Calculate for the first-excited state #psi_1(x)# of the simple harmonic oscillator the value of the Heisenberg Uncertainty relation?

#psi_1(x) = ((momega)/(piℏ))^(1//4)sqrt((2momega)/ℏ)xe^(-momegax^2//2ℏ)#

This was a question on my Physics exam. I know how to do it now, so I plan to post it. :)

1 Answer
Truong-Son N.
Jan 5, 2018

From the fruits of our labor,

#DeltaxDeltap = 3ℏ//2#

CAUTION: EXTREMELY LONG ANSWER!

First of all, the uncertainty principle is

#DeltaxDeltap >= ℏ//2#,

so we ought to get a value #>= ℏ//2#.

Next, the uncertainties are defined as follows:

#DeltaA = sqrt(<< A^2 >> - << A >>^2)#, #" "bb((1))#

where #<< A >># is the expectation value, or average value, of the observable #A#.

An expectation value in one dimension is given by:

#<< A >> = << psi | A | psi >> = int_("allspace") psi^"*" hatA psi dx# #" "bb((2))#

where #hatA# is the operator corresponding to the observable #A#.

So, we need to calculate #<< x^2 >>#, #<< x >>^2#, #<< p^2 >>#, and #<< p >>^2#, given that

#psi_1(x) = ((momega)/(piℏ))^(1//4)sqrt((2momega)/ℏ)xe^(-momegax^2//2ℏ)#,

a real-valued function.

POSITION UNCERTAINTY

We have that #psi^"*" = psi# and so, from #(2)#:

#<< x^2 >> = int_(-oo)^(oo) psi^"*" x^2 psi dx#

#= ((momega)/(piℏ))^(1//2)((2momega)/ℏ)int_(-oo)^(oo) x^4 e^(-momegax^2//ℏ)dx#

#= 2/(sqrtpi)((momega)/(ℏ))^(3//2)int_(-oo)^(oo) x^4 e^(-momegax^2//ℏ)dx#

A nice way to solve this is to start from this one (which is the only one I know of these kinds of integrals):

#int_(-oo)^(oo) e^(-momegax^2//ℏ)dx#

From Leibniz's integral rule for constant integration bounds (which was not taught in the class, by the way),

#d/(dalpha)int_(-oo)^(oo) e^(-alphax^2)dx = int_(-oo)^(oo) (del)/(delalpha)[e^(-alphax^2)]dx#

#= -int_(-oo)^(oo) x^2e^(-alphax^2)dx#

This integral can be evaluated in polar coordinates to be #-(pi/alpha)^(1//2)#. As a result, with #alpha = momega//ℏ#,

#color(green)(int_(-oo)^(oo) x^2 e^(-momegax^2//ℏ)dx) = -d/(dalpha)[-(pi/alpha)^(1//2)]#

#= sqrtpi cdot -1/2alpha^(-3//2)#

#= color(green)((sqrtpi)/(2alpha^(3//2)))##" "bb((3))#

Following the sequence once more,

#d/(dalpha)int_(-oo)^(oo) x^2e^(-alphax^2)dx = int_(-oo)^(oo) (del)/(delalpha)[x^2e^(-alphax^2)]dx#

#= -int_(-oo)^(oo) x^4e^(-alphax^2)dx#

Therefore,

#color(green)(int_(-oo)^(oo) x^4e^(-alphax^2)dx) = -d/(dalpha)[(sqrtpi)/(2alpha^(3//2))]#

#= -(sqrt(pi)/2 cdot -3/2 alpha^(-5//2))#

#= color(green)((3sqrtpi)/(4alpha^(5//2)))##" "bb((4))#

And so, using result #(4)#,

#color(green)(<< x^2 >>) = 2/(sqrtpi)((momega)/(ℏ))^(3//2)int_(-oo)^(oo) x^4 e^(-momegax^2//ℏ)dx#

#= 2/(sqrtpi)((momega)/(ℏ))^(3//2) cdot (3sqrtpi)/(4((momega)/(ℏ))^(5//2))#

#= 2/cancel(sqrtpi)cancel(((momega)/(ℏ))^(3//2)) cdot (3cancel(sqrtpi))/(4((momega)/(ℏ))^(cancel(5//2)))#

#= color(green)(3/2 (ℏ)/(momega))# #" "bb((5))#

Next, the average position is:

#<< x >> = int_(-oo)^(oo) psi^"*"xpsidx#

#= 2/(sqrtpi)((momega)/(ℏ))^(3//2)int_(-oo)^(oo) x^3 e^(-momegax^2//ℏ)dx#

The integrand is an odd function times an even function, which is then odd. The integral of an odd function over a symmetric interval is zero, so #<< x >> = 0#.

Therefore, using relation #(1)# and result #(5)#:

#barul|stackrel(" ")(" "color(green)(Deltax) = sqrt(<< x^2 >> - << x >>^2) = sqrt(<< x^2 >>) = color(green)(sqrt(3/2 (ℏ)/(momega)))" ")|# #" "bb((6))#

MOMENTUM UNCERTAINTY

Similarly, we now need the analogous values for the momentum. The operator for momentum needed, which is #hatp = -iℏ(del)/(delx)#. So, #hatp^2 = -ℏ^2(del^2)/(delx^2)#.

From #(2)#:

#<< p^2 >> = -ℏ^2int_(-oo)^(oo) psi^"*"(del^2psi)/(delx^2)dx#

The next thing we should evaluate then, is the second derivative of #psi#.

#(del^2psi)/(delx^2) = ((momega)/(piℏ))^(1//4)sqrt((2momega)/ℏ)d^2/(dx^2)[xe^(-momegax^2//2ℏ)]#

The derivative of #e^(-alphax^2//2)# for simplicity is #-alphaxe^(-alphax^2//2)#, and so, from a lot of product rule...

#=> (alpha/pi)^(1//4)sqrt(2alpha) cdot d/(dx)[-alphax^2e^(-alphax^2//2) + e^(-alphax^2//2)]#

#= (alpha/pi)^(1//4)sqrt(2alpha) cdot [-alpha(-alphax^3e^(-alphax^2//2) + 2xe^(-alphax^2//2)) - alphaxe^(-alphax^2//2)]#

#= (alpha/pi)^(1//4)sqrt(2alpha) cdot (alpha^2x^2 - 3alpha)xe^(-alphax^2//2)#

Let's not put the constants back in yet. Continuing, we got:

#(del^2psi)/(delx^2) = (alpha/pi)^(1//4)sqrt(2alpha) cdot (alpha^2x^2 - 3alpha)xe^(-alphax^2//2)# #" "bb((7))#

From this result #(7)#, the integral becomes, after some simplification:

#<< p^2 >> = -ℏ^2(alpha/pi)^(1//2)(2alpha) int_(-oo)^(oo) xe^(-alphax^2//2) cdot (alpha^2x^2 - 3alpha)e^(-alphax^2//2)dx#

#= -(2ℏ^2)/sqrtpi alpha^(3//2) int_(-oo)^(oo) (alpha^2x^2 - 3alpha)x^2e^(-alphax^2)dx#

#= -(2ℏ^2)/sqrtpi alpha^(3//2) {alpha^2int_(-oo)^(oo) x^4e^(-alphax^2)dx - 3alphaint_(-oo)^(oo)x^2e^(-alphax^2)dx}#

Fortunately we have basically already done these integrals. From results #(3)# and #(4)#, we then get:

#= -(2ℏ^2)/sqrtpi alpha^(3//2) {alpha^2((3sqrtpi)/(4alpha^(5//2))) - 3alpha((sqrtpi)/(2alpha^(3//2)))}#

#= -(2ℏ^2)/sqrtpi alpha^(cancel(3//2)) {((3sqrtpi)/(4cancel(alpha^(1//2)))) - 3((sqrtpi)/(2cancel(alpha^(1//2))))}#

#= -(2ℏ^2)/cancelsqrtpi alpha (-(3cancelsqrtpi)/4)#

#= ℏ^2 cdot alpha (3/2)#

Recalling that #alpha = momega//ℏ#, we get:

#color(green)(<< p^2 >> = 3/2momegaℏ)# #" "bb((8))#

Last one!! The average momentum from #(2)# is:

#<< p >> = int_(-oo)^(oo) psi^"*"hatppsidx#

where #hatp = -iℏ(del)/(delx)# as stated before.

#<< p >> = -iℏ((momega)/(piℏ))^(1//2)((2momega)/ℏ)int_(-oo)^(oo) xe^(-momegax^2//2ℏ) d/dx[xe^(-momegax^2//2ℏ)]dx#

But there is a nice trick here. The derivative of #psi#, which in this case is an odd function (the odd #x# times the even #e^(-alphax^2//2)# equals an odd function), becomes even.

The result is that the integrand is an odd function times an even function, giving another odd function to integrate over a symmetric interval.

So, #<< p >> = 0#. As a result, using #(8)# in #(1)#, we get:

#barul|stackrel(" ")(" "color(green)(Deltap) = sqrt(<< p^2 >> - << p >>^2) = sqrt(<< p^2 >>) = color(green)(sqrt(3/2 momegaℏ))" ")|# #" "bb((9))#

FINAL RESULT!

FINALLY, we can then evaluate the uncertainty relation! Using results #(6)# and #(9)#:

#color(blue)ul(DeltaxDeltap) = sqrt(3/2 ℏ/cancel(momega)) cdot sqrt(3/2 cancel(momega)ℏ)#

#= color(blue)ul(3ℏ//2)#

And this is indeed #>= ℏ//2#, as the uncertainty principle requires!

Related questions
Impact of this question
10255 views around the world
You can reuse this answer
Creative Commons License