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Schrodinger equation for a free particle in 3d - Quantum Mechanics

I've got the solution to the question but I just need more detail. I can't work out the first step of the solution to the second step. IMG_20190903_093320.jpg IMG_20190903_092948.jpg
Original post by GeolPhysics
I've got the solution to the question but I just need more detail. I can't work out the first step of the solution to the second step. IMG_20190903_093320.jpg IMG_20190903_092948.jpg

I found out from Physics Forum that it's the derivative of the exponential.
Original post by GeolPhysics
I found out from Physics Forum that it's the derivative of the exponential.


Glad that you found out. In QM, it is quite often that you need to differentiate exponential and trigonometric functions. You also need to know how to rationalize the imaginary i when it is in the denominator.
You also need to know how to do completing the square.
Original post by Eimmanuel
Glad that you found out. In QM, it is quite often that you need to differentiate exponential and trigonometric functions. You also need to know how to rationalize the imaginary i when it is in the denominator.
You also need to know how to do completing the square.

Hi Eimmanuel, I still seem to be stuck on this part of the question. The other guy said to do the second derivative of the right hand side of the first starred line but I ended up with a weird answer. See the two stars that I put next to the equations. How can the first line's right hand equation's derivation end up like what's on the second line's right hand side. I'm very stuck. Please help.
Original post by GeolPhysics
Hi Eimmanuel, I still seem to be stuck on this part of the question. The other guy said to do the second derivative of the right hand side of the first starred line but I ended up with a weird answer. See the two stars that I put next to the equations. How can the first line's right hand equation's derivation end up like what's on the second line's right hand side. I'm very stuck. Please help.


Can you show what you get for the second derivative of the wavefunction?

You ought to know that the energy of a free particle is
E=pp2m E = \dfrac{\vec{p}\cdot \vec{p}}{2m}

This is coming from your course on classical mechanics.

Writing reply using a phone is pain, I would write a more detail reply when I am back home if you are still stuck.
Original post by Eimmanuel
Can you show what you get for the second derivative of the wavefunction?

You ought to know that the energy of a free particle is
E=pp2m E = \dfrac{\vec{p}\cdot \vec{p}}{2m}

This is coming from your course on classical mechanics.

Writing reply using a phone is pain, I would write a more detail reply when I am back home if you are still stuck.

I got this for the second derivative but I know it's wrong.I tried getting the first derivative by the e^u=e^u . u' law. IMG_20190908_172241.jpg Someone on Physics Forums told me to go back to the wavefunction equation and do the product of exponentials. I've not been taught that and I don't know how to do that. I definitely need to brush up on first year Calculus which I wasn't very strong in.
(edited 4 years ago)
Original post by GeolPhysics
I got this for the second derivative but I know it's wrong.I tried getting the first derivative by the e^u=e^u . u' law. IMG_20190908_172241.jpg Someone on Physics Forums told me to go back to the wavefunction equation and do the product of exponentials. I've not been taught that and I don't know how to do that. I definitely need to brush up on first year Calculus which I wasn't very strong in.


Indeed, you need to brush up your calculus.


Note that the wavefunction of the free particle is


ψ(x,t)=cexp[i(pxEt)] \psi \left( \mathbf{x},t \right)=c\exp \left[ \frac{i}{\hbar }\left( \mathbf{p}\cdot \mathbf{x}-Et \right) \right]


and when we expand the dot product of momentum and position vector, we have


ψ(x,t)=cexp[i(pxx+pyy+pzzEt)] \psi \left( \mathbf{x},t \right)=c\exp \left[ \frac{i}{\hbar }\left( {{p}_{x}}x+{{p}_{y}}y+{{p}_{z}}z-Et \right) \right]


Let consider only the partial derivative with respect to x only


22m2x2ψ(x,t)=22mx[xψ(x,t)] -\frac{{{\hbar }^{2}}}{2m}\frac{{{\partial }^{2}}}{\partial {{x}^{2}}}\psi \left( \mathbf{x},t \right)=-\frac{{{\hbar }^{2}}}{2m}\frac{\partial }{\partial x}\left[ \frac{\partial }{\partial x}\psi \left( \mathbf{x},t \right) \right]



free_particle_wavefunction01.JPG

I would let you continue and you would get


22m2x2ψ(x,t)=22m(ipx)2cexp[i(pxx+pyy+pzzEt)] -\frac{{{\hbar }^{2}}}{2m}\frac{{{\partial }^{2}}}{\partial {{x}^{2}}}\psi \left( \mathbf{x},t \right)=-\frac{{{\hbar }^{2}}}{2m}{{\left( \frac{i}{\hbar }{{p}_{x}} \right)}^{2}}c\exp \left[ \frac{i}{\hbar }\left( {{p}_{x}}x+{{p}_{y}}y+{{p}_{z}}z-Et \right) \right]


Do the partial derivative with respect to y and z, we would obtain a similar result.


22m2y2ψ(x,t)=22m(ipy)2cexp[i(pxx+pyy+pzzEt)]-\frac{{{\hbar }^{2}}}{2m}\frac{{{\partial }^{2}}}{\partial {{y}^{2}}}\psi \left( \mathbf{x},t \right)=-\frac{{{\hbar }^{2}}}{2m}{{\left( \frac{i}{\hbar }{{p}_{y}} \right)}^{2}}c\exp \left[ \frac{i}{\hbar }\left( {{p}_{x}}x+{{p}_{y}}y+{{p}_{z}}z-Et \right) \right]



22m2z2ψ(x,t)=22m(ipz)2cexp[i(pxx+pyy+pzzEt)]-\frac{{{\hbar }^{2}}}{2m}\frac{{{\partial }^{2}}}{\partial {{z}^{2}}}\psi \left( \mathbf{x},t \right)=-\frac{{{\hbar }^{2}}}{2m}{{\left( \frac{i}{\hbar }{{p}_{z}} \right)}^{2}}c\exp \left[ \frac{i}{\hbar }\left( {{p}_{x}}x+{{p}_{y}}y+{{p}_{z}}z-Et \right) \right]



Combine the result, we would get

22m(2x2+2y2+2z2)ψ(x,t)=22m((ipx)2+(ipy)2+(ipz)2)ψ(x,t)-\frac{{{\hbar }^{2}}}{2m}\left( \frac{{{\partial }^{2}}}{\partial {{x}^{2}}}+\frac{{{\partial }^{2}}}{\partial {{y}^{2}}}+\frac{{{\partial }^{2}}}{\partial {{z}^{2}}} \right)\psi \left( \mathbf{x},t \right)=-\frac{{{\hbar }^{2}}}{2m}\left( {{\left( \frac{i}{\hbar }{{p}_{x}} \right)}^{2}}+{{\left( \frac{i}{\hbar }{{p}_{y}} \right)}^{2}}+{{\left( \frac{i}{\hbar }{{p}_{z}} \right)}^{2}} \right)\psi \left( \mathbf{x},t \right)

After some simplification, you should get the kinetic energy of the free particle as


EK=px2+py2+pz22m{{E}_{K}}=\frac{p_{x}^{2}+p_{y}^{2}+p_{z}^{2}}{2m}


There is some incomplete info in post #5, the energy expression that I wrote in the post is kinetic energy NOT total energy.


[22m(2x2+2y2+2z2)+V0]ψ(x,t)=(px2+py2+pz22m+V0)ψ(x,t)\left[ -\frac{{{\hbar }^{2}}}{2m}\left( \frac{{{\partial }^{2}}}{\partial {{x}^{2}}}+\frac{{{\partial }^{2}}}{\partial {{y}^{2}}}+\frac{{{\partial }^{2}}}{\partial {{z}^{2}}} \right)+{{V}_{0}} \right]\psi \left( \mathbf{x},t \right)=\left( \frac{p_{x}^{2}+p_{y}^{2}+p_{z}^{2}}{2m}+{{V}_{0}} \right)\psi \left( \mathbf{x},t \right)


The total energy is

E=px2+py2+pz22m+V0E=\frac{p_{x}^{2}+p_{y}^{2}+p_{z}^{2}}{2m}+{{V}_{0}}
Original post by Eimmanuel
Indeed, you need to brush up your calculus.


Note that the wavefunction of the free particle is


ψ(x,t)=cexp[i(pxEt)] \psi \left( \mathbf{x},t \right)=c\exp \left[ \frac{i}{\hbar }\left( \mathbf{p}\cdot \mathbf{x}-Et \right) \right]


and when we expand the dot product of momentum and position vector, we have


ψ(x,t)=cexp[i(pxx+pyy+pzzEt)] \psi \left( \mathbf{x},t \right)=c\exp \left[ \frac{i}{\hbar }\left( {{p}_{x}}x+{{p}_{y}}y+{{p}_{z}}z-Et \right) \right]


Let consider only the partial derivative with respect to x only


22m2x2ψ(x,t)=22mx[xψ(x,t)] -\frac{{{\hbar }^{2}}}{2m}\frac{{{\partial }^{2}}}{\partial {{x}^{2}}}\psi \left( \mathbf{x},t \right)=-\frac{{{\hbar }^{2}}}{2m}\frac{\partial }{\partial x}\left[ \frac{\partial }{\partial x}\psi \left( \mathbf{x},t \right) \right]



free_particle_wavefunction01.JPG

I would let you continue and you would get


22m2x2ψ(x,t)=22m(ipx)2cexp[i(pxx+pyy+pzzEt)] -\frac{{{\hbar }^{2}}}{2m}\frac{{{\partial }^{2}}}{\partial {{x}^{2}}}\psi \left( \mathbf{x},t \right)=-\frac{{{\hbar }^{2}}}{2m}{{\left( \frac{i}{\hbar }{{p}_{x}} \right)}^{2}}c\exp \left[ \frac{i}{\hbar }\left( {{p}_{x}}x+{{p}_{y}}y+{{p}_{z}}z-Et \right) \right]


Do the partial derivative with respect to y and z, we would obtain a similar result.


22m2y2ψ(x,t)=22m(ipy)2cexp[i(pxx+pyy+pzzEt)]-\frac{{{\hbar }^{2}}}{2m}\frac{{{\partial }^{2}}}{\partial {{y}^{2}}}\psi \left( \mathbf{x},t \right)=-\frac{{{\hbar }^{2}}}{2m}{{\left( \frac{i}{\hbar }{{p}_{y}} \right)}^{2}}c\exp \left[ \frac{i}{\hbar }\left( {{p}_{x}}x+{{p}_{y}}y+{{p}_{z}}z-Et \right) \right]



22m2z2ψ(x,t)=22m(ipz)2cexp[i(pxx+pyy+pzzEt)]-\frac{{{\hbar }^{2}}}{2m}\frac{{{\partial }^{2}}}{\partial {{z}^{2}}}\psi \left( \mathbf{x},t \right)=-\frac{{{\hbar }^{2}}}{2m}{{\left( \frac{i}{\hbar }{{p}_{z}} \right)}^{2}}c\exp \left[ \frac{i}{\hbar }\left( {{p}_{x}}x+{{p}_{y}}y+{{p}_{z}}z-Et \right) \right]



Combine the result, we would get

22m(2x2+2y2+2z2)ψ(x,t)=22m((ipx)2+(ipy)2+(ipz)2)ψ(x,t)-\frac{{{\hbar }^{2}}}{2m}\left( \frac{{{\partial }^{2}}}{\partial {{x}^{2}}}+\frac{{{\partial }^{2}}}{\partial {{y}^{2}}}+\frac{{{\partial }^{2}}}{\partial {{z}^{2}}} \right)\psi \left( \mathbf{x},t \right)=-\frac{{{\hbar }^{2}}}{2m}\left( {{\left( \frac{i}{\hbar }{{p}_{x}} \right)}^{2}}+{{\left( \frac{i}{\hbar }{{p}_{y}} \right)}^{2}}+{{\left( \frac{i}{\hbar }{{p}_{z}} \right)}^{2}} \right)\psi \left( \mathbf{x},t \right)

After some simplification, you should get the kinetic energy of the free particle as


EK=px2+py2+pz22m{{E}_{K}}=\frac{p_{x}^{2}+p_{y}^{2}+p_{z}^{2}}{2m}


There is some incomplete info in post #5, the energy expression that I wrote in the post is kinetic energy NOT total energy.


[22m(2x2+2y2+2z2)+V0]ψ(x,t)=(px2+py2+pz22m+V0)ψ(x,t)\left[ -\frac{{{\hbar }^{2}}}{2m}\left( \frac{{{\partial }^{2}}}{\partial {{x}^{2}}}+\frac{{{\partial }^{2}}}{\partial {{y}^{2}}}+\frac{{{\partial }^{2}}}{\partial {{z}^{2}}} \right)+{{V}_{0}} \right]\psi \left( \mathbf{x},t \right)=\left( \frac{p_{x}^{2}+p_{y}^{2}+p_{z}^{2}}{2m}+{{V}_{0}} \right)\psi \left( \mathbf{x},t \right)


The total energy is

E=px2+py2+pz22m+V0E=\frac{p_{x}^{2}+p_{y}^{2}+p_{z}^{2}}{2m}+{{V}_{0}}


That's fine. I understand it now. There's a line in the solution on the sheet that says, Identical expressions... The equation line after that, is this how you get the left hand side equation to equal the right hand side? IMG_20190918_095940.jpg
Original post by GeolPhysics
That's fine. I understand it now. There's a line in the solution on the sheet that says, Identical expressions... The equation line after that, is this how you get the left hand side equation to equal the right hand side? IMG_20190918_095940.jpg



You need to really brush up your calculus. When we differentiate an exponential function, we do the following:



xexp[f(x,t)]=[xf(x,t)]×exp[f(x,t)] \dfrac{\partial }{\partial x}\exp \left[ \operatorname{f}\left( x,t \right) \right]=\left[ \dfrac{\partial }{\partial x}\operatorname{f}\left( x,t \right) \right]\times \exp \left[ \operatorname{f}\left( x,t \right) \right]



We need to bring down the function [i(pxEt)] \left[ \frac{i}{\hbar }\left( \mathbf{p}\cdot \mathbf{x}-Et \right) \right] in the power of the exponential function and then do the differentiation NOT performing the differentiation in the power as you did in the working.

Make sure when you read my writing, you pay attention to all the brackets.

In fact, the given solution is already pretty explicit in the working already. The only problem that I am seeing is your calculus skill.
I've got a new question where I'm stuck on in a thread titled 'What's z* in this equation?' could you please help?

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