side, ψ*φ | from -∞ to ∞, is zero because ψ and φ are square integrable, which means they go to zero faster than 1/√x as x goes to ±∞.

So, we have ∫ψ*(dφ/dx) dx = - ∫(dψ/dx)* φ dx.

This shows that the operator d/dx is not Hermitian because its action on the inner product does not satisfy the property of a Hermitian operator.

Knowee AI · Accepted Answer

To show that the operator A = d/dx is not Hermitian, we need to show that it does not satisfy the property of a Hermitian operator. A Hermitian operator A satisfies the property that the inner product of (ψ, Aφ) is equal to the complex conjugate of the inner product of (Aψ, φ) for all ψ and φ in the space of square integrable functions.

The inner product of two functions f and g in the space of square integrable functions is defined as the integral over all space of the complex conjugate of f times g.

So, we need to show that ∫ψ*(dφ/dx) dx is not equal to the complex conjugate of ∫(dψ/dx)* φ dx.

Let's compute these two quantities:

∫ψ*(dφ/dx) dx = ψ*φ | from -∞ to ∞ - ∫(dψ/dx)* φ dx

The first term on the right hand side, ψ*φ | from -∞ to ∞, is zero because ψ and φ are square integrable, which means they go to zero faster than 1/√x as x goes to ±∞.

So, we have ∫ψ*(dφ/dx) dx = - ∫(dψ/dx)* φ dx.

This shows that the operator d/dx is not Hermitian because its action on the inner product does not satisfy the property of a Hermitian operator.

a) Show that the operator ˆA = ddx is not Hermitian in the space of square integrable functions.

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