# Didn't understand what Sakurai meant by this.

1. Mar 12, 2014

### M. next

In Sakurai (Quantum Mechanics) (See attached).

What is meant by "which is what we expect by substitution of ∇' "? Do they mean that I have to substitute it above and I should get equation (2.6.30)? Because I tried but couldn't simplify further.

Thanks in advance.

#### Attached Files:

• ###### Sakurai.PNG
File size:
10.8 KB
Views:
120
2. Mar 12, 2014

### Simon Bridge

You are right - the text kinda begs the question "substitute into what?" there... going only by that page, my reading is similar to yours - with a different target:

Whenever you see a $\nabla^\prime$ in equ. 2.7.30, you substitute as equ. 2.7.32
You also need to use $\rho=|\psi|^2$ and look at the "as before" to see what else to do before you get equ 2.7.31.

Notice the mass seems to appear out of nowhere - he's using another relation not on that page.

It could be that he is referring to an earlier step where the substitution could have been made.
What is the significance of 2.7.32

3. Mar 12, 2014

### M. next

Thank you for your reply. Here we are talking about gauge transformation. I guess he is trying to make us substitute the ∇′ -(ie/hbar*c) into 2.7.30 in order the same form as if we substitute only ∇′ into 2.7.30

But the thing is am not geting the same form, or am not knowing how to proceed and what to eliminate.

4. Mar 12, 2014

### Bill_K

Sakurai has a habit of stating things upside down and backwards. :uhh:

What he means here is that if you take the previous continuity equation Eq.(2.4.15) and expression for probability current Eq.(2.4.16) and make the substitution Eq.(2.7.32) you get the same contintuity equation, Eq.(2.7.30) but now with j given by Eq.(2.7.31).

5. Mar 12, 2014

### kith

What he means is that if you know the probability current in the field-free case you can get the probability current including the field simply by the substitution 2.6.32.

So there are two ways to obtain the correct probability current:
1) Derive the continuity equation from the Schrödinger equation including the correct Hamiltonian. This is what he does up until 2.6.31.
2) Take the field-free probability current as given and use the substitution 2.6.32.

Now why do we expect the substitution to work? This is essentially the substitution of the canonical momentum with the kinematical momentum again (diveded by iħ).

/edit: Apparently, I used an old version of the thread and Bill already answered. Also we seem to use different versions of Sakurai in this thread.

6. Mar 12, 2014

### Bill_K

S'okay, happens a lot. A good habit is to hit the "Preview" button as the very last thing just before posting, as this will pick up the current version of the thread.

The Sakurai I'm using says Second Edition, with a date of 2011.

7. Mar 12, 2014

### M. next

Thanks guys!

Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook