How Can I Show That \( \exp(-iaP/\hbar) = U \)?

In summary, the conversation discusses how to show that the exponential of the momentum operator P, e^(-iaP/h), is equal to the operator U, which is defined as U*Psi(x) = Psi(x-a). The suggested approach is to use the definition of the exponential of an operator and combine it with the expression for P in the x-basis to calculate e^(-iaP/h)*Psi(x). This results in a Taylor series which is not immediately recognizable, but it is a series for a shifted Psi(x) about the point a.
  • #1
naggy
60
0

Homework Statement



If [tex]U[/tex] is an operator so [tex]U\Psi(x)[/tex] = [tex]\Psi(x-a)[/tex].

How can I show that [tex]exp(-iaP/h) = U[/tex]

where P is the momentum operator [tex]P = -ih(d/dx)[/tex]

Homework Equations



Not sure

The Attempt at a Solution



What I do know that if I have a function F of an operator then

[tex]F(P)\psi[/tex] = [tex]$\sum_{i} c_iF(\lambda_i)\psi_i[/tex]

where [tex]\lambda_i[/tex] are the eigenvalues of [tex]P[/tex]

and [tex]c_i = <\psi_i,\psi>[/tex]

can I somehow relate all of this to the operator [tex]U[/tex]
 
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  • #2
I think what you'll want to do is combine the fact that (by the definition of the exponential of an operator)

[tex]e^{-iaP/\hbar}=\sum_{n=0}^{\infty}\frac{(-1)^na^nP^n}{\hbar^n n!}[/tex]

with your expression for [itex]P[/itex] in the x-basis to calculate [itex]e^{-iaP/\hbar}\Psi(x)[/itex]...your result should remind you of a Taylor series!:wink:
 
  • #3
gabbagabbahey said:
I think what you'll want to do is combine the fact that (by the definition of the exponential of an operator)

[tex]e^{-iaP/\hbar}=\sum_{n=0}^{\infty}\frac{(-1)^na^nP^n}{\hbar^n n!}[/tex]

with your expression for [itex]P[/itex] in the x-basis to calculate [itex]e^{-iaP/\hbar}\Psi(x)[/itex]...your result should remind you of a Taylor series!:wink:

Ok so:

[tex]e^{-iaP/\hbar}= e^{-aD_x} = 1 - aD_x + \frac{a^2D_x^2}{2!} - \frac{a^3D_x^3}{3!} + ... [/tex]

and applying this to [tex]\psi(x)[/tex] I can´t really say that I recognise the series, but I suppose it is the series for shifted [tex]\psi(x)[/tex] about a
 

FAQ: How Can I Show That \( \exp(-iaP/\hbar) = U \)?

What is the momentum operator?

The momentum operator is a mathematical representation of the physical quantity of momentum in quantum mechanics. It is denoted by the symbol p and is defined as the product of the mass and velocity of a particle.

What is the role of the momentum operator in quantum mechanics?

The momentum operator plays a crucial role in quantum mechanics as it is used to describe the momentum of a particle in terms of its wave function. It is also used to calculate the uncertainty in the momentum of a particle, which is a fundamental aspect of quantum mechanics.

How is the momentum operator related to the position operator?

The momentum operator and the position operator are closely related in quantum mechanics. They both represent physical quantities and are related through the Heisenberg uncertainty principle, which states that the product of the uncertainties in position and momentum of a particle must be greater than or equal to a certain value.

How is the momentum operator used in solving quantum mechanical problems?

The momentum operator is used in the Schrödinger equation, which is the fundamental equation of quantum mechanics. It is also used to calculate the expectation value of the momentum of a particle in a given quantum state, which helps in solving various quantum mechanical problems.

Can the momentum operator be measured directly?

No, the momentum operator cannot be measured directly. In quantum mechanics, measurements are made in terms of observables, which are related to the Hermitian operators such as the momentum operator. The measurement of the momentum of a particle is indirect and involves the use of mathematical operations on the wave function.

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